2 Limits 极限

本页译自 OpenStax《Calculus Volume 1》第 2 章 Limits。公式经 MathJax 渲染，自定义宏已注入。

Chapter Outline 本章大纲

• 2.1 A Preview of Calculus
• 2.2 The Limit of a Function
• 2.3 The Limit Laws
• 2.4 Continuity
• 2.5 The Precise Definition of a Limit

• 2.1 微积分学预览
• 2.2 函数的极限
• 2.3 极限法则
• 2.4 连续性
• 2.5 极限的精确定义

2.1 A Preview of Calculus 2.1 微积分学预览

• 2.1.1 Describe the tangent problem and how it led to the idea of a derivative.
• 2.1.2 Explain how the idea of a limit is involved in solving the tangent problem.
• 2.1.3 Recognize a tangent to a curve at a point as the limit of secant lines.
• 2.1.4 Identify instantaneous velocity as the limit of average velocity over a small time interval.
• 2.1.5 Describe the area problem and how it was solved by the integral.
• 2.1.6 Explain how the idea of a limit is involved in solving the area problem.
• 2.1.7 Recognize how the ideas of limit, derivative, and integral led to the studies of infinite series and multivariable calculus.

• 2.1.1 描述切线问题以及它如何引出了导数的概念。
• 2.1.2 解释极限思想是如何参与解决切线问题的。
• 2.1.3 将曲线在一点处的切线识别为割线的极限。
• 2.1.4 将瞬时速度识别为在很短时间区间内平均速度的极限。
• 2.1.5 描述面积问题以及它是如何被积分解决的。
• 2.1.6 解释极限思想是如何参与解决面积问题的。
• 2.1.7 认识极限、导数与积分的思想如何引出了无穷级数与多元微积分的研究。

As we embark on our study of calculus, we shall see how its development arose from common solutions to practical problems in areas such as engineering physics—like the space travel problem posed in the chapter opener. Two key problems led to the initial formulation of calculus: (1) the tangent problem, or how to determine the slope of a line tangent to a curve at a point; and (2) the area problem, or how to determine the area under a curve.

The Tangent Problem and Differential Calculus 切线问题与微分学

Rate of change is one of the most critical concepts in calculus. We begin our investigation of rates of change by looking at the graphs of the three lines $f(x) = -2x - 3,g(x) = \frac{1}{2}x + 1,$ and $h(x) = 2,$ shown in Figure 2.2.

As we move from left to right along the graph of $f(x) = -2x - 3,$ we see that the graph decreases at a constant rate. For every 1 unit we move to the right along the x-axis, the y-coordinate decreases by 2 units. This rate of change is determined by the slope (−2) of the line. Similarly, the slope of 1/2 in the function $g(x)$ tells us that for every change in x of 1 unit there is a corresponding change in y of 1/2 unit. The function $h(x) = 2$ has a slope of zero, indicating that the values of the function remain constant. We see that the slope of each linear function indicates the rate of change of the function.

Compare the graphs of these three functions with the graph of $k(x) = x^{2}$ (Figure 2.3). The graph of $k(x) = x^{2}$ starts from the left by decreasing rapidly, then begins to decrease more slowly and level off, and then finally begins to increase—slowly at first, followed by an increasing rate of increase as it moves toward the right. Unlike a linear function, no single number represents the rate of change for this function. We quite naturally ask: How do we measure the rate of change of a nonlinear function?

We can approximate the rate of change of a function $f(x)$ at a point $\left( {a,f(a)} \right)$ on its graph by taking another point $\left( {x,f(x)} \right)$ on the graph of $f(x),$ drawing a line through the two points, and calculating the slope of the resulting line. Such a line is called a secant line. Figure 2.4 shows a secant line to a function $f(x)$ at a point $\left( {a,f(a)} \right).$

We formally define a secant line as follows:

The secant to the function $f(x)$ through the points $\left( {a,f(a)} \right)$ and $\left( {x,f(x)} \right)$ is the line passing through these points. Its slope is given by

$$m_{\text{sec}} = \frac{f(x) - f(a)}{x - a}.$$ (2.1)

The accuracy of approximating the rate of change of the function with a secant line depends on how close x is to a. As we see in Figure 2.5, if x is closer to a, the slope of the secant line is a better measure of the rate of change of $f(x)$ at a.

The secant lines themselves approach a line that is called the tangent to the function $f(x)$ at a (Figure 2.6). The slope of the tangent line to the graph at a measures the rate of change of the function at a. This value also represents the derivative of the function $f(x)$ at a, or the rate of change of the function at a. This derivative is denoted by $f^{\prime}(a).$ Differential calculus is the field of calculus concerned with the study of derivatives and their applications.

For an interactive demonstration of the slope of a secant line that you can manipulate yourself, visit this applet (*Note:* this site requires a Java browser plugin): Math Insight.

Example 2.1 illustrates how to find slopes of secant lines. These slopes estimate the slope of the tangent line or, equivalently, the rate of change of the function at the point at which the slopes are calculated.

Finding Slopes of Secant Lines 求割线的斜率

Estimate the slope of the tangent line (rate of change) to $f(x) = x^{2}$ at $x = 1$ by finding slopes of secant lines through $\left( {1,1} \right)$ and each of the following points on the graph of $f(x) = x^{2}.$

1. $\left( {2,4} \right)$

2. $\left( {\frac{3}{2},\frac{9}{4}} \right)$

Solution 解答

Use the formula for the slope of a secant line from the definition.

1. $m_{\text{sec}} = \frac{4 - 1}{2 - 1} = 3$

2. $m_{\text{sec}} = \frac{\frac{9}{4} - 1}{\frac{3}{2} - 1} = \frac{5}{2} = 2.5$

The point in part b. is closer to the point $\left( {1,1} \right),$ so the slope of 2.5 is closer to the slope of the tangent line. A good estimate for the slope of the tangent would be in the range of 2 to 2.5 (Figure 2.7).

Estimate the slope of the tangent line (rate of change) to $f(x) = x^{2}$ at $x = 1$ by finding the slope of the secant line through $\left( {1,1} \right)$ and the point $\left( {\frac{5}{4},\frac{25}{16}} \right)$ on the graph of $f(x) = x^{2}.$

We continue our investigation by exploring a related question. Keeping in mind that velocity may be thought of as the rate of change of position, suppose that we have a function, $s(t),$ that gives the position of an object along a coordinate axis at any given time t. Can we use these same ideas to create a reasonable definition of the instantaneous velocity at a given time $t = a?$ We start by approximating the instantaneous velocity with an average velocity. First, recall that the speed of an object traveling at a constant rate is the ratio of the distance traveled to the length of time it has traveled. We define the average velocity of an object over a time period to be the change in its position divided by the length of the time period.

Let $s(t)$ be the position of an object moving along a coordinate axis at time t. The average velocity of the object over a time interval $\left\lbrack {a,t} \right\rbrack$ where $a < t$ (or $\left\lbrack {t,a} \right\rbrack$ if $\left. t < a \right)$ is

$$v_{\text{ave}} = \frac{s(t) - s(a)}{t - a}.$$ (2.2)

As t is chosen closer to a, the average velocity becomes closer to the instantaneous velocity. Note that finding the average velocity of a position function over a time interval is essentially the same as finding the slope of a secant line to a function. Furthermore, to find the slope of a tangent line at a point a, we let the x-values approach a in the slope of the secant line. Similarly, to find the instantaneous velocity at time a, we let the t-values approach a in the average velocity. This process of letting x or t approach a in an expression is called taking a limit. Thus, we may define the instantaneous velocity as follows.

For a position function $s(t),$ the instantaneous velocity at a time $t = a$ is the value that the average velocities approach on intervals of the form $\left\lbrack {a,t} \right\rbrack$ and $\left\lbrack {t,a} \right\rbrack$ as the values of t become closer to a, provided such a value exists.

Example 2.2 illustrates this concept of limits and average velocity.

Finding Average Velocity 求平均速度

A rock is dropped from a height of 64 ft. It is determined that its height (in feet) above ground t seconds later (for $\left. 0 \leq t \leq 2 \right)$ is given by $s(t) = -16t^{2} + 64.$ Find the average velocity of the rock over each of the given time intervals. Use this information to guess the instantaneous velocity of the rock at time $t = 0.5.$

1. $\left\lbrack {0.49,0.5} \right\rbrack$

2. $\left\lbrack {0.5,0.51} \right\rbrack$

Solution 解答

Substitute the data into the formula for the definition of average velocity.

1. $v_{\text{ave}} = \frac{s(0.5) - s(0.49)}{0.5 - 0.49} = -15.84$

2. $v_{\text{ave}} = \frac{s(0.51) - s(0.5)}{0.51 - 0.5} = -16.16$

The instantaneous velocity is somewhere between −15.84 and −16.16 ft/sec. A good guess might be −16 ft/sec.

An object moves along a coordinate axis so that its position at time t is given by $s(t) = t^{3}.$ Estimate its instantaneous velocity at time $t = 2$ by computing its average velocity over the time interval $\left\lbrack {2,2.001} \right\rbrack.$

The Area Problem and Integral Calculus 面积问题与积分学

We now turn our attention to a classic question from calculus. Many quantities in physics—for example, quantities of work—may be interpreted as the area under a curve. This leads us to ask the question: How can we find the area between the graph of a function and the x-axis over an interval (Figure 2.8)?

As in the answer to our previous questions on velocity, we first try to approximate the solution. We approximate the area by dividing up the interval $\left\lbrack {a,b} \right\rbrack$ into smaller intervals in the shape of rectangles. The approximation of the area comes from adding up the areas of these rectangles (Figure 2.9).

As the widths of the rectangles become smaller (approach zero), the sums of the areas of the rectangles approach the area between the graph of $f(x)$ and the x-axis over the interval $\left\lbrack {a,b} \right\rbrack.$ Once again, we find ourselves taking a limit. Limits of this type serve as a basis for the definition of the definite integral. Integral calculus is the study of integrals and their applications.

Estimation Using Rectangles 用矩形估计

Estimate the area between the x-axis and the graph of $f(x) = x^{2} + 1$ over the interval $\left\lbrack {0,3} \right\rbrack$ by using the three rectangles shown in Figure 2.10.

Solution 解答

The areas of the three rectangles are 1 unit², 2 unit², and 5 unit². Using these rectangles, our area estimate is 8 unit².

Estimate the area between the x-axis and the graph of $f(x) = x^{2} + 1$ over the interval $\left\lbrack {0,3} \right\rbrack$ by using the three rectangles shown here:

Other Aspects of Calculus 微积分的其他方面

So far, we have studied functions of one variable only. Such functions can be represented visually using graphs in two dimensions; however, there is no good reason to restrict our investigation to two dimensions. Suppose, for example, that instead of determining the velocity of an object moving along a coordinate axis, we want to determine the velocity of a rock fired from a catapult at a given time, or of an airplane moving in three dimensions. We might want to graph real-value functions of two variables or determine volumes of solids of the type shown in Figure 2.11. These are only a few of the types of questions that can be asked and answered using multivariable calculus. Informally, multivariable calculus can be characterized as the study of the calculus of functions of two or more variables. However, before exploring these and other ideas, we must first lay a foundation for the study of calculus in one variable by exploring the concept of a limit.

Section 2.1 Exercises 2.1 节习题

For the following exercises, points $P\left( {1,2} \right)$ and $Q\left( {x,y} \right)$ are on the graph of the function $f(x) = x^{2} + 1.$

1.

[T] Complete the following table with the appropriate values: y-coordinate of Q, the point $Q\left( {x,y} \right),$ and the slope of the secant line passing through points P and Q. Round your answer to eight significant digits.

2.

Use the values in the right column of the table in the preceding exercise to guess the value of the slope of the line tangent to f at $x = 1.$

3.

Use the value in the preceding exercise to find an equation of the tangent line at point P. Graph $f(x)$ and the tangent line.

For the following exercises, points $P\left( {1,1} \right)$ and $Q\left( {x,y} \right)$ are on the graph of the function $f(x) = x^{3}.$

4.

[T] Complete the following table with the appropriate values: y-coordinate of Q, the point $Q\left( {x,y} \right),$ and the slope of the secant line passing through points P and Q. Round your answer to eight significant digits.

5.

Use the values in the right column of the table in the preceding exercise to guess the value of the slope of the tangent line to f at $x = 1.$

6.

Use the value in the preceding exercise to find an equation of the tangent line at point P. Graph $f(x)$ and the tangent line.

For the following exercises, points $P\left( {4,2} \right)$ and $Q\left( {x,y} \right)$ are on the graph of the function $f(x) = \sqrt{x}.$

7.

[T] Complete the following table with the appropriate values: y-coordinate of Q, the point $Q\left( {x,y} \right),$ and the slope of the secant line passing through points P and Q. Round your answer to eight significant digits.

8.

Use the values in the right column of the table in the preceding exercise to guess the value of the slope of the tangent line to f at $x = 4.$

9.

Use the value in the preceding exercise to find an equation of the tangent line at point P.

For the following exercises, points $P\left( {1.5,0} \right)$ and $Q\left( {\varphi,y} \right)$ are on the graph of the function $f(\varphi) = \text{cos}\mspace{2mu}\left( {\pi\varphi} \right).$

10.

[T] Complete the following table with the appropriate values: y-coordinate of Q, the point $Q\left( {\varphi,y} \right),$ and the slope of the secant line passing through points P and Q. Round your answer to eight significant digits.

11.

Use the values in the right column of the table in the preceding exercise to guess the value of the slope of the tangent line to f at $\varphi = 1.5.$

12.

Use the value in the preceding exercise to find an equation of the tangent line at point P.

For the following exercises, points $P\left( {-1,-1} \right)$ and $Q\left( {x,y} \right)$ are on the graph of the function $f(x) = \frac{1}{x}.$

13.

[T] Complete the following table with the appropriate values: y-coordinate of Q, the point $Q\left( {x,y} \right),$ and the slope of the secant line passing through points P and Q. Round your answer to eight significant digits.

14.

Use the values in the right column of the table in the preceding exercise to guess the value of the slope of the line tangent to f at $x = -1.$

15.

Use the value in the preceding exercise to find an equation of the tangent line at point P.

For the following exercises, the position function of a ball dropped from the top of a 200-meter tall building is given by $s(t) = 200 - 4.9t^{2},$ where position s is measured in meters and time t is measured in seconds. Round your answer to eight significant digits.

16.

[T] Compute the average velocity of the ball over the given time intervals.

1. $\left\lbrack {4.99,5} \right\rbrack$

2. $\left\lbrack {5,5.01} \right\rbrack$

3. $\left\lbrack {4.999,5} \right\rbrack$

4. $\left\lbrack {5,5.001} \right\rbrack$

17.

Use the preceding exercise to guess the instantaneous velocity of the ball at $t = 5$ sec.

For the following exercises, consider a stone tossed into the air from ground level with an initial velocity of 15 m/sec. Its height in meters at time t seconds is $h(t) = 15t - 4.9t^{2}.$

18.

[T] Compute the average velocity of the stone over the given time intervals.

1. $\left\lbrack {1,1.05} \right\rbrack$

2. $\left\lbrack {1,1.01} \right\rbrack$

3. $\left\lbrack {1,1.005} \right\rbrack$

4. $\left\lbrack {1,1.001} \right\rbrack$

19.

Use the preceding exercise to guess the instantaneous velocity of the stone at $t = 1$ sec.

For the following exercises, consider a rocket shot into the air that then returns to Earth. The height of the rocket in meters is given by $h(t) = 600 + 78.4t - 4.9t^{2},$ where t is measured in seconds.

20.

[T] Compute the average velocity of the rocket over the given time intervals.

1. $\left\lbrack {9,9.01} \right\rbrack$

2. $\left\lbrack {8.99,9} \right\rbrack$

3. $\left\lbrack {9,9.001} \right\rbrack$

4. $\left\lbrack {8.999,9} \right\rbrack$

21.

Use the preceding exercise to guess the instantaneous velocity of the rocket at $t = 9$ sec.

For the following exercises, consider an athlete running a 40-m dash. The position of the athlete is given by $d(t) = \frac{t^{3}}{6} + 4t,$ where d is the position in meters and t is the time elapsed, measured in seconds.

22.

[T] Compute the average velocity of the runner over the given time intervals.

1. $\left\lbrack {1.95,2.05} \right\rbrack$

2. $\left\lbrack {1.995,2.005} \right\rbrack$

3. $\left\lbrack {1.9995,2.0005} \right\rbrack$

4. $\left\lbrack {2,2.00001} \right\rbrack$

23.

Use the preceding exercise to guess the instantaneous velocity of the runner at $t = 2$ sec.

For the following exercises, consider the function $f(x) = |x|.$

24.

Sketch the graph of f over the interval $\left\lbrack {-1,2} \right\rbrack$ and shade the region above the x-axis.

25.

Use the preceding exercise to find the aproximate value of the area between the x-axis and the graph of f over the interval $\left\lbrack {-1,2} \right\rbrack$ using rectangles. For the rectangles, use the square units, and approximate both above and below the lines. Use geometry to find the exact answer.

For the following exercises, consider the function $f(x) = \sqrt{1 - x^{2}}.$ (*Hint*: This is the upper half of a circle of radius 1 positioned at $\left( {0,0} \right)\text{.)}$

26.

Sketch the graph of f over the interval $\left\lbrack {-1,1} \right\rbrack.$

27.

Use the preceding exercise to find the aproximate area between the x-axis and the graph of f over the interval $\left\lbrack {-1,1} \right\rbrack$ using rectangles. For the rectangles, use squares 0.4 by 0.4 units, and approximate both above and below the lines. Use geometry to find the exact answer.

For the following exercises, consider the function $f(x) = \text{−}x^{2} + 1.$

28.

Sketch the graph of f over the interval $\left\lbrack {-1,1} \right\rbrack.$

29.

Approximate the area of the region between the x-axis and the graph of f over the interval $\left\lbrack {-1,1} \right\rbrack.$

2.2 The Limit of a Function 2.2 函数的极限

• 2.2.1 Using correct notation, describe the limit of a function.
• 2.2.2 Use a table of values to estimate the limit of a function or to identify when the limit does not exist.
• 2.2.3 Use a graph to estimate the limit of a function or to identify when the limit does not exist.
• 2.2.4 Define one-sided limits and provide examples.
• 2.2.5 Explain the relationship between one-sided and two-sided limits.
• 2.2.6 Using correct notation, describe an infinite limit.
• 2.2.7 Define a vertical asymptote.

• 2.2.1 使用正确的记号描述函数的极限。
• 2.2.2 使用数值表估计函数的极限，或判断极限不存在的情形。
• 2.2.3 使用图像估计函数的极限，或判断极限不存在的情形。
• 2.2.4 定义单侧极限并举例说明。
• 2.2.5 解释单侧极限与双侧极限之间的关系。
• 2.2.6 使用正确的记号描述无穷极限。
• 2.2.7 定义垂直渐近线。

The concept of a limit or limiting process, essential to the understanding of calculus, has been around for thousands of years. In fact, early mathematicians used a limiting process to obtain better and better approximations of areas of circles. Yet, the formal definition of a limit—as we know and understand it today—did not appear until the late 19th century. We therefore begin our quest to understand limits, as our mathematical ancestors did, by using an intuitive approach. At the end of this chapter, armed with a conceptual understanding of limits, we examine the formal definition of a limit.

We begin our exploration of limits by taking a look at the graphs of the functions

$$f(x) = \frac{x^{2} - 4}{x - 2},\ g(x) = \frac{\left| {x - 2} \right|}{x - 2},\ \text{and}\ h(x) = \frac{1}{\left( {x - 2} \right)^{2}},$$

which are shown in Figure 2.12. In particular, let’s focus our attention on the behavior of each graph at and around $x = 2.$

Each of the three functions is undefined at $x = 2,$ but if we make this statement and no other, we give a very incomplete picture of how each function behaves in the vicinity of $x = 2.$ To express the behavior of each graph in the vicinity of 2 more completely, we need to introduce the concept of a limit.

Intuitive Definition of a Limit 极限的直观定义

Let’s first take a closer look at how the function $f(x) = {{(x^{2} - 4)}\text{/}{(x - 2)}}$ behaves around $x = 2$ in Figure 2.12. As the values of x approach 2 from either side of 2, the values of $y = f(x)$ approach 4. Mathematically, we say that the limit of $f(x)$ as x approaches 2 is 4. Symbolically, we express this limit as

$$\underset{x\rightarrow 2}{\text{lim}}f(x) = 4.$$

From this very brief informal look at one limit, let’s start to develop an intuitive definition of the limit. We can think of the limit of a function at a number a as being the one real number L that the functional values approach as the x-values approach a, provided such a real number L exists. Stated more carefully, we have the following definition:

Let $f(x)$ be a function defined at all values in an open interval containing a, with the possible exception of a itself, and let L be a real number. If all values of the function $f(x)$ approach the real number L as the values of $x\left( {\neq a} \right)$ approach the number a, then we say that the limit of $f(x)$ as x approaches a is L. (More succinct, as x gets closer to a, $f(x)$ gets closer and stays close to L.) Symbolically, we express this idea as

$$\underset{x\rightarrow a}{\text{lim}}f(x) = L.$$ (2.3)

We can estimate limits by constructing tables of functional values and by looking at their graphs. This process is described in the following Problem-Solving Strategy.

Evaluating a Limit Using a Table of Functional Values 用函数值表求极限

1. To evaluate $\underset{x\rightarrow a}{\text{lim}}f(x),$ we begin by completing a table of functional values. We should choose two sets of x-values—one set of values approaching a and less than a, and another set of values approaching a and greater than a. Table 2.1 demonstrates what your tables might look like.

Table 2.1 Table of Functional Values for $\underset{x\rightarrow a}{\text{lim}}f(x)$

2. Next, let’s look at the values in each of the $f(x)$ columns and determine whether the values seem to be approaching a single value as we move down each column. In our columns, we look at the sequence $f\left( {a - 0.1} \right),f\left( {a - 0.01} \right),f\left( {a - 0.001} \right).,f\left( {a - 0.0001} \right),$ and so on, and $f\left( {a + 0.1} \right),f\left( {a + 0.01} \right),f\left( {a + 0.001} \right),f\left( {a + 0.0001} \right),$ and so on. (*Note*: Although we have chosen the x-values $a \pm 0.1,a \pm 0.01,a \pm 0.001,a \pm 0.0001,$ and so forth, and these values will probably work nearly every time, on very rare occasions we may need to modify our choices.)

3. If both columns approach a common y-value L, we state $\underset{x\rightarrow a}{\text{lim}}f(x) = L.$ We can use the following strategy to confirm the result obtained from the table or as an alternative method for estimating a limit.

4. Using a graphing calculator or computer software that allows us to graph functions, we can plot the function $f(x),$ making sure the functional values of $f(x)$ for x-values near a are in our window. We can use the trace feature to move along the graph of the function and watch the y-value readout as the x-values approach a. If the y-values approach L as our x-values approach a from both directions, then $\underset{x\rightarrow a}{\text{lim}}f(x) = L.$ We may need to zoom in on our graph and repeat this process several times.

We apply this Problem-Solving Strategy to compute a limit in Example 2.4.

Evaluating a Limit Using a Table of Functional Values 1 用函数值表求极限 1

Evaluate $\underset{x\rightarrow 0}{\text{lim}}\frac{\text{sin}\mspace{2mu} x}{x}$ using a table of functional values.

Solution 解答

We have calculated the values of $f(x) = {{(\text{sin}\mspace{2mu} x)}\text{/}x}$ for the values of x listed in Table 2.2.

Table 2.2 Table of Functional Values for $\underset{x\rightarrow 0}{\text{lim}}\frac{\text{sin}\mspace{2mu} x}{x}$

Note: The values in this table were obtained using a calculator and using all the places given in the calculator output.

As we read down each $\frac{\left( {\text{sin}\mspace{2mu} x} \right)}{x}$ column, we see that the values in each column appear to be approaching one. Thus, it is fairly reasonable to conclude that $\underset{x\rightarrow 0}{\text{lim}}\frac{\text{sin}\mspace{2mu} x}{x} = 1.$ A calculator or computer-generated graph of $f(x) = \frac{\left( {\text{sin}\mspace{2mu} x} \right)}{x}$ would be similar to that shown in Figure 2.13, and it confirms our estimate.

Evaluating a Limit Using a Table of Functional Values 2 用函数值表求极限 2

Evaluate $\underset{x\rightarrow 4}{\text{lim}}\frac{\sqrt{x} - 2}{x - 4}$ using a table of functional values.

Solution 解答

As before, we use a table—in this case, Table 2.3—to list the values of the function for the given values of x.

Table 2.3 Table of Functional Values for $\underset{x\rightarrow 4}{\text{lim}}\frac{\sqrt{x} - 2}{x - 4}$

After inspecting this table, we see that the functional values less than 4 appear to be decreasing toward 0.25 whereas the functional values greater than 4 appear to be increasing toward 0.25. We conclude that $\underset{x\rightarrow 4}{\text{lim}}\frac{\sqrt{x} - 2}{x - 4} = 0.25.$ We confirm this estimate using the graph of $f(x) = \frac{\sqrt{x} - 2}{x - 4}$ shown in Figure 2.14.

Estimate $\underset{x\rightarrow 1}{\text{lim}}\frac{\frac{1}{x} - 1}{x - 1}$ using a table of functional values. Use a graph to confirm your estimate.

At this point, we see from Example 2.4 and Example 2.5 that it may be just as easy, if not easier, to estimate a limit of a function by inspecting its graph as it is to estimate the limit by using a table of functional values. In Example 2.6, we evaluate a limit exclusively by looking at a graph rather than by using a table of functional values.

Evaluating a Limit Using a Graph 用图像求极限

For $g(x)$ shown in Figure 2.15, evaluate $\underset{x\rightarrow-1}{\text{lim}}g(x).$

Solution 解答

Despite the fact that $g(-1) = 4,$ as the x-values approach −1 from either side, the $g(x)$ values approach 3. Therefore, $\underset{x\rightarrow-1}{\text{lim}}g(x) = 3.$ Note that we can determine this limit without even knowing the algebraic expression of the function.

Based on Example 2.6, we make the following observation: It is possible for the limit of a function to exist at a point, and for the function to be defined at this point, but the limit of the function and the value of the function at the point may be different.

Use the graph of $h(x)$ in Figure 2.16 to evaluate $\underset{x\rightarrow 2}{\text{lim}}h(x),$ if possible.

Looking at a table of functional values or looking at the graph of a function provides us with useful insight into the value of the limit of a function at a given point. However, these techniques rely too much on guesswork. We eventually need to develop alternative methods of evaluating limits. These new methods are more algebraic in nature and we explore them in the next section; however, at this point we introduce two special limits that are foundational to the techniques to come.

Two Important Limits 两个重要极限

Let a be a real number and c be a constant.

1. $$\underset{x\rightarrow a}{\text{lim}}x = a$$ (2.4)

2. $$\underset{x\rightarrow a}{\text{lim}}c = c$$ (2.5)

We can make the following observations about these two limits.

1. For the first limit, observe that as x approaches a, so does $f(x),$ because $f(x) = x.$ Consequently, $\underset{x\rightarrow a}{\text{lim}}x = a.$

2. For the second limit, consider Table 2.4.

Table 2.4 Table of Functional Values for $\underset{x\rightarrow a}{\text{lim}}c = c$

Observe that for all values of x (regardless of whether they are approaching a), the values $f(x)$ remain constant at c. We have no choice but to conclude $\underset{x\rightarrow a}{\text{lim}}c = c.$

The Existence of a Limit 极限的存在性

As we consider the limit in the next example, keep in mind that for the limit of a function to exist at a point, the functional values must approach a single real-number value at that point. If the functional values do not approach a single value, then the limit does not exist.

Evaluating a Limit That Fails to Exist 求一个不存在的极限

Evaluate $\underset{x\rightarrow 0}{\text{lim}}\text{sin}\mspace{2mu}\left( {1\text{/}\textit{x}} \right)$ using a table of values.

Solution 解答

Table 2.5 lists values for the function $\text{sin}({1\text{/}{x)}}$ for the given values of x.

Table 2.5 Table of Functional Values for $\underset{x\rightarrow 0}{\text{lim}}\text{sin}\mspace{2mu}\left( \frac{1}{x} \right)$

After examining the table of functional values, we can see that the y-values do not seem to approach any one single value. It appears the limit does not exist. Before drawing this conclusion, let’s take a more systematic approach. Take the following sequence of x-values approaching 0:

$$\frac{2}{\pi},\frac{2}{3\pi},\frac{2}{5\pi},\frac{2}{7\pi},\frac{2}{9\pi},\frac{2}{11\pi}\text{,….}$$

The corresponding y-values are

$$1,-1,1,-1,1,-1\text{,….}$$

At this point we can indeed conclude that $\underset{x\rightarrow 0}{\text{lim}}\text{sin}\mspace{2mu}\left( {1\text{/}\textit{x}} \right)$ does not exist. (Mathematicians frequently abbreviate "does not exist" as DNE. Thus, we would write $\underset{x\rightarrow 0}{\text{lim}}\text{sin}\mspace{2mu}\left( {1\text{/}\textit{x}} \right)$ DNE.) The graph of $f(x) = \text{sin}\mspace{2mu}\left( {1\text{/}x} \right)$ is shown in Figure 2.17 and it gives a clearer picture of the behavior of $\text{sin}({1\text{/}{x)}}$ as x approaches 0. You can see that $\text{sin}({1\text{/}{\textit{x})}}$ oscillates ever more wildly between −1 and 1 as x approaches 0.

Use a table of functional values to evaluate $\underset{x\rightarrow 2}{\text{lim}}\frac{\left| {x^{2} - 4} \right|}{x - 2},$ if possible.

One-Sided Limits 单侧极限

Sometimes indicating that the limit of a function fails to exist at a point does not provide us with enough information about the behavior of the function at that particular point. To see this, we now revisit the function $g(x) = {\left| {x - 2} \right|\text{/}\left( {x - 2} \right)}$ introduced at the beginning of the section (see Figure 2.12(b)). As we pick values of x close to 2, $g(x)$ does not approach a single value, so the limit as x approaches 2 does not exist—that is, $\underset{x\rightarrow 2}{\text{lim}}g(x)$ DNE. However, this statement alone does not give us a complete picture of the behavior of the function around the x-value 2. To provide a more accurate description, we introduce the idea of a one-sided limit. For all values to the left of 2 (or *the negative side of* 2), $g(x) = -1.$ Thus, as x approaches 2 from the left, $g(x)$ approaches −1. Mathematically, we say that the limit as x approaches 2 from the left is −1. Symbolically, we express this idea as

$$\underset{x\rightarrow 2^{-}}{\text{lim}}g(x) = -1.$$

Similarly, as x approaches 2 from the right (or *from the positive side*), $g(x)$ approaches 1. Symbolically, we express this idea as

$$\underset{x\rightarrow 2^{+}}{\text{lim}}g(x) = 1.$$

We can now present an informal definition of one-sided limits.

We define two types of one-sided limits.

Limit from the left: Let $f(x)$ be a function defined at all values in an open interval of the form (c, a), and let L be a real number. If the values of the function $f(x)$ approach the real number L as the values of x (where $x < \textit{a}\text{)}$ approach the number a, then we say that L is the limit of $f(x)$ as x approaches a from the left. Symbolically, we express this idea as

$$\underset{x\rightarrow a^{-}}{\text{lim}}f(x) = L.$$ (2.6)

Limit from the right: Let $f(x)$ be a function defined at all values in an open interval of the form $\left( {a,c} \right),$ and let L be a real number. If the values of the function $f(x)$ approach the real number L as the values of x (where $x > \textit{a}\text{)}$ approach the number a, then we say that L is the limit of $f(x)$ as x approaches a from the right. Symbolically, we express this idea as

$$\underset{x\rightarrow a^{+}}{\text{lim}}f(x) = L.$$ (2.7)

Evaluating One-Sided Limits 求单侧极限

For the function $f(x) = \begin{cases}{x + 1} & {\text{if}\ x < 2} \\{x^{2} - 4} & {\text{if}\ x \geq 2}\end{cases},$ evaluate each of the following limits.

1. $\underset{x\rightarrow 2^{-}}{\text{lim}}f(x)$

2. $\underset{x\rightarrow 2^{+}}{\text{lim}}f(x)$

Solution 解答

We can use tables of functional values again Table 2.6. Observe that for values of x less than 2, we use $f(x) = x + 1$ and for values of x greater than 2, we use $f(x) = x^{2} - 4.$

Table 2.6 Table of Functional Values for $f(x) = \left\{ \begin{array}{l}{x + 1\ \text{if}\ x < 2} \\{x^{2} - 4\ \text{if}\ x \geq 2}\end{array} \right.$

Based on this table, we can conclude that a. $\underset{x\rightarrow 2^{-}}{\text{lim}}f(x) = 3$ and b. $\underset{x\rightarrow 2^{+}}{\text{lim}}f(x) = 0.$ Therefore, the (two-sided) limit of $f(x)$ does not exist at $x = 2.$ Figure 2.18 shows a graph of $f(x)$ and reinforces our conclusion about these limits.

Use a table of functional values to estimate the following limits, if possible.

1. $\underset{x\rightarrow 2^{-}}{\text{lim}}\frac{\left| {x^{2} - 4} \right|}{x - 2}$

2. $\underset{x\rightarrow 2^{+}}{\text{lim}}\frac{\left| {x^{2} - 4} \right|}{x - 2}$

Let us now consider the relationship between the limit of a function at a point and the limits from the right and left at that point. It seems clear that if the limit from the right and the limit from the left have a common value, then that common value is the limit of the function at that point. Similarly, if the limit from the left and the limit from the right take on different values, the limit of the function does not exist. These conclusions are summarized in Relating One-Sided and Two-Sided Limits.

Relating One-Sided and Two-Sided Limits 单侧极限与双侧极限的关系

Let $f(x)$ be a function defined at all values in an open interval containing a, with the possible exception of a itself, and let L be a real number. Then,

$$\underset{x\rightarrow a}{\text{lim}}f(x) = L\ \text{if and only if}\ \underset{x\rightarrow a^{-}}{\text{lim}}f(x) = L\ \text{and}\ \underset{x\rightarrow a^{+}}{\text{lim}}f(x) = L.$$

Infinite Limits 无穷极限

Evaluating the limit of a function at a point or evaluating the limit of a function from the right and left at a point helps us to characterize the behavior of a function around a given value. As we shall see, we can also describe the behavior of functions that do not have finite limits.

We now turn our attention to $h(x) = {1\text{/}{{(x - 2)}^{2},}}$ the third and final function introduced at the beginning of this section (see Figure 2.12(c)). From its graph we see that as the values of x approach 2, the values of $h(x) = {1\text{/}{(x - 2)}^{2}}$ become larger and larger and, in fact, become infinite. Mathematically, we say that the limit of $h(x)$ as x approaches 2 is positive infinity. Symbolically, we express this idea as

$$\underset{x\rightarrow 2}{\text{lim}}h(x) = \text{+}\infty.$$

More generally, we define infinite limits as follows:

We define three types of infinite limits.

Infinite limits from the left: Let $f(x)$ be a function defined at all values in an open interval of the form $\left( {b,a} \right).$

1. If the values of $f(x)$ increase without bound as the values of x (where $x < \textit{a}\text{)}$ approach the number a, then we say that the limit as x approaches a from the left is positive infinity and we write

$$\underset{x\rightarrow a^{-}}{\text{lim}}f(x) = \text{+}\infty.$$ (2.8)

2. If the values of $f(x)$ decrease without bound as the values of x (where $x < \textit{a}\text{)}$ approach the number a, then we say that the limit as x approaches a from the left is negative infinity and we write

$$\underset{x\rightarrow a^{-}}{\text{lim}}f(x) = \text{−}\infty.$$ (2.9)

Infinite limits from the right: Let $f(x)$ be a function defined at all values in an open interval of the form $\left( {a,c} \right).$

1. If the values of $f(x)$ increase without bound as the values of x (where $x > \textit{a}\text{)}$ approach the number a, then we say that the limit as x approaches a from the right is positive infinity and we write

$$\underset{x\rightarrow a^{+}}{\text{lim}}f(x) = \text{+}\infty.$$ (2.10)

2. If the values of $f(x)$ decrease without bound as the values of x (where $x > \textit{a}\text{)}$ approach the number a, then we say that the limit as x approaches a from the right is negative infinity and we write

$$\underset{x\rightarrow a^{+}}{\text{lim}}f(x) = \text{−}\infty.$$ (2.11)

Two-sided infinite limit: Let $f(x)$ be defined for all $x \neq a$ in an open interval containing a.

1. If the values of $f(x)$ increase without bound as the values of x (where $x \neq \textit{a}\text{)}$ approach the number a, then we say that the limit as x approaches a is positive infinity and we write

$$\underset{x\rightarrow a}{\text{lim}}f(x) = \text{+}\infty.$$ (2.12)

2. If the values of $f(x)$ decrease without bound as the values of x (where $x \neq \textit{a}\text{)}$ approach the number a, then we say that the limit as x approaches a is negative infinity and we write

$$\underset{x\rightarrow a}{\text{lim}}f(x) = \text{−}\infty.$$ (2.13)

It is important to understand that when we write statements such as $\underset{x\rightarrow a}{\text{lim}}f(x) = \text{+}\infty$ or $\underset{x\rightarrow a}{\text{lim}}f(x) = \text{−}\infty$ we are describing the behavior of the function, as we have just defined it. We are not asserting that a limit exists. For the limit of a function $f(x)$ to exist at a, it must approach a real number L as x approaches a. That said, if, for example, $\underset{x\rightarrow a}{\text{lim}}f(x) = \text{+}\infty,$ we always write $\underset{x\rightarrow a}{\text{lim}}f(x) = \text{+}\infty$ rather than $\underset{x\rightarrow a}{\text{lim}}f(x)$ DNE.

Recognizing an Infinite Limit 识别无穷极限

Evaluate each of the following limits, if possible. Use a table of functional values and graph $f(x) = {1\text{/}x}$ to confirm your conclusion.

1. $\underset{x\rightarrow 0^{-}}{\text{lim}}\frac{1}{x}$

2. $\underset{x\rightarrow 0^{+}}{\text{lim}}\frac{1}{x}$

3. $\underset{x\rightarrow 0}{\text{lim}}\frac{1}{x}$

Solution 解答

Begin by constructing a table of functional values.

Table 2.7 Table of Functional Values for $f(x) = \frac{1}{x}$

1. The values of $1\text{/}x$ decrease without bound as x approaches 0 from the left. We conclude that

$$\underset{x\rightarrow 0^{-}}{\text{lim}}\frac{1}{x} = \text{−}\infty.$$

2. The values of $1\text{/}x$ increase without bound as x approaches 0 from the right. We conclude that

$$\underset{x\rightarrow 0^{+}}{\text{lim}}\frac{1}{x} = \text{+}\infty.$$

3. Since $\underset{x\rightarrow 0^{-}}{\text{lim}}\frac{1}{x} = \text{−}\infty$ and $\underset{x\rightarrow 0^{+}}{\text{lim}}\frac{1}{x} = \text{+}\infty$ have different values, we conclude that

$$\underset{x\rightarrow 0}{\text{lim}}\frac{1}{x}\ \text{DNE.}$$

The graph of $f(x) = {1\text{/}x}$ in Figure 2.19 confirms these conclusions.

Evaluate each of the following limits, if possible. Use a table of functional values and graph $f(x) = {1\text{/}x^{2}}$ to confirm your conclusion.

1. $\underset{x\rightarrow 0^{-}}{\text{lim}}\frac{1}{x^{2}}$

2. $\underset{x\rightarrow 0^{+}}{\text{lim}}\frac{1}{x^{2}}$

3. $\underset{x\rightarrow 0}{\text{lim}}\frac{1}{x^{2}}$

It is useful to point out that functions of the form $f(x) = {1\text{/}\left( {x - a} \right)^{n}},$ where n is a positive integer, have infinite limits as x approaches a from either the left or right (Figure 2.20). These limits are summarized in Infinite Limits from Positive Integers.

Infinite Limits from Positive Integers 正整数的无穷极限

If n is a positive even integer, then

$$\underset{x\rightarrow a}{\text{lim}}\frac{1}{\left( {x - a} \right)^{n}} = \text{+}\infty.$$

If n is a positive odd integer, then

$$\underset{x\rightarrow a^{+}}{\text{lim}}\frac{1}{\left( {x - a} \right)^{n}} = \text{+}\infty$$

and

$$\underset{x\rightarrow a^{-}}{\text{lim}}\frac{1}{\left( {x - a} \right)^{n}} = \text{−}\infty.$$

We should also point out that in the graphs of $f(x) = {1\text{/}{{(x - a)}^{n},}}$ points on the graph having x-coordinates very near to a are very close to the vertical line $x = a.$ That is, as x approaches a, the points on the graph of $f(x)$ are closer to the line $x = a.$ The line $x = a$ is called a vertical asymptote of the graph. We formally define a vertical asymptote as follows:

Let $f(x)$ be a function. If any of the following conditions hold, then the line $x = a$ is a vertical asymptote of $f(x).$

$$\begin{array}{clc} {\underset{x\rightarrow a^{-}}{\text{lim}}f(x)} & = & {\text{+}\infty\ \text{or}\ \text{−∞}} \\ {\underset{x\rightarrow a^{+}}{\text{lim}}f(x)} & = & {\text{+}\infty\ \text{or}\ \text{−∞}} \\ & {\ \text{or}} & \\ {\underset{x\rightarrow a}{\text{lim}}f(x)} & = & {\text{+}\infty\ \text{or}\ \text{−∞}} \end{array}$$

Finding a Vertical Asymptote 求垂直渐近线

Evaluate each of the following limits using Infinite Limits from Positive Integers. Identify any vertical asymptotes of the function $f(x) = {1\text{/}\left( {x + 3} \right)^{4}}.$

1. $\underset{x\rightarrow-3^{-}}{\text{lim}}\frac{1}{\left( {x + 3} \right)^{4}}$

2. $\underset{x\rightarrow-3^{+}}{\text{lim}}\frac{1}{\left( {x + 3} \right)^{4}}$

3. $\underset{x\rightarrow-3}{\text{lim}}\frac{1}{\left( {x + 3} \right)^{4}}$

Solution 解答

We can use Infinite Limits from Positive Integers directly.

1. $\underset{x\rightarrow-3^{-}}{\text{lim}}\frac{1}{\left( {x + 3} \right)^{4}} = \text{+}\infty$

2. $\underset{x\rightarrow-3^{+}}{\text{lim}}\frac{1}{\left( {x + 3} \right)^{4}} = \text{+}\infty$

3. $\underset{x\rightarrow-3}{\text{lim}}\frac{1}{\left( {x + 3} \right)^{4}} = \text{+}\infty$

The function $f(x) = {1\text{/}\left( {x + 3} \right)^{4}}$ has a vertical asymptote of $x = -3.$

Evaluate each of the following limits. Identify any vertical asymptotes of the function $f(x) = \frac{1}{\left( {x - 2} \right)^{3}}.$

1. $\underset{x\rightarrow 2^{-}}{\text{lim}}\frac{1}{\left( {x - 2} \right)^{3}}$

2. $\underset{x\rightarrow 2^{+}}{\text{lim}}\frac{1}{\left( {x - 2} \right)^{3}}$

3. $\underset{x\rightarrow 2}{\text{lim}}\frac{1}{\left( {x - 2} \right)^{3}}$

In the next example we put our knowledge of various types of limits to use to analyze the behavior of a function at several different points.

Behavior of a Function at Different Points 函数在各点的行为

Use the graph of $f(x)$ in Figure 2.21 to determine each of the following values:

1. $\underset{x\rightarrow-4^{-}}{\text{lim}}f(x);\underset{x\rightarrow-4^{+}}{\text{lim}}f(x);\underset{x\rightarrow-4}{\text{lim}}f(x);f(-4)$

2. $\underset{x\rightarrow-2^{-}}{\text{lim}}f(x);\underset{x\rightarrow-2^{+}}{\text{lim}}f(x);\underset{x\rightarrow-2}{\text{lim}}f(x);f(-2)$

3. $\underset{x\rightarrow 1^{-}}{\text{lim}}f(x);\underset{x\rightarrow 1^{+}}{\text{lim}}f(x);\underset{x\rightarrow 1}{\text{lim}}f(x);f(1)$

4. $\underset{x\rightarrow 3^{-}}{\text{lim}}f(x);\underset{x\rightarrow 3^{+}}{\text{lim}}f(x);\underset{x\rightarrow 3}{\text{lim}}f(x);f(3)$

Solution 解答

Using Infinite Limits from Positive Integers and the graph for reference, we arrive at the following values:

1. $\underset{x\rightarrow-4^{-}}{\text{lim}}f(x) = 0;\underset{x\rightarrow-4^{+}}{\text{lim}}f(x) = 0;\underset{x\rightarrow-4}{\text{lim}}f(x) = 0;f(-4) = 0$

2. $\underset{x\rightarrow-2^{-}}{\text{lim}}f(x) = 3.;\underset{x\rightarrow-2^{+}}{\text{lim}}f(x) = 3;\underset{x\rightarrow-2}{\text{lim}}f(x) = 3;f(-2)$ is undefined

3. $\underset{x\rightarrow 1^{-}}{\text{lim}}f(x) = 6;\underset{x\rightarrow 1^{+}}{\text{lim}}f(x) = 3;\underset{x\rightarrow 1}{\text{lim}}f(x)$ DNE; $f(1) = 6$

4. $\underset{x\rightarrow 3^{-}}{\text{lim}}f(x) = \text{−}\infty;\underset{x\rightarrow 3^{+}}{\text{lim}}f(x) = \text{−}\infty;\underset{x\rightarrow 3}{\text{lim}}f(x) = \text{−}\infty;f(3)$ is undefined

Evaluate $\underset{x\rightarrow 1}{\text{lim}}f(x)$ for $f(x)$ shown here:

Chapter Opener: Einstein’s Equation 章首图：爱因斯坦方程

In the chapter opener we mentioned briefly how Albert Einstein showed that a limit exists to how fast any object can travel. Given Einstein’s equation for the mass of a moving object, what is the value of this bound?

Solution 解答

Our starting point is Einstein’s equation for the mass of a moving object,

$$m = \frac{m_{0}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}},$$

where $m_{0}$ is the object’s mass at rest, v is its speed, and c is the speed of light. To see how the mass changes at high speeds, we can graph the ratio of masses $m\text{/}m_{0}$ as a function of the ratio of speeds, $v\text{/}c$ (Figure 2.23).

We can see that as the ratio of speeds approaches 1—that is, as the speed of the object approaches the speed of light—the ratio of masses increases without bound. In other words, the function has a vertical asymptote at ${v\text{/}c} = 1.$ We can try a few values of this ratio to test this idea.

Table 2.8 Ratio of Masses and Speeds for a Moving Object

Thus, according to Table 2.8, if an object with mass 100 kg is traveling at 0.9999c, its mass becomes 7071 kg. Since no object can have an infinite mass, we conclude that no object can travel at or more than the speed of light.

Section 2.2 Exercises 2.2 节习题

For the following exercises, consider the function $f(x) = \frac{x^{2} - 1}{\left| {x - 1} \right|}.$

30.

\[T\] Complete the following table for the function. Round your solutions to four decimal places.

31.

What do your results in the preceding exercise indicate about the two-sided limit $\underset{x\rightarrow 1}{\text{lim}}f(x)?$ Explain your response.

For the following exercises, consider the function $f(x) = \left( {1 + x} \right)^{1\text{/}x}.$

32.

\[T\] Make a table showing the values of f for $x = -0.01,-0.001,-0.0001,-0.00001$ and for $x = 0.01,0.001,0.0001,0.00001.$ Round your solutions to five decimal places.

33.

What does the table of values in the preceding exercise indicate about the function $f(x) = \left( {1 + x} \right)^{1\text{/}x}?$

34.

To which mathematical constant does the limit in the preceding exercise appear to be getting closer?

In the following exercises, use the given values to set up a table to evaluate the limits. Round your solutions to eight decimal places.

35.

\[T\] $\underset{x\rightarrow 0}{\text{lim}}\frac{\text{sin}\mspace{2mu} 2x}{x};\ \pm 0.1,\pm 0.01,\pm 0.001,\pm.0001$

36.

\[T\] $\underset{x\rightarrow 0}{\text{lim}}\frac{\text{sin}\mspace{2mu} 3x}{x}\operatorname{}$ ±0.1, ±0.01, ±0.001, ±0.0001

37.

Use the preceding two exercises to conjecture (guess) the value of the following limit: $\underset{x\rightarrow 0}{\text{lim}}\frac{\text{sin}\mspace{2mu} ax}{x}$ for a, a positive real value.

\[T\] In the following exercises, set up a table of values to find the indicated limit. Round to eight digits.

38.

$\underset{x\rightarrow 2}{\text{lim}}\frac{x^{2} - 4}{x^{2} + x - 6}$

39.

$\underset{x\rightarrow 1}{\text{lim}}\left( {1 - 2x} \right)$

40.

$\underset{x\rightarrow 0}{\text{lim}}\frac{5}{1 - e^{1\text{/}x}}$

41.

$\underset{z\rightarrow 0}{\text{lim}}\frac{z - 1}{z^{2}\left( {z + 3} \right)}$

42.

$\underset{t\rightarrow 0^{+}}{\text{lim}}\frac{\text{cos}\mspace{2mu} t}{t}$

43.

$\underset{x\rightarrow 2}{\text{lim}}\frac{1 - \frac{2}{x}}{x^{2} - 4}$

\[T\] In the following exercises, set up a table of values and round to eight significant digits. Based on the table of values, make a guess about what the limit is. Then, use a calculator to graph the function and determine the limit. Was the conjecture correct? If not, why does the method of tables fail?

44.

$\underset{\theta\rightarrow 0}{\text{lim}}\text{sin}\mspace{2mu}\left( \frac{\pi}{\theta} \right)$

45.

$\underset{\alpha\rightarrow 0^{+}}{\text{lim}}\frac{1}{\alpha}\mspace{2mu}\text{cos}\mspace{2mu}\left( \frac{\pi}{\alpha} \right)$

In the following exercises, consider the graph of the function $y = f(x)$ shown here. Which of the statements about $y = f(x)$ are true and which are false? Explain why a statement is false.

46.

$\underset{x\rightarrow 10}{\text{lim}}f(x) = 0$

47.

$\underset{x\rightarrow-2^{+}}{\text{lim}}f(x) = 3$

48.

$\underset{x\rightarrow-8}{\text{lim}}f(x) = f(-8)$

49.

$\underset{x\rightarrow 6}{\text{lim}}f(x) = 5$

In the following exercises, use the following graph of the function $y = f(x)$ to find the values, if possible. Estimate when necessary.

50.

$\underset{x\rightarrow 1^{-}}{\text{lim}}f(x)$

51.

$\underset{x\rightarrow 1^{+}}{\text{lim}}f(x)$

52.

$\underset{x\rightarrow 1}{\text{lim}}f(x)$

53.

$\underset{x\rightarrow 2}{\text{lim}}f(x)$

54.

$f(1)$

In the following exercises, use the graph of the function $y = f(x)$ shown here to find the values, if possible. Estimate when necessary.

55.

$\underset{x\rightarrow 0^{-}}{\text{lim}}f(x)$

56.

$\underset{x\rightarrow 0^{+}}{\text{lim}}f(x)$

57.

$\underset{x\rightarrow 0}{\text{lim}}f(x)$

58.

$\underset{x\rightarrow 2}{\text{lim}}f(x)$

In the following exercises, use the graph of the function $y = f(x)$ shown here to find the values, if possible. Estimate when necessary.

59.

$\underset{x\rightarrow-2^{-}}{\text{lim}}f(x)$

60.

$\underset{x\rightarrow-2^{+}}{\text{lim}}f(x)$

61.

$\underset{x\rightarrow-2}{\text{lim}}f(x)$

62.

$\underset{x\rightarrow 2^{-}}{\text{lim}}f(x)$

63.

$\underset{x\rightarrow 2^{+}}{\text{lim}}f(x)$

64.

$\underset{x\rightarrow 2}{\text{lim}}f(x)$

In the following exercises, use the graph of the function $y = g(x)$ shown here to find the values, if possible. Estimate when necessary.

65.

$\underset{x\rightarrow 0^{-}}{\text{lim}}g(x)$

66.

$\underset{x\rightarrow 0^{+}}{\text{lim}}g(x)$

67.

$\underset{x\rightarrow 0}{\text{lim}}g(x)$

In the following exercises, use the graph of the function $y = h(x)$ shown here to find the values, if possible. Estimate when necessary.

68.

$\underset{x\rightarrow 0^{-}}{\text{lim}}h(x)$

69.

$\underset{x\rightarrow 0^{+}}{\text{lim}}h(x)$

70.

$\underset{x\rightarrow 0}{\text{lim}}h(x)$

In the following exercises, use the graph of the function $y = f(x)$ shown here to find the values, if possible. Estimate when necessary.

71.

$\underset{x\rightarrow 0^{-}}{\text{lim}}f(x)$

72.

$\underset{x\rightarrow 0^{+}}{\text{lim}}f(x)$

73.

$\underset{x\rightarrow 0}{\text{lim}}f(x)$

74.

$\underset{x\rightarrow 1}{\text{lim}}f(x)$

75.

$\underset{x\rightarrow 2}{\text{lim}}f(x)$

In the following exercises, sketch the graph of a function with the given properties.

76.

$\underset{x\rightarrow 2}{\text{lim}}f(x) = 1,\underset{x\rightarrow 4^{-}}{\text{lim}}f(x) = 3,\underset{x\rightarrow 4^{+}}{\text{lim}}f(x) = 6,f(4)$ is not defined.

77.

$As~x\rightarrow - \infty~,~f(x)\rightarrow 0,\underset{x\rightarrow-1^{-}}{\text{lim}}f(x) = \text{−}\infty,$ $\underset{x\rightarrow-1^{+}}{\text{lim}}f(x) = \infty,\underset{x\rightarrow 0}{\text{lim}}f(x) = f(0),f(0) = 1,~As~x\rightarrow\infty,~f(x)\rightarrow\text{−}\infty$

78.

$As~x\rightarrow - \infty,~f(x)\rightarrow 2,\underset{x\rightarrow 3^{-}}{\text{lim}}f(x) = \text{−}\infty,$ $\underset{x\rightarrow 3^{+}}{\text{lim}}f(x) = \infty,~As~x\rightarrow\infty,~f(x)\rightarrow 2,f(0) = \frac{-1}{3}$

79.

$As~x\rightarrow - \infty,~f(x)\rightarrow 2,\underset{x\rightarrow-2}{\text{lim}}f(x) = \text{−}\infty,$ $As~x\rightarrow\infty,~f(x)\rightarrow 2,f(0) = 0$

80.

$As~x\rightarrow - \infty,~f(x)\rightarrow 0,\underset{x\rightarrow-1^{-}}{\text{lim}}f(x) = \infty,\underset{x\rightarrow-1^{+}}{\text{lim}}f(x) = \text{−}\infty,$ $f(0) = -1,\underset{x\rightarrow 1^{-}}{\text{lim}}f(x) = \text{−}\infty,\underset{x\rightarrow 1^{+}}{\text{lim}}f(x) = \infty,~As~x\rightarrow\infty,~f(x)\rightarrow 0$

81.

Shock waves arise in many physical applications, ranging from supernovas to detonation waves. A graph of the density of a shock wave with respect to distance, x, is shown here. We are mainly interested in the location of the front of the shock, labeled $x_{\text{SF}}$ in the diagram.

1. Evaluate $\underset{x\rightarrow x_{SF}{}^{+}}{\text{lim}}\rho(x).$

2. Evaluate $\underset{x\rightarrow x_{SF}{}^{-}}{\text{lim}}\rho(x).$

3. Evaluate $\underset{x\rightarrow x_{SF}}{\text{lim}}\rho(x).$ Explain the physical meanings behind your answers.

82.

A track coach uses a camera with a fast shutter to estimate the position of a runner with respect to time. A table of the values of position of the athlete versus time is given here, where x is the position in meters of the runner and t is time in seconds. What is $\underset{t\rightarrow 2}{\text{lim}}x(t)?$ What does it mean physically?

2.3 The Limit Laws 2.3 极限法则

• 2.3.1 Recognize the basic limit laws.
• 2.3.2 Use the limit laws to evaluate the limit of a function.
• 2.3.3 Evaluate the limit of a function by factoring.
• 2.3.4 Use the limit laws to evaluate the limit of a polynomial or rational function.
• 2.3.5 Evaluate the limit of a function by factoring or by using conjugates.
• 2.3.6 Evaluate the limit of a function by using the squeeze theorem.

• 2.3.1 识别基本极限法则。
• 2.3.2 运用极限法则求函数的极限。
• 2.3.3 通过因式分解求函数的极限。
• 2.3.4 运用极限法则求多项式或有理函数的极限。
• 2.3.5 通过因式分解或乘以共轭求函数的极限。
• 2.3.6 运用夹逼定理求函数的极限。

In the previous section, we evaluated limits by looking at graphs or by constructing a table of values. In this section, we establish laws for calculating limits and learn how to apply these laws. In the Student Project at the end of this section, you have the opportunity to apply these limit laws to derive the formula for the area of a circle by adapting a method devised by the Greek mathematician Archimedes. We begin by restating two useful limit results from the previous section. These two results, together with the limit laws, serve as a foundation for calculating many limits.

Evaluating Limits with the Limit Laws 运用极限法则求极限

The first two limit laws were stated in Two Important Limits and we repeat them here. These basic results, together with the other limit laws, allow us to evaluate limits of many algebraic functions.

Basic Limit Results 基本极限结论

For any real number a and any constant c,

$$\underset{x\rightarrow a}{\text{lim}}x = a$$ (2.14)

$$\underset{x\rightarrow a}{\text{lim}}c = c$$ (2.15)

Evaluating a Basic Limit 求一个基本极限

Evaluate each of the following limits using Basic Limit Results.

1. $\underset{x\rightarrow 2}{\text{lim}}x$

2. $\underset{x\rightarrow 2}{\text{lim}}5$

Solution 解答

1. The limit of x as x approaches a is a: $\underset{x\rightarrow 2}{\text{lim}}x = 2.$

2. The limit of a constant is that constant: $\underset{x\rightarrow 2}{\text{lim}}5 = 5.$

We now take a look at the limit laws, the individual properties of limits. The proofs that these laws hold are omitted here.

Limit Laws 极限法则

Let $f(x)$ and $g(x)$ be defined for all $x \neq a$ over some open interval containing a. Assume that L and M are real numbers such that $\underset{x\rightarrow a}{\text{lim}}f(x) = L$ and $\underset{x\rightarrow a}{\text{lim}}g(x) = M.$ Let c be a constant. Then, each of the following statements holds:

Sum law for limits: $\underset{x\rightarrow a}{\text{lim}}\left( {f(x) + g(x)} \right) = \underset{x\rightarrow a}{\text{lim}}f(x) + \underset{x\rightarrow a}{\text{lim}}g(x) = L + M$

Difference law for limits: $\underset{x\rightarrow a}{\text{lim}}\left( {f(x) - g(x)} \right) = \underset{x\rightarrow a}{\text{lim}}f(x) - \underset{x\rightarrow a}{\text{lim}}g(x) = L - M$

Constant multiple law for limits: $\underset{x\rightarrow a}{\text{lim}}cf(x) = c \cdot \underset{x\rightarrow a}{\text{lim}}f(x) = cL$

Product law for limits: $\underset{x\rightarrow a}{\text{lim}}\left( {f(x) \cdot g(x)} \right) = \underset{x\rightarrow a}{\text{lim}}f(x) \cdot \underset{x\rightarrow a}{\text{lim}}g(x) = L \cdot M$

Quotient law for limits: $\underset{x\rightarrow a}{\text{lim}}\frac{f(x)}{g(x)} = \frac{\underset{x\rightarrow a}{\text{lim}}f(x)}{\underset{x\rightarrow a}{\text{lim}}g(x)} = \frac{L}{M}$ for $M \neq 0$

Power law for limits: $\underset{x\rightarrow a}{\text{lim}}\left( {f(x)} \right)^{n} = \left( {\underset{x\rightarrow a}{\text{lim}}f(x)} \right)^{n} = L^{n}$ for every positive integer n.

Root law for limits: $\underset{x\rightarrow a}{\text{lim}}\sqrt[n]{f(x)} = \sqrt[n]{\underset{x\rightarrow a}{\text{lim}}f(x)} = \sqrt[n]{L}$ for all L if n is odd and for $L \geq 0$ if n is even and $f(x) \geq 0$.

We now practice applying these limit laws to evaluate a limit.

Evaluating a Limit Using Limit Laws 运用极限法则求极限

Use the limit laws to evaluate $\underset{x\rightarrow-3}{\text{lim}}\left( {4x + 2} \right).$

Solution 解答

Let’s apply the limit laws one step at a time to be sure we understand how they work. We need to keep in mind the requirement that, at each application of a limit law, the new limits must exist for the limit law to be applied.

$\begin{array}{clccl} {\underset{x\rightarrow-3}{\text{lim}}\left( {4x + 2} \right)} & {= \underset{x\rightarrow-3}{\text{lim}}4x + \underset{x\rightarrow-3}{\text{lim}}2} & & & \text{Apply the sum law.} \\ & {= 4 \cdot \underset{x\rightarrow-3}{\text{lim}}x + \underset{x\rightarrow-3}{\text{lim}}2} & & & \text{Apply the constant multiple law.} \\ & {= 4 \cdot (-3) + 2 = -10.} & & & \text{Apply the basic limit results and simplify.} \end{array}$

Using Limit Laws Repeatedly 反复运用极限法则

Use the limit laws to evaluate $\underset{x\rightarrow 2}{\text{lim}}\frac{2x^{2} - 3x + 1}{x^{3} + 4}.$

Solution 解答

To find this limit, we need to apply the limit laws several times. Again, we need to keep in mind that as we rewrite the limit in terms of other limits, each new limit must exist for the limit law to be applied.

$\begin{array}{clccl} & & & & \\ & & & & \\ {\underset{x\rightarrow 2}{\text{lim}}\frac{2x^{2} - 3x + 1}{x^{3} + 4}} & {= \frac{\underset{x\rightarrow 2}{\text{lim}}\left( {2x^{2} - 3x + 1} \right)}{\underset{x\rightarrow 2}{\text{lim}}\left( {x^{3} + 4} \right)}} & & & {\text{Apply the quotient law, making sure that.}\ (2)^{3} + 4 \neq 0} \\ & {= \frac{2 \cdot \underset{x\rightarrow 2}{\text{lim}}x^{2} - 3 \cdot \underset{x\rightarrow 2}{\text{lim}}x + \underset{x\rightarrow 2}{\text{lim}}1}{\underset{x\rightarrow 2}{\text{lim}}x^{3} + \underset{x\rightarrow 2}{\text{lim}}4}} & & & \text{Apply the sum law and constant multiple law.} \\ & {= \frac{2 \cdot \left( {\underset{x\rightarrow 2}{\text{lim}}x} \right)^{2} - 3 \cdot \underset{x\rightarrow 2}{\text{lim}}x + \underset{x\rightarrow 2}{\text{lim}}1}{\left( {\underset{x\rightarrow 2}{\text{lim}}x} \right)^{3} + \underset{x\rightarrow 2}{\text{lim}}4}} & & & \text{Apply the power law.} \\ & {= \frac{2(4) - 3(2) + 1}{(2)^{3} + 4} = \frac{1}{4}.} & & & \text{Apply the basic limit laws and simplify.} \end{array}$

Use the limit laws to evaluate $\underset{x\rightarrow 6}{\text{lim}}\left( {2x - 1} \right)\sqrt{x + 4}.$ In each step, indicate the limit law applied.

Limits of Polynomial and Rational Functions 多项式与有理函数的极限

By now you have probably noticed that, in each of the previous examples, it has been the case that $\underset{x\rightarrow a}{\text{lim}}f(x) = f(a).$ This is not always true, but it does hold for all polynomials for any choice of a and for all rational functions at all values of a for which the rational function is defined.

Limits of Polynomial and Rational Functions 多项式与有理函数的极限

Let $p(x)$ and $q(x)$ be polynomial functions. Let a be a real number. Then,

$$\underset{x\rightarrow a}{\text{lim}}p(x) = p(a)$$ $$\underset{x\rightarrow a}{\text{lim}}\frac{p(x)}{q(x)} = \frac{p(a)}{q(a)}\ \text{when}\ q(a) \neq 0.$$

To see that this theorem holds, consider the polynomial $p(x) = c_{n}x^{n} + c_{n - 1}x^{n - 1} + \cdots + c_{1}x + c_{0}.$ By applying the sum, constant multiple, and power laws, we end up with

$$\begin{array}{cl} {\underset{x\rightarrow a}{\text{lim}}p(x)} & {= \underset{x\rightarrow a}{\text{lim}}\left( {c_{n}x^{n} + c_{n - 1}x^{n - 1} + \cdots + c_{1}x + c_{0}} \right)} \\ & {= c_{n}\left( {\underset{x\rightarrow a}{\text{lim}}x} \right)^{n} + c_{n - 1}\left( {\underset{x\rightarrow a}{\text{lim}}x} \right)^{n - 1} + \cdots + c_{1}\left( {\underset{x\rightarrow a}{\text{lim}}x} \right) + \underset{x\rightarrow a}{\text{lim}}c_{0}} \\ & {= c_{n}a^{n} + c_{n - 1}a^{n - 1} + \cdots + c_{1}a + c_{0}} \\ & {= p(a).} \end{array}$$

It now follows from the quotient law that if $p(x)$ and $q(x)$ are polynomials for which $q(a) \neq 0,$ then

$$\underset{x\rightarrow a}{\text{lim}}\frac{p(x)}{q(x)} = \frac{p(a)}{q(a)}.$$

Example 2.16 applies this result.

Evaluating a Limit of a Rational Function 求有理函数的极限

Evaluate the $\underset{x\rightarrow 3}{\text{lim}}\frac{2x^{2} - 3x + 1}{5x + 4}.$

Solution 解答

Since 3 is in the domain of the rational function $f(x) = \frac{2x^{2} - 3x + 1}{5x + 4},$ we can calculate the limit by substituting 3 for x into the function. Thus,

$$\underset{x\rightarrow 3}{\text{lim}}\frac{2x^{2} - 3x + 1}{5x + 4} = \frac{10}{19}.$$

Evaluate $\underset{x\rightarrow-2}{\text{lim}}\left( {3x^{3} - 2x + 7} \right).$

Additional Limit Evaluation Techniques 附加的极限求值技巧

As we have seen, we may evaluate easily the limits of polynomials and limits of some (but not all) rational functions by direct substitution. However, as we saw in the introductory section on limits, it is certainly possible for $\underset{x\rightarrow a}{\text{lim}}f(x)$ to exist when $f(a)$ is undefined. The following observation allows us to evaluate many limits of this type:

If for all $x \neq a,f(x) = g(x)$ over some open interval containing a, then $\underset{x\rightarrow a}{\text{lim}}f(x) = \underset{x\rightarrow a}{\text{lim}}g(x).$

To understand this idea better, consider the limit $\underset{x\rightarrow 1}{\text{lim}}\frac{x^{2} - 1}{x - 1}.$

The function

$$\begin{array}{cl} {f(x)} & {= \frac{x^{2} - 1}{x - 1}} \\ & {= \frac{\left( {x - 1} \right)\left( {x + 1} \right)}{x - 1}} \end{array}$$

and the function $g(x) = x + 1$ are identical for all values of $x \neq 1.$ The graphs of these two functions are shown in Figure 2.24.

We see that

$$\begin{array}{cl} {\underset{x\rightarrow 1}{\text{lim}}\frac{x^{2} - 1}{x - 1}} & {= \underset{x\rightarrow 1}{\text{lim}}\frac{\left( {x - 1} \right)\left( {x + 1} \right)}{x - 1}} \\ & {= \underset{x\rightarrow 1}{\text{lim}}\left( {x + 1} \right)} \\ & {= 2.} \end{array}$$

The limit has the form $\underset{x\rightarrow a}{\text{lim}}\frac{f(x)}{g(x)},$ where $\underset{x\rightarrow a}{\text{lim}}f(x) = 0$ and $\underset{x\rightarrow a}{\text{lim}}g(x) = 0.$ (In this case, we say that ${f(x)}\text{/}{g(x)}$ has the indeterminate form ${0\text{/}0}\text{.)}$ The following Problem-Solving Strategy provides a general outline for evaluating limits of this type.

Calculating a Limit When $f(x)\text{/}g(x)$ has the Indeterminate Form 0/0 当 $f(x)\text{/}g(x)$ 呈未定式 0/0 时求极限

1. First, we need to make sure that our function has the appropriate form and cannot be evaluated immediately using the limit laws.

2. We then need to find a function that is equal to $h(x) = {{f(x)}\text{/}{g(x)}}$ for all $x \neq a$ over some interval containing a. To do this, we may need to try one or more of the following steps:

1. If $f(x)$ and $g(x)$ are polynomials, we should factor each function and cancel out any common factors.

2. If the numerator or denominator contains a difference involving a square root, we should try multiplying the numerator and denominator by the conjugate of the expression involving the square root.

3. If ${f(x)}\text{/}{g(x)}$ is a complex fraction, we begin by simplifying it.

3. Last, we apply the limit laws.

The next examples demonstrate the use of this Problem-Solving Strategy. Example 2.17 illustrates the factor-and-cancel technique; Example 2.18 shows multiplying by a conjugate. In Example 2.19, we look at simplifying a complex fraction.

Evaluating a Limit by Factoring and Canceling 通过因式分解与约分求极限

Evaluate $\underset{x\rightarrow 3}{\text{lim}}\frac{x^{2} - 3x}{2x^{2} - 5x - 3}.$

Solution 解答

Step 1. The function $f(x) = \frac{x^{2} - 3x}{2x^{2} - 5x - 3}$ is undefined for $x = 3.$ In fact, if we substitute 3 into the function we get ${0\text{/}0},$ which is indeterminate. Factoring and canceling is a good strategy:

$$\underset{x\rightarrow 3}{\text{lim}}\frac{x^{2} - 3x}{2x^{2} - 5x - 3} = \underset{x\rightarrow 3}{\text{lim}}\frac{x\left( {x - 3} \right)}{\left( {x - 3} \right)\left( {2x + 1} \right)}$$

Step 2. For all $x \neq 3,\frac{x^{2} - 3x}{2x^{2} - 5x - 3} = \frac{x}{2x + 1}.$ Therefore,

$$\underset{x\rightarrow 3}{\text{lim}}\frac{x\left( {x - 3} \right)}{\left( {x - 3} \right)\left( {2x + 1} \right)} = \underset{x\rightarrow 3}{\text{lim}}\frac{x}{2x + 1}.$$

Step 3. Evaluate using the limit laws:

$$\underset{x\rightarrow 3}{\text{lim}}\frac{x}{2x + 1} = \frac{3}{7}.$$

Evaluate $\underset{x\rightarrow-3}{\text{lim}}\frac{x^{2} + 4x + 3}{x^{2} - 9}.$

Evaluating a Limit by Multiplying by a Conjugate 通过乘以共轭求极限

Evaluate $\underset{x\rightarrow-1}{\text{lim}}\frac{\sqrt{x + 2} - 1}{x + 1}.$

Solution 解答

Step 1. $\frac{\sqrt{x + 2} - 1}{x + 1}$ has the form $0\text{/}0$ at −1. Let’s begin by multiplying by $\sqrt{x + 2} + 1,$ the conjugate of $\sqrt{x + 2} - 1,$ on the numerator and denominator:

$$\underset{x\rightarrow-1}{\text{lim}}\frac{\sqrt{x + 2} - 1}{x + 1} = \underset{x\rightarrow-1}{\text{lim}}\frac{\sqrt{x + 2} - 1}{x + 1} \cdot \frac{\sqrt{x + 2} + 1}{\sqrt{x + 2} + 1}.$$

Step 2. We then multiply out the numerator. We don’t multiply out the denominator because we are hoping that the $\left( {x + 1} \right)$ in the denominator cancels out in the end:

$$= \underset{x\rightarrow-1}{\text{lim}}\frac{x + 1}{\left( {x + 1} \right)\left( {\sqrt{x + 2} + 1} \right)}.$$

Step 3. Then we cancel:

$$= \underset{x\rightarrow-1}{\text{lim}}\frac{1}{\sqrt{x + 2} + 1}.$$

Step 4. Last, we apply the limit laws:

$$\underset{x\rightarrow-1}{\text{lim}}\frac{1}{\sqrt{x + 2} + 1} = \frac{1}{2}.$$

Evaluate $\underset{x\rightarrow 5}{\text{lim}}\frac{\sqrt{x - 1} - 2}{x - 5}.$

Evaluating a Limit by Simplifying a Complex Fraction 通过化简繁分数求极限

Evaluate $\underset{x\rightarrow 1}{\text{lim}}\frac{\frac{1}{x + 1} - \frac{1}{2}}{x - 1}.$

Solution 解答

Step 1. $\frac{\frac{1}{x + 1} - \frac{1}{2}}{x - 1}$ has the form $0\text{/}0$ at 1. We simplify the algebraic fraction by multiplying by ${{2\left( {x + 1} \right)}\text{/}{2\left( {x + 1} \right)}}:$

$$\underset{x\rightarrow 1}{\text{lim}}\frac{\frac{1}{x + 1} - \frac{1}{2}}{x - 1} = \underset{x\rightarrow 1}{\text{lim}}\frac{\frac{1}{x + 1} - \frac{1}{2}}{x - 1} \cdot \frac{2\left( {x + 1} \right)}{2\left( {x + 1} \right)}.$$

Step 2. Next, we multiply through the numerators. Do not multiply the denominators because we want to be able to cancel the factor $\left( {x - 1} \right)\text{:}$

$$= \underset{x\rightarrow 1}{\text{lim}}\frac{2 - \left( {x + 1} \right)}{2\left( {x - 1} \right)\left( {x + 1} \right)}.$$

Step 3. Then, we simplify the numerator:

$$= \underset{x\rightarrow 1}{\text{lim}}\frac{- x + 1}{2\left( {x - 1} \right)\left( {x + 1} \right)}.$$

Step 4. Now we factor out −1 from the numerator:

$$= \underset{x\rightarrow 1}{\text{lim}}\frac{- \left( {x - 1} \right)}{2\left( {x - 1} \right)\left( {x + 1} \right)}.$$

Step 5. Then, we cancel the common factors of $\left( {x - 1} \right)\text{:}$

$$= \underset{x\rightarrow 1}{\text{lim}}\frac{-1}{2\left( {x + 1} \right)}.$$

Step 6. Last, we evaluate using the limit laws:

$$\underset{x\rightarrow 1}{\text{lim}}\frac{-1}{2(x + 1)} = - \frac{1}{4}.$$

Evaluate $\underset{x\rightarrow-3}{\text{lim}}\frac{\frac{1}{x + 2} + 1}{x + 3}.$

Example 2.20 does not fall neatly into any of the patterns established in the previous examples. However, with a little creativity, we can still use these same techniques.

Evaluating a Limit When the Limit Laws Do Not Apply 当极限法则不适用时求极限

Evaluate $\underset{x\rightarrow 0}{\text{lim}}\left( {\frac{1}{x} + \frac{5}{x\left( {x - 5} \right)}} \right).$

Solution 解答

Both $1\text{/}x$ and $5\text{/}{x\left( {x - 5} \right)}$ fail to have a limit at zero. Since neither of the two functions has a limit at zero, we cannot apply the sum law for limits; we must use a different strategy. In this case, we find the limit by performing addition and then applying one of our previous strategies. Observe that

$$\quad\begin{array}{cl} {\frac{1}{x} + \frac{5}{x\left( {x - 5} \right)}} & {= \frac{x - 5 + 5}{x\left( {x - 5} \right)}} \\ & {= \frac{x}{x\left( {x - 5} \right)}.} \end{array}$$

Thus,

$$\begin{array}{cl} {\underset{x\rightarrow 0}{\text{lim}}\left( {\frac{1}{x} + \frac{5}{x\left( {x - 5} \right)}} \right)} & {= \underset{x\rightarrow 0}{\text{lim}}\frac{x}{x\left( {x - 5} \right)}} \\ & {= \underset{x\rightarrow 0}{\text{lim}}\frac{1}{x - 5}} \\ & {= - \frac{1}{5}.} \end{array}$$

Evaluate $\underset{x\rightarrow 3}{\text{lim}}\left( {\frac{1}{x - 3} - \frac{4}{x^{2} - 2x - 3}} \right).$

Let’s now revisit one-sided limits. Simple modifications in the limit laws allow us to apply them to one-sided limits. For example, to apply the limit laws to a limit of the form $\underset{x\rightarrow a^{-}}{\text{lim}}h(x),$ we require the function $h(x)$ to be defined over an open interval of the form $\left( {b,a} \right);$ for a limit of the form $\underset{x\rightarrow a^{+}}{\text{lim}}h(x),$ we require the function $h(x)$ to be defined over an open interval of the form $\left( {a,c} \right).$ Example 2.21 illustrates this point.

Evaluating a One-Sided Limit Using the Limit Laws 运用极限法则求单侧极限

Evaluate each of the following limits, if possible.

1. $\underset{x\rightarrow 3^{-}}{\text{lim}}\sqrt{x - 3}$

2. $\underset{x\rightarrow 3^{+}}{\text{lim}}\sqrt{x - 3}$

Solution 解答

Figure 2.25 illustrates the function $f(x) = \sqrt{x - 3}$ and aids in our understanding of these limits.

1. The function $f(x) = \sqrt{x - 3}$ is defined over the interval $\left\lbrack {3,\text{+}\infty} \right).$ Since this function is not defined to the left of 3, we cannot apply the limit laws to compute $\underset{x\rightarrow 3^{-}}{\text{lim}}\sqrt{x - 3}.$ In fact, since $f(x) = \sqrt{x - 3}$ is undefined to the left of 3, $\underset{x\rightarrow 3^{-}}{\text{lim}}\sqrt{x - 3}$ does not exist.

2. Since $f(x) = \sqrt{x - 3}$ is defined to the right of 3, the limit laws do apply to $\underset{x\rightarrow 3^{+}}{\text{lim}}\sqrt{x - 3}.$ By applying these limit laws we obtain $\underset{x\rightarrow 3^{+}}{\text{lim}}\sqrt{x - 3} = 0.$

In Example 2.22 we look at one-sided limits of a piecewise-defined function and use these limits to draw a conclusion about a two-sided limit of the same function.

Evaluating a Two-Sided Limit Using the Limit Laws 运用极限法则求双侧极限

For $f(x) = \begin{cases} {4x - 3} & {\text{if}\ x < 2} \\ \left( {x - 3} \right)^{2} & {\text{if}\ x \geq 2} \end{cases},$ evaluate each of the following limits:

1. $\underset{x\rightarrow 2^{-}}{\text{lim}}f(x)$

2. $\underset{x\rightarrow 2^{+}}{\text{lim}}f(x)$

3. $\underset{x\rightarrow 2}{\text{lim}}f(x)$

Solution 解答

Figure 2.26 illustrates the function $f(x)$ and aids in our understanding of these limits.

1. Since $f(x) = 4x - 3$ for all x in $\left( {\text{−}\infty,2} \right),$ replace $f(x)$ in the limit with $4x - 3$ and apply the limit laws:

$$\underset{x\rightarrow 2^{-}}{\text{lim}}f(x) = \underset{x\rightarrow 2^{-}}{\text{lim}}\left( {4x - 3} \right) = 5.$$

2. Since $f(x) = \left( {x - 3} \right)^{2}$ for all x in $\left( {2,\text{+}\infty} \right),$ replace $f(x)$ in the limit with $\left( {x - 3} \right)^{2}$ and apply the limit laws:

$$\underset{x\rightarrow 2^{+}}{\text{lim}}f(x) = \underset{x\rightarrow 2^{+}}{\text{lim}}\left( {x - 3} \right)^{2} = 1.$$

3. Since $\underset{x\rightarrow 2^{-}}{\text{lim}}f(x) = 5$ and $\underset{x\rightarrow 2^{+}}{\text{lim}}f(x) = 1,$ we conclude that $\underset{x\rightarrow 2}{\text{lim}}f(x)$ does not exist.

Graph $f(x) = \left\{ \begin{matrix} {- x - 2\ \text{if}\ x < \text{−}1} \\ {2\ \text{if}\ x = -1} \\ {x^{3}\ \text{if}\ x > \text{−}1} \end{matrix} \right.$ and evaluate $\underset{x\rightarrow-1^{-}}{\text{lim}}f(x).$

We now turn our attention to evaluating a limit of the form $\underset{x\rightarrow a}{\text{lim}}\frac{f(x)}{g(x)},$ where $\underset{x\rightarrow a}{\text{lim}}f(x) = K,$ where $K \neq 0$ and $\underset{x\rightarrow a}{\text{lim}}g(x) = 0.$ That is, ${f(x)}\text{/}{g(x)}$ has the form ${K\text{/}0},K \neq 0$ at a.

Evaluating a Limit of the Form $K\text{/}0,K \neq 0$ Using the Limit Laws 运用极限法则求形如 $K\text{/}0,K \neq 0$ 的极限

Evaluate $\underset{x\rightarrow 2^{-}}{\text{lim}}\frac{x - 3}{x^{2} - 2x}.$

Solution 解答

Step 1. After substituting in $x = 2,$ we see that this limit has the form $-1\text{/}0.$ That is, as x approaches 2 from the left, the numerator approaches −1; and the denominator approaches 0. Consequently, the magnitude of $\frac{x - 3}{x(x - 2)}$ becomes infinite. To get a better idea of what the limit is, we need to factor the denominator:

$$\underset{x\rightarrow 2^{-}}{\text{lim}}\frac{x - 3}{x^{2} - 2x} = \underset{x\rightarrow 2^{-}}{\text{lim}}\frac{x - 3}{x\left( {x - 2} \right)}.$$

Step 2. Since $x - 2$ is the only part of the denominator that is zero when 2 is substituted, we then separate ${1\text{/}{(x - 2}})$ from the rest of the function:

$$= \underset{x\rightarrow 2^{-}}{\text{lim}}\frac{x - 3}{x} \cdot \frac{1}{x - 2}.$$

Step 3. $\underset{x\rightarrow 2^{-}}{\text{lim}}\frac{x - 3}{x} = - \frac{1}{2}$ and $\underset{x\rightarrow 2^{-}}{\text{lim}}\frac{1}{x - 2} = \text{−}\infty.$ Therefore, the product of ${(x - 3)}\text{/}x$ and ${1\text{/}{(x - 2}})$ has a limit of $\text{+∞:}$

$$\underset{x\rightarrow 2^{-}}{\text{lim}}\frac{x - 3}{x^{2} - 2x} = \text{+}\infty.$$

Evaluate $\underset{x\rightarrow 1}{\text{lim}}\frac{x + 2}{\left( {x - 1} \right)^{2}}.$

The Squeeze Theorem 夹逼定理

The techniques we have developed thus far work very well for algebraic functions, but we are still unable to evaluate limits of very basic trigonometric functions. The next theorem, called the squeeze theorem, proves very useful for establishing basic trigonometric limits. This theorem allows us to calculate limits by “squeezing” a function, with a limit at a point a that is unknown, between two functions having a common known limit at a. Figure 2.27 illustrates this idea.

The Squeeze Theorem 夹逼定理

Let $f(x),g(x),$ and $h(x)$ be defined for all $x \neq a$ over an open interval containing a. If

$$f(x) \leq g(x) \leq h(x)$$

for all $x \neq a$ in an open interval containing a and

$$\underset{x\rightarrow a}{\text{lim}}f(x) = L = \underset{x\rightarrow a}{\text{lim}}h(x)$$

where L is a real number, then $\underset{x\rightarrow a}{\text{lim}}g(x) = L.$

Applying the Squeeze Theorem 应用夹逼定理

Apply the squeeze theorem to evaluate $\underset{x\rightarrow 0}{\text{lim}}x\mspace{2mu}\text{cos}\mspace{2mu} x.$

Solution 解答

Because $-1 \leq \text{cos}\mspace{2mu} x \leq 1$ for all x, we have $\left. - \middle| x \middle| \leq x\mspace{2mu}\text{cos}\mspace{2mu} x \leq \middle| x \right|$. Since $\underset{x\rightarrow 0}{\text{lim}}\left( \left. - \middle| x \right| \right) = 0 = \underset{x\rightarrow 0}{\text{lim}}|x|,$ from the squeeze theorem, we obtain $\underset{x\rightarrow 0}{\text{lim}}x\mspace{2mu}\text{cos}\mspace{2mu} x = 0.$ The graphs of $\left. f(x) = - \middle| x \middle| ,g(x) = x\mspace{2mu}\text{cos}\mspace{2mu} x, \right.$ and $\left. h(x) = \middle| x \right|$ are shown in Figure 2.28.

Use the squeeze theorem to evaluate $\underset{x\rightarrow 0}{\text{lim}}x^{2}\text{sin}\frac{1}{x}.$

We now use the squeeze theorem to tackle several very important limits. Although this discussion is somewhat lengthy, these limits prove invaluable for the development of the material in both the next section and the next chapter. The first of these limits is $\underset{\theta\rightarrow 0}{\text{lim}}\text{sin}\mspace{2mu}\theta.$ Consider the unit circle shown in Figure 2.29. In the figure, we see that $\text{sin}\mspace{2mu}\theta$ is the y-coordinate on the unit circle and it corresponds to the line segment shown in blue. The radian measure of angle θ is the length of the arc it subtends on the unit circle. Therefore, we see that for $0 < \theta < \frac{\pi}{2},0 < \text{sin}\mspace{2mu}\theta < \theta.$

Because $\underset{\theta\rightarrow 0^{+}}{\text{lim}}0 = 0$ and $\underset{\theta\rightarrow 0^{+}}{\text{lim}}\theta = 0,$ by using the squeeze theorem we conclude that

$$\underset{\theta\rightarrow 0^{+}}{\text{lim}}\text{sin}\mspace{2mu}\theta = 0.$$

To see that $\underset{\theta\rightarrow 0^{-}}{\text{lim}}\text{sin}\mspace{2mu}\theta = 0$ as well, observe that for $- \frac{\pi}{2} < \theta < 0,0 < \text{−}\theta < \frac{\pi}{2}$ and hence, $0 < \text{sin}\mspace{2mu}\left( {- \theta} \right) < \text{−}\theta.$ Consequently, $0 < - \text{sin}\mspace{2mu}\theta < \text{−}\theta.$ It follows that $0 > \text{sin}\mspace{2mu}\theta > \theta.$ An application of the squeeze theorem produces the desired limit. Thus, since $\underset{\theta\rightarrow 0^{+}}{\text{lim}}\text{sin}\mspace{2mu}\theta = 0$ and $\underset{\theta\rightarrow 0^{-}}{\text{lim}}\text{sin}\mspace{2mu}\theta = 0,$

$$\underset{\theta\rightarrow 0}{\text{lim}}\mspace{7mu}\text{sin}\mspace{2mu}\theta = 0.$$ (2.16)

Next, using the identity $\text{cos}\mspace{2mu}\theta = \sqrt{1 - \text{sin}^{2}\theta}$ for $- \frac{\pi}{2} < \theta < \frac{\pi}{2},$ we see that

$$\underset{\theta\rightarrow 0}{\text{lim}}\mspace{7mu}\text{cos}\mspace{7mu}\theta = \underset{\theta\rightarrow 0}{\text{lim}}\sqrt{1 - \text{sin}^{2}\theta} = 1.$$ (2.17)

We now take a look at a limit that plays an important role in later chapters—namely, $\underset{\theta\rightarrow 0}{\text{lim}}\frac{\text{sin}\mspace{2mu}\theta}{\theta}.$ To evaluate this limit, we use the unit circle in Figure 2.30. Notice that this figure adds one additional triangle to Figure 2.30. We see that the length of the side opposite angle θ in this new triangle is $\text{tan}\mspace{2mu}\theta.$ Thus, we see that for $0 < \theta < \frac{\pi}{2},\text{sin}\mspace{2mu}\theta < \theta < \text{tan}\mspace{2mu}\theta.$

By dividing by $\text{sin}\mspace{2mu}\theta$ in all parts of the inequality, we obtain

$$1 < \frac{\theta}{\text{sin}\mspace{2mu}\theta} < \frac{1}{\text{cos}\mspace{2mu}\theta}.$$

Equivalently, we have

$$1 > \frac{\text{sin}\mspace{2mu}\theta}{\theta} > \text{cos}\mspace{2mu}\theta.$$

Since $\underset{\theta\rightarrow 0^{+}}{\text{lim}}1 = 1 = \underset{\theta\rightarrow 0^{+}}{\text{lim}}\text{cos}\mspace{2mu}\theta,$ we conclude that $\underset{\theta\rightarrow 0^{+}}{\text{lim}}\frac{\text{sin}\mspace{2mu}\theta}{\theta} = 1.$ By applying a manipulation similar to that used in demonstrating that $\underset{\theta\rightarrow 0^{-}}{\text{lim}}\text{sin}\mspace{2mu}\theta = 0,$ we can show that $\underset{\theta\rightarrow 0^{-}}{\text{lim}}\frac{\text{sin}\mspace{2mu}\theta}{\theta} = 1.$ Thus,

$$\underset{\theta\rightarrow 0}{\text{lim}}\frac{\text{sin}\mspace{2mu}\theta}{\theta} = 1.$$ (2.18)

In Example 2.25 we use this limit to establish $\underset{\theta\rightarrow 0}{\text{lim}}\frac{1 - \text{cos}\mspace{2mu}\theta}{\theta} = 0.$ This limit also proves useful in later chapters.

Evaluating an Important Trigonometric Limit 求一个重要的三角函数极限

Evaluate $\underset{\theta\rightarrow 0}{\text{lim}}\frac{1 - \text{cos}\mspace{2mu}\theta}{\theta}.$

Solution 解答

In the first step, we multiply by the conjugate so that we can use a trigonometric identity to convert the cosine in the numerator to a sine:

$$\begin{array}{cl} {\underset{\theta\rightarrow 0}{\text{lim}}\frac{1 - \text{cos}\mspace{2mu}\theta}{\theta}} & {= \underset{\theta\rightarrow 0}{\text{lim}}\frac{1 - \text{cos}\mspace{2mu}\theta}{\theta} \cdot \frac{1 + \text{cos}\mspace{2mu}\theta}{1 + \text{cos}\mspace{2mu}\theta}} \\ & {= \underset{\theta\rightarrow 0}{\text{lim}}\frac{1 - \text{cos}^{2}\theta}{\theta\left( {1 + \text{cos}\mspace{2mu}\theta} \right)}} \\ & {= \underset{\theta\rightarrow 0}{\text{lim}}\frac{\text{sin}^{2}\theta}{\theta\left( {1 + \text{cos}\mspace{2mu}\theta} \right)}} \\ & {= \underset{\theta\rightarrow 0}{\text{lim}}\frac{\text{sin}\mspace{2mu}\theta}{\theta} \cdot \frac{\text{sin}\mspace{2mu}\theta}{1 + \text{cos}\mspace{2mu}\theta}} \\ & {= 1 \cdot \frac{0}{2} = 0.} \end{array}$$

Therefore,

$$\underset{\theta\rightarrow 0}{\text{lim}}\frac{1 - \text{cos}\mspace{2mu}\theta}{\theta} = 0.$$ (2.19)

Evaluate $\underset{\theta\rightarrow 0}{\text{lim}}\frac{1 - \text{cos}\mspace{2mu}\theta}{\text{sin}\mspace{2mu}\theta}.$

Deriving the Formula for the Area of a Circle 推导圆的面积公式

Some of the geometric formulas we take for granted today were first derived by methods that anticipate some of the methods of calculus. The Greek mathematician Archimedes (ca. 287−212; BCE) was particularly inventive, using polygons inscribed within circles to approximate the area of the circle as the number of sides of the polygon increased. He never came up with the idea of a limit, but we can use this idea to see what his geometric constructions could have predicted about the limit.

We can estimate the area of a circle by computing the area of an inscribed regular polygon. Think of the regular polygon as being made up of n triangles. By taking the limit as the vertex angle of these triangles goes to zero, you can obtain the area of the circle. To see this, carry out the following steps:

1. Express the height h and the base b of the isosceles triangle in Figure 2.31 in terms of $\theta$ and r.

2. Using the expressions that you obtained in step 1, express the area of the isosceles triangle in terms of θ and r.

(Substitute $(1\mspace{2mu}\text{/}\mspace{2mu} 2)\mspace{2mu}\text{sin}\mspace{2mu}\theta$ for $\text{sin}(\theta\mspace{2mu}\text{/}\mspace{2mu} 2)\mspace{2mu}\text{cos}(\theta\mspace{2mu}\text{/}\mspace{2mu} 2)$ in your expression.)

3. If an n-sided regular polygon is inscribed in a circle of radius r, find a relationship between θ and n. Solve this for n. Keep in mind there are 2π radians in a circle. (Use radians, not degrees.)

4. Find an expression for the area of the n-sided polygon in terms of r and θ.

5. To find a formula for the area of the circle, find the limit of the expression in step 4 as θ goes to zero. (Hint: $\underset{\theta\rightarrow 0}{\text{lim}}\frac{\left( {\text{sin}\mspace{2mu}\theta} \right)}{\theta} = 1\text{).}$

The technique of estimating areas of regions by using polygons is revisited in Introduction to Integration.

Section 2.3 Exercises 2.3 节习题

In the following exercises, use the limit laws to evaluate each limit. Justify each step by indicating the appropriate limit law(s).

83.

$\underset{x\rightarrow 0}{\text{lim}}\left( {4x^{2} - 2x + 3} \right)$

84.

$\underset{x\rightarrow 1}{\text{lim}}\frac{x^{3} + 3x^{2} + 5}{4 - 7x}$

85.

$\underset{x\rightarrow-2}{\text{lim}}\sqrt{x^{2} - 6x + 3}$

86.

$\underset{x\rightarrow-1}{\text{lim}}\left( {9x + 1} \right)^{2}$

In the following exercises, use direct substitution to evaluate each limit.

87.

$\underset{x\rightarrow 7}{\text{lim}}x^{2}$

88.

$\underset{x\rightarrow-2}{\text{lim}}\left( {4x^{2} - 1} \right)$

89.

$\underset{x\rightarrow 0}{\text{lim}}\frac{1}{1 + \text{sin}\mspace{2mu} x}$

90.

$\underset{x\rightarrow 2}{\text{lim}}e^{2x - x^{2}}$

91.

$\underset{x\rightarrow 1}{\text{lim}}\frac{2 - 7x}{x + 6}$

92.

$\underset{x\rightarrow 3}{\text{lim}}\text{ln}e^{3x}$

In the following exercises, use direct substitution to show that each limit leads to the indeterminate form ${0\text{/}0}.$ Then, evaluate the limit.

93.

$\underset{x\rightarrow 4}{\text{lim}}\frac{x^{2} - 16}{x - 4}$

94.

$\underset{x\rightarrow 2}{\text{lim}}\frac{x - 2}{x^{2} - 2x}$

95.

$\underset{x\rightarrow 6}{\text{lim}}\frac{3x - 18}{2x - 12}$

96.

$\underset{h\rightarrow 0}{\text{lim}}\frac{\left( {1 + h} \right)^{2} - 1}{h}$

97.

$\underset{t\rightarrow 9}{\text{lim}}\frac{t - 9}{\sqrt{t} - 3}$

98.

$\underset{h\rightarrow 0}{\text{lim}}\frac{\frac{1}{a + h} - \frac{1}{a}}{h},$ where a is a non-zero real-valued constant

99.

$\underset{\theta\rightarrow\pi}{\text{lim}}\frac{\text{sin}\mspace{2mu}\theta}{\text{tan}\mspace{2mu}\theta}$

100.

$\underset{x\rightarrow 1}{\text{lim}}\frac{x^{3} - 1}{x^{2} - 1}$

101.

$\underset{x\rightarrow 1\text{/}2}{\text{lim}}\frac{2x^{2} + 3x - 2}{2x - 1}$

102.

$\underset{x\rightarrow-3}{\text{lim}}\frac{\sqrt{x + 4} - 1}{x + 3}$

In the following exercises, use direct substitution to obtain an undefined expression. Then, use the method of Example 2.23 to simplify the function to help determine the limit.

103.

$\underset{x\rightarrow-2^{-}}{\text{lim}}\frac{2x^{2} + 7x - 4}{x^{2} + x - 2}$

104.

$\underset{x\rightarrow-2^{+}}{\text{lim}}\frac{2x^{2} + 7x - 4}{x^{2} + x - 2}$

105.

$\underset{x\rightarrow 1^{-}}{\text{lim}}\frac{2x^{2} + 7x - 4}{x^{2} + x - 2}$

106.

$\underset{x\rightarrow 1^{+}}{\text{lim}}\frac{2x^{2} + 7x - 4}{x^{2} + x - 2}$

In the following exercises, assume that $\underset{x\rightarrow 6}{\text{lim}}f(x) = 4,\underset{x\rightarrow 6}{\text{lim}}g(x) = 9,$ and $\underset{x\rightarrow 6}{\text{lim}}h(x) = 6.$ Use these three facts and the limit laws to evaluate each limit.

107.

$\underset{x\rightarrow 6}{\text{lim}}2f(x)g(x)$

108.

$\underset{x\rightarrow 6}{\text{lim}}\frac{g(x) - 1}{f(x)}$

109.

$\underset{x\rightarrow 6}{\text{lim}}\left( {f(x) + \frac{1}{3}g(x)} \right)$

110.

$\underset{x\rightarrow 6}{\text{lim}}\frac{\left( {h(x)} \right)^{3}}{2}$

111.

$\underset{x\rightarrow 6}{\text{lim}}\sqrt{g(x) - f(x)}$

112.

$\underset{x\rightarrow 6}{\text{lim}}x \cdot h(x)$

113.

$\underset{x\rightarrow 6}{\text{lim}}\left\lbrack {\left( {x + 1} \right) \cdot f(x)} \right\rbrack$

114.

$\underset{x\rightarrow 6}{\text{lim}}\left( {f(x) \cdot g(x) - h(x)} \right)$

\[T\] In the following exercises, use a calculator to draw the graph of each piecewise-defined function and study the graph to evaluate the given limits.

115.

$f(x) = \begin{cases} {x^{2},} & {x \leq 3} \\ {x + 4,} & {x > 3} \end{cases}$

1. $\underset{x\rightarrow 3^{-}}{\text{lim}}f(x)$

2. $\underset{x\rightarrow 3^{+}}{\text{lim}}f(x)$

116.

$g(x) = \begin{cases} {x^{3} - 1,} & {x \leq 0} \\ {1,} & {x > 0} \end{cases}$

1. $\underset{x\rightarrow 0^{-}}{\text{lim}}g(x)$

2. $\underset{x\rightarrow 0^{+}}{\text{lim}}g(x)$

117.

$h(x) = \begin{cases} {x^{2} - 2x + 1,} & {x < 2} \\ {3 - x,} & {x \geq 2} \end{cases}$

1. $\underset{x\rightarrow 2^{-}}{\text{lim}}h(x)$

2. $\underset{x\rightarrow 2^{+}}{\text{lim}}h(x)$

In the following exercises, use the following graphs and the limit laws to evaluate each limit.

118.

$\underset{x\rightarrow-3^{+}}{\text{lim}}\left( {f(x) + g(x)} \right)$

119.

$\underset{x\rightarrow-3^{-}}{\text{lim}}\left( {f(x) - 3g(x)} \right)$

120.

$\underset{x\rightarrow 0}{\text{lim}}\frac{f(x)g(x)}{3}$

121.

$\underset{x\rightarrow-5}{\text{lim}}\frac{2 + g(x)}{f(x)}$

122.

$\underset{x\rightarrow 1}{\text{lim}}\left( {f(x)} \right)^{2}$

123.

$\underset{x\rightarrow 1}{\text{lim}}\sqrt[3]{f(x) - g(x)}$

124.

$\underset{x\rightarrow-7}{\text{lim}}\left( {x \cdot g(x)} \right)$

125.

$\underset{x\rightarrow-9}{\text{lim}}\left\lbrack {x \cdot f(x) + 2 \cdot g(x)} \right\rbrack$

For the following problems, evaluate the limit using the squeeze theorem. Use a calculator to graph the functions $f(x),g(x),$ and $h(x)$ when possible.

126.

\[T\] True or False? If $2x - 1 \leq g(x) \leq x^{2} - 2x + 3,$ then $\underset{x\rightarrow 2}{\text{lim}}g(x) = 0.$

127.

\[T\] $\underset{\theta\rightarrow 0}{\text{lim}}\theta^{2}\text{cos}\mspace{2mu}\left( \frac{1}{\theta} \right)$

128.

$\underset{x\rightarrow 0}{\text{lim}}f(x),$ where $f(x) = \begin{cases} {0,} & {x\ \text{rational}} \\ {x^{2},} & {x\ \text{irrational}} \end{cases}$

129.

\[T\] In physics, the magnitude of an electric field generated by a point charge at a distance r in vacuum is governed by Coulomb’s law: $E(r) = \frac{q}{4\pi\varepsilon_{0}r^{2}},$ where E represents the magnitude of the electric field, q is the charge of the particle, r is the distance between the particle and where the strength of the field is measured, and $\frac{1}{4\pi\varepsilon_{0}}$ is Coulomb’s constant: $8.988\ \times\ 10^{9}\ \text{N} \cdot \text{m}^{2}\text{/}\text{C}^{2}.$

1. Use a graphing calculator to graph $E(r)$ given that the charge of the particle is $q = 10^{-10}.$

2. Evaluate $\underset{r\rightarrow 0^{+}}{\text{lim}}E(r).$ What is the physical meaning of this quantity? Is it physically relevant? Why are you evaluating from the right?

130.

\[T\] The density of an object is given by its mass divided by its volume: $\rho = {m\text{/}V}.$

1. Use a calculator to plot the volume as a function of density $\left( {V = {m\text{/}\rho}} \right),$ assuming you are examining something of mass 8 kg ($m = 8\text{).}$

2. Evaluate $\underset{\rho\rightarrow 0^{+}}{\text{lim}}V(\rho)$ and explain the physical meaning.

2.4 Continuity 2.4 连续性

• 2.4.1 Explain the three conditions for continuity at a point.
• 2.4.2 Describe three kinds of discontinuities.
• 2.4.3 Define continuity on an interval.
• 2.4.4 State the theorem for limits of composite functions.
• 2.4.5 Provide an example of the intermediate value theorem.

• 2.4.1 解释函数在一点处连续的三个条件。
• 2.4.2 描述三种间断类型。
• 2.4.3 定义函数在某区间上的连续性。
• 2.4.4 陈述复合函数极限的定理。
• 2.4.5 给出介值定理的一个例子。

Many functions have the property that their graphs can be traced with a pencil without lifting the pencil from the page. Such functions are called continuous. Other functions have points at which a break in the graph occurs, but satisfy this property over intervals contained in their domains. They are continuous on these intervals and are said to have a discontinuity at a point where a break occurs.

We begin our investigation of continuity by exploring what it means for a function to have continuity at a point. Intuitively, a function is continuous at a particular point if there is no break in its graph at that point.

Continuity at a Point 在一点处的连续性

Before we look at a formal definition of what it means for a function to be continuous at a point, let’s consider various functions that fail to meet our intuitive notion of what it means to be continuous at a point. We then create a list of conditions that prevent such failures.

Our first function of interest is shown in Figure 2.32. We see that the graph of $f(x)$ has a hole at a. In fact, $f(a)$ is undefined. At the very least, for $f(x)$ to be continuous at a, we need the following condition:

$$\text{i.}\ f{(a)\ \text{is defined.}}$$

However, as we see in Figure 2.33, this condition alone is insufficient to guarantee continuity at the point a. Although $f(a)$ is defined, the function has a gap at a. In this example, the gap exists because $\underset{x\rightarrow a}{\text{lim}}f(x)$ does not exist. We must add another condition for continuity at a—namely,

$$\text{ii.}\ \underset{x\rightarrow a}{\text{lim}}f{(x)\ \text{exists.}}$$

However, as we see in Figure 2.34, these two conditions by themselves do not guarantee continuity at a point. The function in this figure satisfies both of our first two conditions, but is still not continuous at a. We must add a third condition to our list:

$$\text{iii.}\ \underset{x\rightarrow a}{\text{lim}}f(x) = f(a).$$

Now we put our list of conditions together and form a definition of continuity at a point.

A function $f(x)$ is continuous at a point a if and only if the following three conditions are satisfied:

1. $f(a)$ is defined

2. $\underset{x\rightarrow a}{\text{lim}}f(x)$ exists

3. $\underset{x\rightarrow a}{\text{lim}}f(x) = f(a)$

A function is discontinuous at a point a if it fails to be continuous at a.

The following procedure can be used to analyze the continuity of a function at a point using this definition.

Determining Continuity at a Point 判定在一点处的连续性

1. Check to see if $f(a)$ is defined. If $f(a)$ is undefined, we need go no further. The function is not continuous at a. If $f(a)$ is defined, continue to step 2.

2. Compute $\underset{x\rightarrow a}{\text{lim}}f(x).$ In some cases, we may need to do this by first computing $\underset{x\rightarrow a^{-}}{\text{lim}}f(x)$ and $\underset{x\rightarrow a^{+}}{\text{lim}}f(x).$ If $\underset{x\rightarrow a}{\text{lim}}f(x)$ does not exist (that is, it is not a real number), then the function is not continuous at a and the problem is solved. If $\underset{x\rightarrow a}{\text{lim}}f(x)$ exists, then continue to step 3.

3. Compare $f(a)$ and $\underset{x\rightarrow a}{\text{lim}}f(x).$ If $\underset{x\rightarrow a}{\text{lim}}f(x) \neq f(a),$ then the function is not continuous at a. If $\underset{x\rightarrow a}{\text{lim}}f(x) = f(a),$ then the function is continuous at a.

The next three examples demonstrate how to apply this definition to determine whether a function is continuous at a given point. These examples illustrate situations in which each of the conditions for continuity in the definition succeed or fail.

Determining Continuity at a Point, Condition 1 判定在一点处的连续性，条件 1

Using the definition, determine whether the function $f(x) = {{(x^{2} - 4)}\text{/}{(x - 2}})$ is continuous at $x = 2.$ Justify the conclusion.

Solution 解答

Let’s begin by trying to calculate $f(2).$ We can see that $f(2) = {0\text{/}0},$ which is undefined. Therefore, $f(x) = \frac{x^{2} - 4}{x - 2}$ is discontinuous at 2 because $f(2)$ is undefined. The graph of $f(x)$ is shown in Figure 2.35.

Determining Continuity at a Point, Condition 2 判定在一点处的连续性，条件 2

Using the definition, determine whether the function $f(x) = \begin{cases}

{- x^{2} + 4} & {\text{if}\ x \leq 3} \\

{4x - 8} & {\text{if}\ x > 3}

\end{cases}$ is continuous at $x = 3.$ Justify the conclusion.

Solution 解答

Let’s begin by trying to calculate $f(3).$

$$f(3) = - {(3^{2})} + 4 = -5.$$

Thus, $f(3)$ is defined. Next, we calculate $\underset{x\rightarrow 3}{\text{lim}}f(x).$ To do this, we must compute $\underset{x\rightarrow 3^{-}}{\text{lim}}f(x)$ and $\underset{x\rightarrow 3^{+}}{\text{lim}}f(x)\text{:}$

$$\underset{x\rightarrow 3^{-}}{\text{lim}}f(x) = - {(3^{2})} + 4 = -5$$

and

$$\underset{x\rightarrow 3^{+}}{\text{lim}}f(x) = 4(3) - 8 = 4.$$

Therefore, $\underset{x\rightarrow 3}{\text{lim}}f(x)$ does not exist. Thus, $f(x)$ is not continuous at 3. The graph of $f(x)$ is shown in Figure 2.36.

Determining Continuity at a Point, Condition 3 判定在一点处的连续性，条件 3

Using the definition, determine whether the function $f(x) = \left\{ \begin{array}{cl}

\frac{\text{sin}\mspace{2mu} x}{x} & {\text{if}\ x \neq 0} \\

1 & {\text{if}\ x = 0}

\end{array} \right.$ is continuous at $x = 0.$

Solution 解答

First, observe that

$$f(0) = 1.$$

Next,

$$\underset{x\rightarrow 0}{\text{lim}}f(x) = \underset{x\rightarrow 0}{\text{lim}}\frac{\text{sin}\mspace{2mu} x}{x} = 1.$$

Last, compare $f(0)$ and $\underset{x\rightarrow 0}{\text{lim}}f(x).$ We see that

$$f(0) = 1 = \underset{x\rightarrow 0}{\text{lim}}f(x).$$

Since all three of the conditions in the definition of continuity are satisfied, $f(x)$ is continuous at $x = 0.$

Using the definition, determine whether the function $f(x) = \left\{ \begin{array}{cl}

{2x + 1} & {\text{if}\ x < 1} \\

2 & {\text{if}\ x = 1} \\

{- x + 4} & {\text{if}\ x > 1}

\end{array} \right.$ is continuous at $x = 1.$ If the function is not continuous at 1, indicate the condition for continuity at a point that fails to hold.

By applying the definition of continuity and previously established theorems concerning the evaluation of limits, we can state the following theorem.

Continuity of Polynomials and Rational Functions 多项式与有理函数的连续性

Polynomials and rational functions are continuous at every point in their domains.

Proof 证明

Previously, we showed that if $p(x)$ and $q(x)$ are polynomials, $\underset{x\rightarrow a}{\text{lim}}p(x) = p(a)$ for every polynomial $p(x)$ and $\underset{x\rightarrow a}{\text{lim}}\frac{p(x)}{q(x)} = \frac{p(a)}{q(a)}$ as long as $q(a) \neq 0.$ Therefore, polynomials and rational functions are continuous on their domains.

□

We now apply Continuity of Polynomials and Rational Functions to determine the points at which a given rational function is continuous.

Continuity of a Rational Function 有理函数的连续性

For what values of x is $f(x) = \frac{x + 1}{x - 5}$ continuous?

Solution 解答

The rational function $f(x) = \frac{x + 1}{x - 5}$ is continuous for every value of x except $x = 5.$

For what values of x is $f(x) = 3x^{4} - 4x^{2}$ continuous?

Types of Discontinuities 间断的类型

As we have seen in Example 2.26 and Example 2.27, discontinuities take on several different appearances. We classify the types of discontinuities we have seen thus far as removable discontinuities, infinite discontinuities, or jump discontinuities. Intuitively, a removable discontinuity is a discontinuity for which there is a hole in the graph, a jump discontinuity is a noninfinite discontinuity for which the sections of the function do not meet up, and an infinite discontinuity is a discontinuity located at a vertical asymptote. Figure 2.37 illustrates the differences in these types of discontinuities. Although these terms provide a handy way of describing three common types of discontinuities, keep in mind that not all discontinuities fit neatly into these categories.

These three discontinuities are formally defined as follows:

If $f(x)$ is discontinuous at a, then

1. $f$ has a removable discontinuity at a if $\underset{x\rightarrow a}{\text{lim}}f(x)$ exists. (Note: When we state that $\underset{x\rightarrow a}{\text{lim}}f(x)$ exists, we mean that $\underset{x\rightarrow a}{\text{lim}}f(x) = L,$ where L is a real number.)

2. $f$ has a jump discontinuity at a if $\underset{x\rightarrow a^{-}}{\text{lim}}f(x)$ and $\underset{x\rightarrow a^{+}}{\text{lim}}f(x)$ both exist, but $\underset{x\rightarrow a^{-}}{\text{lim}}f(x) \neq \underset{x\rightarrow a^{+}}{\text{lim}}f(x).$ (Note: When we state that $\underset{x\rightarrow a^{-}}{\text{lim}}f(x)$ and $\underset{x\rightarrow a^{+}}{\text{lim}}f(x)$ both exist, we mean that both are real-valued and that neither take on the values ±∞.)

3. $f$ has an infinite discontinuity at a if $\underset{x\rightarrow a^{-}}{\text{lim}}f(x) = \text{±}\infty$ and/or $\underset{x\rightarrow a^{+}}{\text{lim}}f(x) = \text{±}\infty.$

Classifying a Discontinuity 间断的分类

In Example 2.26, we showed that $f(x) = \frac{x^{2} - 4}{x - 2}$ is discontinuous at $x = 2.$ Classify this discontinuity as removable, jump, or infinite.

Solution 解答

To classify the discontinuity at 2 we must evaluate $\underset{x\rightarrow 2}{\text{lim}}f(x)\text{:}$

$$\begin{array}{cl}

{\underset{x\rightarrow 2}{\text{lim}}f(x)} & {= \underset{x\rightarrow 2}{\text{lim}}\frac{x^{2} - 4}{x - 2}} \\

& {= \underset{x\rightarrow 2}{\text{lim}}\frac{\left( {x - 2} \right)\left( {x + 2} \right)}{x - 2}} \\

& {= \underset{x\rightarrow 2}{\text{lim}}\left( {x + 2} \right)} \\

& {= 4.}

\end{array}$$

Since f is discontinuous at 2 and $\underset{x\rightarrow 2}{\text{lim}}f(x)$ exists, f has a removable discontinuity at $x = 2.$

Classifying a Discontinuity 间断的分类

In Example 2.27, we showed that $f(x) = \left\{ \begin{array}{cl}

{- x^{2} + 4} & {\text{if}\ x \leq 3} \\

{4x - 8} & {\text{if}\ x > 3}

\end{array} \right.$ is discontinuous at $x = 3.$ Classify this discontinuity as removable, jump, or infinite.

Solution 解答

Earlier, we showed that f is discontinuous at 3 because $\underset{x\rightarrow 3}{\text{lim}}f(x)$ does not exist. However, since $\underset{x\rightarrow 3^{-}}{\text{lim}}f(x) = -5$ and $\underset{x\rightarrow 3^{+}}{\text{lim}}f(x) = 4$ both exist, we conclude that the function has a jump discontinuity at 3.

Classifying a Discontinuity 间断的分类

Determine whether $f(x) = \frac{x + 2}{x + 1}$ is continuous at −1. If the function is discontinuous at −1, classify the discontinuity as removable, jump, or infinite.

Solution 解答

The function value $f(-1)$ is undefined. Therefore, the function is not continuous at −1. To determine the type of discontinuity, we must determine the limit at −1. We see that $\underset{x\rightarrow-1^{-}}{\text{lim}}\frac{x + 2}{x + 1} = \text{−}\infty$ and $\underset{x\rightarrow-1^{+}}{\text{lim}}\frac{x + 2}{x + 1} = \text{+}\infty.$ Therefore, the function has an infinite discontinuity at −1.

For $f(x) = \left\{ \begin{array}{cl}

x^{2} & {\text{if}\ x \neq 1} \\

3 & {\text{if}\ x = 1}

\end{array} \right.,$ decide whether f is continuous at 1. If f is not continuous at 1, classify the discontinuity as removable, jump, or infinite.

Continuity over an Interval 区间上的连续性

Now that we have explored the concept of continuity at a point, we extend that idea to continuity over an interval. As we develop this idea for different types of intervals, it may be useful to keep in mind the intuitive idea that a function is continuous over an interval if we can use a pencil to trace the function between any two points in the interval without lifting the pencil from the paper. In preparation for defining continuity on an interval, we begin by looking at the definition of what it means for a function to be continuous from the right at a point and continuous from the left at a point.

A function $f(x)$ is said to be continuous from the right at a if $\underset{x\rightarrow a^{+}}{\text{lim}}f(x) = f(a).$

A function $f(x)$ is said to be continuous from the left at a if $\underset{x\rightarrow a^{-}}{\text{lim}}f(x) = f(a).$

A function is continuous over an open interval if it is continuous at every point in the interval. A function $f(x)$ is continuous over a closed interval of the form $\left\lbrack {a,b} \right\rbrack$ if it is continuous at every point in $\left( {a,b} \right)$ and is continuous from the right at a and is continuous from the left at b. Analogously, a function $f(x)$ is continuous over an interval of the form $\left( {a,b} \right\rbrack$ if it is continuous over $\left( {a,b} \right)$ and is continuous from the left at b. Continuity over other types of intervals are defined in a similar fashion.

Requiring that $\underset{x\rightarrow a^{+}}{\text{lim}}f(x) = f(a)$ and $\underset{x\rightarrow b^{-}}{\text{lim}}f(x) = f(b)$ ensures that we can trace the graph of the function from the point $\left( {a,f(a)} \right)$ to the point $\left( {b,f(b)} \right)$ without lifting the pencil. If, for example, $\underset{x\rightarrow a^{+}}{\text{lim}}f(x) \neq f(a),$ we would need to lift our pencil to jump from $f(a)$ to the graph of the rest of the function over $\left( {a,b} \right\rbrack.$

Continuity on an Interval 区间上的连续性

State the interval(s) over which the function $f(x) = \frac{x - 1}{x^{2} + 2x}$ is continuous.

Solution 解答

Since $f(x) = \frac{x - 1}{x^{2} + 2x}$ is a rational function, it is continuous at every point in its domain. The domain of $f(x)$ is the set $\left( {\text{−}\infty,-2} \right) \cup \left( {-2,0} \right) \cup \left( {0,\text{+}\infty} \right).$ Thus, $f(x)$ is continuous over each of the intervals $\left( {\text{−}\infty,-2} \right),\left( {-2,0} \right),$ and $\left( {0,\text{+}\infty} \right).$

Continuity over an Interval 区间上的连续性

State the interval(s) over which the function $f(x) = \sqrt{4 - x^{2}}$ is continuous.

Solution 解答

From the limit laws, we know that $\underset{x\rightarrow a}{\text{lim}}\sqrt[{}]{4 - x^{2}} = \sqrt{4 - a^{2}}$ for all values of a in $\left( {-2,2} \right).$ We also know that $\underset{x\rightarrow-2^{+}}{\text{lim}}\sqrt{4 - x^{2}} = 0$ exists and $\underset{x\rightarrow 2^{-}}{\text{lim}}\sqrt{4 - x^{2}} = 0$ exists. Therefore, $f(x)$ is continuous over the interval $\left\lbrack {-2,2} \right\rbrack.$

State the interval(s) over which the function $f(x) = \sqrt{x + 3}$ is continuous.

The Composite Function Theorem allows us to expand our ability to compute limits. In particular, this theorem ultimately allows us to demonstrate that trigonometric functions are continuous over their domains.

Composite Function Theorem 复合函数定理

If $f(x)$ is continuous at L and $\underset{x\rightarrow a}{\text{lim}}\mspace{14mu} g(x) = L,$ then

$$\underset{x\rightarrow a}{\text{lim}}\mspace{14mu} f\left( {g(x)} \right) = f\left( {\underset{x\rightarrow a}{\text{lim}}\mspace{14mu} g(x)} \right) = f(L).$$

Before we move on to Example 2.35, recall that earlier, in the section on limit laws, we showed $\underset{x\rightarrow 0}{\text{lim}}\mspace{14mu}\text{cos}\mspace{2mu} x = 1 = \text{cos}\mspace{2mu}(0).$ Consequently, we know that $f(x) = \text{cos}\mspace{2mu} x$ is continuous at 0. In Example 2.35 we see how to combine this result with the composite function theorem.

Limit of a Composite Cosine Function 复合余弦函数的极限

Evaluate $\underset{x\rightarrow\pi\text{/}2}{\text{lim}}\text{cos}\mspace{2mu}\left( {x - \frac{\pi}{2}} \right).$

Solution 解答

The given function is a composite of $\text{cos}\mspace{2mu} x$ and $x - \frac{\pi}{2}.$ Since $\underset{x\rightarrow{\pi\text{/}2}}{\text{lim}}\left( {x - \frac{\pi}{2}} \right) = 0$ and $\text{cos}\mspace{2mu} x$ is continuous at 0, we may apply the composite function theorem. Thus,

$$\underset{x\rightarrow\pi\text{/}2}{\text{lim}}\text{cos}\mspace{2mu}\left( {x - \frac{\pi}{2}} \right) = \text{cos}\mspace{2mu}\left( {\underset{x\rightarrow\pi\text{/}2}{\text{lim}}\left( {x - \frac{\pi}{2}} \right)} \right) = \text{cos}\mspace{2mu}(0) = 1.$$

Evaluate $\underset{x\rightarrow\pi}{\text{lim}}\text{sin}\mspace{2mu}\left( {x - \pi} \right).$

The proof of the next theorem uses the composite function theorem as well as the continuity of $f(x) = \text{sin}\mspace{2mu} x$ and $g(x) = \text{cos}\mspace{2mu} x$ at the point 0 to show that trigonometric functions are continuous over their entire domains.

Continuity of Trigonometric Functions 三角函数的连续性

Trigonometric functions are continuous over their entire domains.

Proof 证明

We begin by demonstrating that $\text{cos}\mspace{2mu} x$ is continuous at every real number. To do this, we must show that $\underset{x\rightarrow a}{\text{lim}}\text{cos}\mspace{2mu} x = \text{cos}\mspace{2mu} a$ for all values of a.

$\begin{array}{clccl}

{\underset{x\rightarrow a}{\text{lim}}\text{cos}\mspace{2mu} x} & {= \underset{x\rightarrow a}{\text{lim}}\text{cos}\mspace{2mu}\left( {\left( {x - a} \right) + a} \right)} & & & {\text{rewrite}\ x = x - a + a} \\

& {= \underset{x\rightarrow a}{\text{lim}}\left( {\text{cos}\mspace{2mu}\left( {x - a} \right)\mspace{2mu}\text{cos}\mspace{2mu} a - \text{sin}\mspace{2mu}\left( {x - a} \right)\mspace{2mu}\text{sin}\mspace{2mu} a} \right)} & & & \text{apply the identity for the cosine of the sum of two angles} \\

& {= \text{cos}\mspace{2mu}\left( {\underset{x\rightarrow a}{\text{lim}}\left( {x - a} \right)} \right)\mspace{2mu}\text{cos}\mspace{2mu} a - \text{sin}\mspace{2mu}\left( {\underset{x\rightarrow a}{\text{lim}}\left( {x - a} \right)} \right)\mspace{2mu}\text{sin}\mspace{2mu} a} & & & {\underset{x\rightarrow a}{\text{lim}}\left( {x - a} \right) = 0,\ \text{and}\ \text{sin}\mspace{2mu} x\ \text{and}\ \text{cos}\mspace{2mu} x\ \text{are continuous at 0}} \\

& {= \text{cos}\mspace{2mu}(0)\mspace{2mu}\text{cos}\mspace{2mu} a - \text{sin}\mspace{2mu}(0)\mspace{2mu}\text{sin}\mspace{2mu} a} & & & \text{evaluate cos(0) and sin(0) and simplify} \\

& {= 1 \cdot \text{cos}\mspace{2mu} a - 0 \cdot \text{sin}\mspace{2mu} a = \text{cos}\mspace{2mu} a.} & & &

\end{array}$

The proof that $\text{sin}\mspace{2mu} x$ is continuous at every real number is analogous. Because the remaining trigonometric functions may be expressed in terms of $\text{sin}\mspace{2mu} x$ and $\text{cos}\mspace{2mu} x,$ their continuity follows from the quotient limit law.

□

As you can see, the composite function theorem is invaluable in demonstrating the continuity of trigonometric functions. As we continue our study of calculus, we revisit this theorem many times.

The Intermediate Value Theorem 介值定理

Functions that are continuous over intervals of the form $\left\lbrack {a,b} \right\rbrack,$ where a and b are real numbers, exhibit many useful properties. Throughout our study of calculus, we will encounter many powerful theorems concerning such functions. The first of these theorems is the Intermediate Value Theorem.

The Intermediate Value Theorem 介值定理

Let f be continuous over a closed, bounded interval $\left\lbrack {a,b} \right\rbrack.$ If z is any real number between $f(a)$ and $f(b),$ then there is a number c in $\left\lbrack {a,b} \right\rbrack$ satisfying $f(c) = z$ in Figure 2.38.

Application of the Intermediate Value Theorem 介值定理的应用

Show that $f(x) = x - \text{cos}\mspace{2mu} x$ has at least one zero.

Solution 解答

Since $f(x) = x - \text{cos}\mspace{2mu} x$ is continuous over $\left( {\text{−}\infty,\text{+}\infty} \right),$ it is continuous over any closed interval of the form $\left\lbrack {a,b} \right\rbrack.$ If you can find an interval $\left\lbrack {a,b} \right\rbrack$ such that $f(a)$ and $f(b)$ have opposite signs, you can use the Intermediate Value Theorem to conclude there must be a real number c in $\left( {a,b} \right)$ that satisfies $f(c) = 0.$ Note that

$$f(0) = 0 - \text{cos}\mspace{2mu}(0) = -1 < 0$$

and

$$f\left( \frac{\pi}{2} \right) = \frac{\pi}{2} - \text{cos}\frac{\pi}{2} = \frac{\pi}{2} > 0.$$

Using the Intermediate Value Theorem, we can see that there must be a real number c in $\left\lbrack {0,{\pi\text{/}2}} \right\rbrack$ that satisfies $f(c) = 0.$ Therefore, $f(x) = x - \text{cos}\mspace{2mu} x$ has at least one zero.

When Can You Apply the Intermediate Value Theorem? 何时可以应用介值定理？

If $f(x)$ is continuous over $\left\lbrack {0,2} \right\rbrack,f(0) > 0$ and $f(2) > 0,$ can we use the Intermediate Value Theorem to conclude that $f(x)$ has no zeros in the interval $\left\lbrack {0,2}\text{]?} \right.$ Explain.

Solution 解答

No. The Intermediate Value Theorem only allows us to conclude that we can find a value between $f(0)$ and $f(2);$ it doesn’t allow us to conclude that we can’t find other values. To see this more clearly, consider the function $f(x) = \left( {x - 1} \right)^{2}.$ It satisfies $f(0) = 1 > 0,f(2) = 1 > 0,$ and $f(1) = 0.$

When Can You Apply the Intermediate Value Theorem? 何时可以应用介值定理？

For $f(x) = {1\text{/}x},f(-1) = -1 < 0$ and $f(1) = 1 > 0.$ Can we conclude that $f(x)$ has a zero in the interval $\left\lbrack {-1,1} \right\rbrack?$

Solution 解答

No. The function is not continuous over $\left\lbrack {-1,1} \right\rbrack.$ The Intermediate Value Theorem does not apply here.

Show that $f(x) = x^{3} - x^{2} - 3x + 1$ has a zero over the interval $\left\lbrack {0,1} \right\rbrack.$

Section 2.4 Exercises 2.4 节习题

For the following exercises, determine the point(s), if any, at which each function is discontinuous. Classify any discontinuity as jump, removable, infinite, or other.

131.

$f(x) = \frac{1}{\sqrt{x}}$

132.

$f(x) = \frac{2}{x^{2} + 1}$

133.

$f(x) = \frac{x}{x^{2} - x}$

134.

$g(t) = t^{-1} + 1$

135.

$f(x) = \frac{5}{e^{x} - 2}$

136.

$f(x) = \frac{\left| {x - 2} \right|}{x - 2}$

137.

$H(x) = \text{tan}\mspace{2mu} 2x$

For the following exercises, decide if the function continuous at the given point. If it is discontinuous, what type of discontinuity is it?

139.

$f(x) = \frac{2x^{2} - 5x + 3}{x - 1}$ at $x = 1$

140.

$h(\theta) = \frac{\text{sin}\mspace{2mu}\theta - \text{cos}\mspace{2mu}\theta}{\text{tan}\mspace{2mu}\theta}$ at $\theta = \pi$

141.

$g(u) = \begin{cases}

\frac{6u^{2} + u - 2}{2u - 1} & {\text{if}\ u \neq \frac{1}{2}} \\

\frac{7}{2} & {\text{if}\ u = \frac{1}{2}}

\end{cases},$ at $u = \frac{1}{2}$

142.

$f(y) = \frac{\text{sin}\mspace{2mu}\left( {\pi y} \right)}{\text{tan}\mspace{2mu}\left( {\pi y} \right)},$ at $y = 1$

143.

$f(x) = \begin{cases}

{x^{2} - e^{x}} & {\text{if}\ x < 0} \\

{x - 1} & {\text{if}\ x \geq 0}

\end{cases},$ at $x = 0$

In the following exercises, find the value(s) of k that makes each function continuous over the given interval.

145.

$f(x) = \begin{cases}

{3x + 2,} & {x < k} \\

{2x - 3,} & {k \leq x \leq 8}

\end{cases}$

146.

$f(\theta) = \left\{ \begin{array}{cl}

{\text{sin}\mspace{2mu}\theta,} & {0 \leq \theta < \frac{\pi}{2}} \\

{\text{cos}\mspace{2mu}\left( {\theta + k} \right),} & {\frac{\pi}{2} \leq \theta \leq \pi}

\end{array} \right.$

147.

$f(x) = \left\{ \begin{array}{cl}

{\frac{x^{2} + 3x + 2}{x + 2},} & {x \neq - 2} \\

{k,} & {x = -2}

\end{array} \right.$

148.

$f(x) = \left\{ \begin{array}{cl}

{e^{kx},} & {0 \leq x < 4} \\

{x + 3,} & {4 \leq x \leq 8}

\end{array} \right.$

149.

$f(x) = \left\{ \begin{array}{cl}

{\sqrt{kx},} & {0 \leq x \leq 3} \\

{x + 1,} & {3 < x \leq 10}

\end{array} \right.$

In the following exercises, use the Intermediate Value Theorem (IVT).

150.

Let $h(x) = \begin{cases}

{3x^{2} - 4,} & {x \leq 2} \\

{5 + 4x,} & {x > 2}

\end{cases}$ Over the interval $\left\lbrack {0,4} \right\rbrack,$ there is no value of x such that $h(x) = 10,$ although $h(0) < 10$ and $h(4) > 10.$ Explain why this does not contradict the IVT.

151.

A particle moving along a line has at each time t a position function $s(t),$ which is continuous. Assume $s(2) = 5$ and $s(5) = 2.$ Another particle moves such that its position is given by $h(t) = s(t) - t.$ Explain why there must be a value c for $2 < c < 5$ such that $h(c) = 0.$

152.

\[T\] Use the statement “The cosine of t is equal to t cubed.”

1. Write a mathematical equation of the statement.

2. Prove that the equation in part a. has at least one real solution.

3. Use a calculator to find an interval of length 0.01 that contains a solution.

153.

Apply the IVT to determine whether $2^{x} = x^{3}$ has a solution in one of the intervals $\left\lbrack {1.25,1.375} \right\rbrack$ or $\left\lbrack {1.375,1.5} \right\rbrack.$ Briefly explain your response for each interval.

154.

Consider the graph of the function $y = f(x)$ shown in the following graph.

1. Find all values for which the function is discontinuous.

2. For each value in part a., state why the formal definition of continuity does not apply.

3. Classify each discontinuity as either jump, removable, or infinite.

155.

Let $f(x) = \left\{ \begin{matrix}

{3x,x > 1} \\

{x^{3},x < 1}

\end{matrix} \right..$

1. Sketch the graph of f.

2. Is it possible to find a value k such that $f(1) = k,$ which makes $f(x)$ continuous for all real numbers? Briefly explain.

156.

Let $f(x) = \frac{x^{4} - 1}{x^{2} - 1}$ for $x \neq - 1,1.$

1. Sketch the graph of f.

2. Is it possible to find values $k_{1}$ and $k_{2}$ such that $f(-1) = k_{1}$ and $f(1) = k_{2},$ and that makes $f(x)$ continuous for all real numbers? Briefly explain.

157.

Sketch the graph of a function $y = f(x)$ with properties i. through vi.

1. The domain of f is $\left( {\text{−}\infty,\text{+}\infty} \right).$

2. f has an infinite discontinuity at $x = -6.$

3. $f(-6) = 3$

4. $\underset{x\rightarrow-3^{-}}{\text{lim}}f(x) = \underset{x\rightarrow-3^{+}}{\text{lim}}f(x) = 2$

5. $f(-3) = 3$

6. f is left continuous but not right continuous at $x = 3.$

158.

Sketch the graph of a function $y = f(x)$ with properties i. through iv.

1. The domain of f is $\left\lbrack {0,5} \right\rbrack.$

2. $\underset{x\rightarrow 1^{+}}{\text{lim}}f(x)$ and $\underset{x\rightarrow 1^{-}}{\text{lim}}f(x)$ exist and are equal.

3. $f(x)$ is left continuous but not continuous at $x = 2,$ and right continuous but not continuous at $x = 3.$

4. $f(x)$ has a removable discontinuity at $x = 1,$ a jump discontinuity at $x = 2,$ and the following limits hold: $\underset{x\rightarrow 3^{-}}{\text{lim}}f(x) = \text{−}\infty$ and $\underset{x\rightarrow 3^{+}}{\text{lim}}f(x) = 2.$

In the following exercises, suppose $y = f(x)$ is defined for all x. For each description, sketch a graph with the indicated property.

159.

Discontinuous at $x = 1$ with $\underset{x\rightarrow-1}{\text{lim}}f(x) = -1$ and $\underset{x\rightarrow 2}{\text{lim}}f(x) = 4$

160.

Discontinuous at $x = 2$ but continuous elsewhere with $\underset{x\rightarrow 0}{\text{lim}}f(x) = \frac{1}{2}$

Determine whether each of the given statements is true. Justify your response with an explanation or counterexample.

161.

$f(t) = \frac{2}{e^{t} - e^{- t}}$ is continuous everywhere.

162.

If the left- and right-hand limits of $f(x)$ as $x\rightarrow a$ exist and are equal, then f cannot be discontinuous at $x = a.$

163.

If a function is not continuous at a point, then it is not defined at that point.

164.

According to the IVT, $\text{cos}\mspace{2mu} x - \text{sin}\mspace{2mu} x - x = 2$ has a solution over the interval $\left\lbrack {-1,1} \right\rbrack.$

165.

If $f(x)$ is continuous such that $f(a)$ and $f(b)$ have opposite signs, then $f(x) = 0$ has exactly one solution in $\left\lbrack {a,b} \right\rbrack.$

166.

The function $f(x) = \frac{x^{2} - 4x + 3}{x^{2} - 1}$ is continuous over the interval $\left\lbrack {0,3} \right\rbrack.$

167.

If $f(x)$ is continuous everywhere and $f(a),f(b) > 0,$ then there is no root of $f(x)$ in the interval $\left\lbrack {a,b} \right\rbrack.$

\[T\] The following problems consider the scalar form of Coulomb’s law, which describes the electrostatic force between two point charges, such as electrons. It is given by the equation $F(r) = k_{e}\frac{\left| {q_{1}q_{2}} \right|}{r^{2}},$ where $k_{e}$ is Coulomb’s constant, $q_{i}$ are the magnitudes of the charges of the two particles, and r is the distance between the two particles.

168.

To simplify the calculation of a model with many interacting particles, after some threshold value $r = R,$ we approximate F as zero.

1. Explain the physical reasoning behind this assumption.

2. What is the force equation?

3. Evaluate the force F using both Coulomb’s law and our approximation, assuming two protons with a charge magnitude of $1.6022\ \times \ 10^{-19}\ \text{coulombs (C)},$ and the Coulomb constant $k_{e} = 8.988\ \times \ 10^{9}{\text{Nm}^{2}\text{/}\text{C}^{2}}$ are 1 m apart. Also, assume $R < 1\ \text{m}.$ How much inaccuracy does our approximation generate? Is our approximation reasonable?

4. Is there any finite value of R for which this system remains continuous at R?

169.

Instead of making the force 0 at R, instead we let the force be 10⁻²⁰ for $r \geq R.$ Assume two protons, which have a magnitude of charge $1.6022\ \times \ 10^{-19}\ \text{C},$ and the Coulomb constant $k_{e} = 8.988\ \times \ 10^{9}{\text{Nm}^{2}\text{/}\text{C}^{2}}.$ Is there a value R that can make this system continuous? If so, find it.

Recall the discussion on spacecraft from the chapter opener. The following problems consider a rocket launch from Earth’s surface. The force of gravity on the rocket is given by $F(d) = - {{mk}\text{/}d^{2}},$ where m is the mass of the rocket, d is the distance of the rocket from the center of Earth, and k is a constant.

170.

\[T\] Determine the value and units of k given that the mass of the rocket is 3 million kg. (*Hint*: The distance from the center of Earth to its surface is 6378 km.)

171.

\[T\] After a certain distance D has passed, the gravitational effect of Earth becomes quite negligible, so we can approximate the force function by $F(d) = \begin{cases}

{- \frac{mk}{d^{2}}} & {\text{if}\ d < D} \\

10,000 & {\text{if}\ d \geq D}

\end{cases}.$ Using the value of k found in the previous exercise, find the necessary condition D such that the force function remains continuous.

172.

As the rocket travels away from Earth’s surface, there is a distance D where the rocket sheds some of its mass, since it no longer needs the excess fuel storage. We can write this function as $F(d) = \left\{ \begin{array}{l}

{- \frac{m_{1}k}{d^{2}}\ \text{if}\ d < D} \\

{- \frac{m_{2}k}{d^{2}}\ \text{if}\ d \geq D}

\end{array} \right..$ Is there a D value such that this function is continuous, assuming $m_{1} \neq m_{2}?$

Prove the following functions are continuous everywhere

173.

$f(\theta) = \text{sin}\mspace{2mu}\theta$

174.

$g(x) = |x|$

175.

Where is $f(x) = \left\{ \begin{array}{l}

{0\ \text{if}\ x\ \text{is irrational}} \\

{1\ \text{if}\ x\ \text{is rational}}

\end{array} \right.$ continuous?

2.5 The Precise Definition of a Limit 2.5 极限的精确定义

• 2.5.1 Describe the epsilon-delta definition of a limit.
• 2.5.2 Apply the epsilon-delta definition to find the limit of a function.
• 2.5.3 Describe the epsilon-delta definitions of one-sided limits and infinite limits.
• 2.5.4 Use the epsilon-delta definition to prove the limit laws.

• 2.5.1 描述极限的 ε-δ 定义。
• 2.5.2 应用 ε-δ 定义求函数极限。
• 2.5.3 描述单侧极限与无穷极限的 ε-δ 定义。
• 2.5.4 使用 ε-δ 定义证明极限法则。

By now you have progressed from the very informal definition of a limit in the introduction of this chapter to the intuitive understanding of a limit. At this point, you should have a very strong intuitive sense of what the limit of a function means and how you can find it. In this section, we convert this intuitive idea of a limit into a formal definition using precise mathematical language. The formal definition of a limit is quite possibly one of the most challenging definitions you will encounter early in your study of calculus; however, it is well worth any effort you make to reconcile it with your intuitive notion of a limit. Understanding this definition is the key that opens the door to a better understanding of calculus.

Quantifying Closeness 量化接近程度

Before stating the formal definition of a limit, we must introduce a few preliminary ideas. Recall that the distance between two points a and b on a number line is given by $\left| {a - b} \right|.$

• The statement $\left| {f(x) - L} \right| < \varepsilon$ may be interpreted as: The distance between $f(x)$ and L is less than ε.
• The statement $0 < \left| {x - a} \right| < \delta$ may be interpreted as: $x \neq a$ and the distance between x and a is less than δ.

• 语句 $\left| {f(x) - L} \right| < \varepsilon$ 可解释为：$f(x)$ 与 $L$ 之间的距离小于 ε。
• 语句 $0 < \left| {x - a} \right| < \delta$ 可解释为：$x \neq a$ 且 $x$ 与 $a$ 之间的距离小于 δ。

It is also important to look at the following equivalences for absolute value:

• The statement $\left| {f(x) - L} \right| < \varepsilon$ is equivalent to the statement $L - \varepsilon < f(x) < L + \varepsilon.$
• The statement $0 < \left| {x - a} \right| < \delta$ is equivalent to the statement $a - \delta < x < a + \delta$ and $x \neq a.$

• 语句 $\left| {f(x) - L} \right| < \varepsilon$ 等价于语句 $L - \varepsilon < f(x) < L + \varepsilon.$
• 语句 $0 < \left| {x - a} \right| < \delta$ 等价于语句 $a - \delta < x < a + \delta$ 且 $x \neq a.$

With these clarifications, we can state the formal epsilon-delta definition of the limit.

Let $f(x)$ be defined for all $x \neq a$ over an open interval containing a. Let L be a real number. Then

$$\underset{x\rightarrow a}{\text{lim}}f(x) = L$$

if, for every $\varepsilon > 0,$ there exists a $\delta > 0,$ such that if $\left. 0 < \middle| x - a \middle| < \delta, \right.$ then $\left| f(x) - L \middle| < \varepsilon. \right.$

This definition may seem rather complex from a mathematical point of view, but it becomes easier to understand if we break it down phrase by phrase. The statement itself involves something called a universal quantifier (for every $\varepsilon > 0\text{),}$ an existential quantifier (there exists a $\delta > 0\text{),}$ and, last, a conditional statement (if $0 < \left| {x - a} \right| < \delta,$ then $\left| {f(x) - L} \right| < \varepsilon\text{).}$ Let’s take a look at Table 2.9, which breaks down the definition and translates each part.

Table 2.9 Translation of the Epsilon-Delta Definition of the Limit

We can get a better handle on this definition by looking at the definition geometrically. Figure 2.39 shows possible values of $\delta$ for various choices of $\varepsilon > 0$ for a given function $f(x),$ a number a, and a limit L at a. Notice that as we choose smaller values of ε (the distance between the function and the limit), we can always find a $\delta$ small enough so that if we have chosen an x value within $\delta$ of a, then the value of $f(x)$ is within ε of the limit L.

Visit the following applet to experiment with finding values of $\delta$ for selected values of ε:

• The epsilon-delta definition of limit

• 极限的 ε–δ 定义

Example 2.39 shows how you can use this definition to prove a statement about the limit of a specific function at a specified value.

Proving a Statement about the Limit of a Specific Function 证明关于特定函数极限的命题

Prove that $\underset{x\rightarrow 1}{\text{lim}}\left( {2x + 1} \right) = 3.$

Solution 解答

Let $\varepsilon > 0.$

The first part of the definition begins “For every $\varepsilon > 0\text{.”}$ This means we must prove that whatever follows is true no matter what positive value of ε is chosen. By stating “Let $\varepsilon > 0\text{,”}$ we signal our intent to do so.

Choose $\delta = \frac{\varepsilon}{2}.$

The definition continues with “there exists a $\delta > 0\text{.”}$ The phrase “there exists” in a mathematical statement is always a signal for a scavenger hunt. In other words, we must go and find $\delta.$ So, where exactly did $\delta = {\varepsilon\text{/}2}$ come from? There are two basic approaches to tracking down $\delta.$ One method is purely algebraic and the other is geometric.

We begin by tackling the problem from an algebraic point of view. Since ultimately we want $\left| {\left( {2x + 1} \right) - 3} \right| < \varepsilon,$ we begin by manipulating this expression: $\left| {\left( {2x + 1} \right) - 3} \right| < \varepsilon$ is equivalent to $\left| {2x - 2} \right| < \varepsilon,$ which in turn is equivalent to $|2|\left| {x - 1} \right| < \varepsilon.$ Last, this is equivalent to $\left| {x - 1} \right| < {\varepsilon\text{/}2}.$ Thus, it would seem that $\delta = {\varepsilon\text{/}2}$ is appropriate.

We may also find $\delta$ through geometric methods. Figure 2.40 demonstrates how this is done.

Assume $0 < \left| {x - 1} \right| < \delta.$ When $\delta$ has been chosen, our goal is to show that if $0 < \left| {x - 1} \right| < \delta,$ then $\left| {\left( {2x + 1} \right) - 3} \right| < \varepsilon.$ To prove any statement of the form “If this, then that,” we begin by assuming “this” and trying to get “that.”

Thus,

$\begin{array}{clccc} \left| {\left( {2x + 1} \right) - 3} \right| & {= \left| {2x - 2} \right|} & & & \text{property of absolute value} \\ & {= \left| {2\left( {x - 1} \right)} \right|} & & & \\ & {= |2|\left| {x - 1} \right|} & & & \left| 2 \middle| = 2 \right. \\ & {= 2\left| {x - 1} \right|} & & & \\ & {< 2 \cdot \delta} & & & {\text{here’s where we use the assumption that}\ 0 < \left| {x - 1} \right| < \delta} \\ & {= 2 \cdot \frac{\varepsilon}{2} = \varepsilon} & & & {\text{here’s where we use our choice of}\ \delta = {\varepsilon\text{/}2}} \end{array}$

Analysis 分析

In this part of the proof, we started with $\left| {\left( {2x + 1} \right) - 3} \right|$ and used our assumption $0 < \left| {x - 1} \right| < \delta$ in a key part of the chain of inequalities to get $\left| {\left( {2x + 1} \right) - 3} \right|$ to be less than ε. We could just as easily have manipulated the assumed inequality $0 < \left| {x - 1} \right| < \delta$ to arrive at $\left| {\left( {2x + 1} \right) - 3} \right| < \varepsilon$ as follows:

$\begin{array}{ll} {0 < |x - 1| < \delta} & {\Rightarrow|x - 1| < \delta} \\ & {\Rightarrow - \delta < x - 1 < \delta} \\ & {\Rightarrow - \frac{\varepsilon}{2} < x - 1 < \frac{\varepsilon}{2}} \\ & {\Rightarrow - \varepsilon < 2x - 2 < \varepsilon} \\ & {\Rightarrow|2x - 2| < \varepsilon} \\ & {\Rightarrow\left| (2x + 1) - 3 \right| < \varepsilon.} \end{array}$

Therefore, $\underset{x\rightarrow 1}{\text{lim}}\left( {2x + 1} \right) = 3.$ (Having completed the proof, we state what we have accomplished.)

After removing all the remarks, here is a final version of the proof:

Let $\varepsilon > 0.$

Choose $\delta = {\varepsilon\text{/}2}.$

Assume $0 < \left| {x - 1} \right| < \delta.$

Thus,

$\begin{array}{cl} \left| {\left( {2x + 1} \right) - 3} \right| & {= \left| {2x - 2} \right|} \\ & {= \left| {2\left( {x - 1} \right)} \right|} \\ & {= |2|\left| {x - 1} \right|} \\ & {= 2\left| {x - 1} \right|} \\ & {< 2 \cdot \delta} \\ & {= 2 \cdot \frac{\varepsilon}{2}} \\ & {= \varepsilon.} \end{array}$

Therefore, $\underset{x\rightarrow 1}{\text{lim}}\left( {2x + 1} \right) = 3.$

The following Problem-Solving Strategy summarizes the type of proof we worked out in Example 2.39.

Proving That $\underset{x\rightarrow a}{\text{lim}}f(x) = L$ for a Specific Function $f(x)$ 证明对特定函数 $f(x)$ 有 $\underset{x\rightarrow a}{\text{lim}}f(x) = L$

1. Let’s begin the proof with the following statement: Let $\varepsilon > 0.$

2. Next, we need to obtain a value for $\delta.$ After we have obtained this value, we make the following statement, filling in the blank with our choice of $\delta$: Choose $\delta = \text{\_\_\_\_\_\_\_.}$

3. The next statement in the proof should be (at this point, we fill in our given value for a):

Assume $0 < \left| {x - a} \right| < \delta.$

4. Next, based on this assumption, we need to show that $\left| {f(x) - L} \right| < \varepsilon,$ where $f(x)$ and L are our function $f(x)$ and our limit L. At some point, we need to use $0 < \left| {x - a} \right| < \delta.$

5. We conclude our proof with the statement: Therefore, $\underset{x\rightarrow a}{\text{lim}}f(x) = L.$

Proving a Statement about a Limit 证明关于极限的命题

Complete the proof that $\underset{x\rightarrow-1}{\text{lim}}\left( {4x + 1} \right) = -3$ by filling in the blanks.

Let ________.

Choose $\delta = \text{\_\_\_\_\_\_\_.}$

Assume $0 < \left| {x - \operatorname{\_\_\_\_\_\_\_}}\text{|} \right. < \delta.$

Thus, $\left| \text{\_\_\_\_\_\_\_\_} - \text{\_\_\_\_\_\_\_\_} \middle| = \operatorname{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}\varepsilon. \right.$

Solution 解答

We begin by filling in the blanks where the choices are specified by the definition. Thus, we have

Let $\varepsilon > 0.$

Choose $\delta = \text{\_\_\_\_\_\_\_.}$

Assume $0 < \left| {x - (-1)} \right| < \delta.$ (or equivalently, $0 < \left| {x + 1} \right| < \delta\text{.)}$

Thus, $\left| {\left( {4x + 1} \right) - (-3)} \right| = \left| {4x + 4} \right| = |4|\left| {x + 1} \right| < 4\delta\text{\_\_\_\_\_\_\_}\varepsilon.$

Focusing on the final line of the proof, we see that we should choose $\delta = \frac{\varepsilon}{4}.$

We now complete the final write-up of the proof:

Let $\varepsilon > 0.$

Choose $\delta = \frac{\varepsilon}{4}.$

Assume $0 < \left| {x - (-1)} \right| < \delta$ (or equivalently, $0 < \left| {x + 1} \right| < \delta\text{.)}$

Thus, $\left| {\left( {4x + 1} \right) - (-3)} \right| = \left| {4x + 4} \right| = |4|\left| {x + 1} \right| < 4\delta = 4\left( {\varepsilon\text{/}4} \right) = \varepsilon.$

Complete the proof that $\underset{x\rightarrow 2}{\text{lim}}\left( {3x - 2} \right) = 4$ by filling in the blanks.

Let ________.

Choose $\delta = \text{\_\_\_\_\_\_\_}.$

Assume $0 < \left| {x - \text{\_\_\_\_}} \right| < \text{\_\_\_\_}.$

Thus,

$\left| {\text{\_\_\_\_\_\_\_} - \text{\_\_\_\_}} \right| = \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}\varepsilon.$

Therefore, $\underset{x\rightarrow 2}{\text{lim}}\left( {3x - 2} \right) = 4.$

In Example 2.39 and Example 2.40, the proofs were fairly straightforward, since the functions with which we were working were linear. In Example 2.41, we see how to modify the proof to accommodate a nonlinear function.

Proving a Statement about the Limit of a Specific Function (Geometric Approach) 证明关于特定函数极限的命题（几何方法）

Prove that $\underset{x\rightarrow 2}{\text{lim}}x^{2} = 4.$

Solution 解答

1. Let $\varepsilon > 0.$ The first part of the definition begins “For every $\varepsilon > 0\text{,”}$ so we must prove that whatever follows is true no matter what positive value of ε is chosen. By stating “Let $\varepsilon > 0\text{,”}$ we signal our intent to do so.

2. Without loss of generality, assume $\varepsilon \leq 4.$ Two questions present themselves: Why do we want $\varepsilon \leq 4$ and why is it okay to make this assumption? In answer to the first question: Later on, in the process of solving for $\delta,$ we will discover that $\delta$ involves the quantity $\sqrt{4 - \varepsilon}.$ Consequently, we need $\varepsilon \leq 4.$ In answer to the second question: If we can find $\delta > 0$ that “works” for $\varepsilon \leq 4,$ then it will “work” for any $\varepsilon > 4$ as well. Keep in mind that, although it is always okay to put an upper bound on ε, it is never okay to put a lower bound (other than zero) on ε.

3. Choose $\delta = \text{min}\left\{ {2 - \sqrt{4 - \varepsilon},\sqrt{4 + \varepsilon} - 2} \right\}.$ Figure 2.41 shows how we made this choice of $\delta.$

4. We must show: If $0 < \left| {x - 2} \right| < \delta,$ then ${|{x^{2} - 4}|} < \varepsilon,$ so we must begin by assuming

$$0 < \left| {x - 2} \right| < \delta.$$

We don’t really need $0 < \left| {x - 2} \right|$ (in other words, $\left. x \neq 2 \right)$ for this proof. Since $0 < \left| {x - 2} \right| < \delta\Rightarrow\left| {x - 2} \right| < \delta,$ it is okay to drop $0 < \left| {x - 2} \right|.$

$$\left| {x - 2} \right| < \delta.$$

Hence,

$$- \delta < x - 2 < \delta.$$

Recall that $\delta = \text{min}\left\{ {2 - \sqrt{4 - \varepsilon},\sqrt{4 + \varepsilon} - 2} \right\}.$ Thus, $\delta \leq 2 - \sqrt{4 - \varepsilon}$ and consequently $- \left( {2 - \sqrt{4 - \varepsilon}} \right) \leq - \delta.$ We also use $\delta \leq \sqrt{4 + \varepsilon} - 2$ here. We might ask at this point: Why did we substitute $2 - \sqrt{4 - \varepsilon}$ for $\delta$ on the left-hand side of the inequality and $\sqrt{4 + \varepsilon} - 2$ on the right-hand side of the inequality? If we look at Figure 2.41, we see that $2 - \sqrt{4 - \varepsilon}$ corresponds to the distance on the left of 2 on the x-axis and $\sqrt{4 + \varepsilon} - 2$ corresponds to the distance on the right. Thus,

$$- \left( {2 - \sqrt{4 - \varepsilon}} \right) \leq - \delta < x - 2 < \delta \leq \sqrt{4 + \varepsilon} - 2.$$

We simplify the expression on the left:

$$-2 + \sqrt{4 - \varepsilon} < x - 2 < \sqrt{4 + \varepsilon} - 2.$$

Then, we add 2 to all parts of the inequality:

$$\sqrt{4 - \varepsilon} < x < \sqrt{4 + \varepsilon}.$$

We square all parts of the inequality. It is okay to do so, since all parts of the inequality are positive:

$$4 - \varepsilon < x^{2} < 4 + \varepsilon.$$

We subtract 4 from all parts of the inequality:

$$- \varepsilon < x^{2} - 4 < \varepsilon.$$

Last,

$${|{x^{2} - 4}|} < \varepsilon.$$

5. Therefore,

$$\underset{x\rightarrow 2}{\text{lim}}x^{2} = 4.$$

Find $\delta$ corresponding to $\varepsilon > 0$ for a proof that $\underset{x\rightarrow 9}{\text{lim}}\sqrt{x} = 3.$

The geometric approach to proving that the limit of a function takes on a specific value works quite well for some functions. Also, the insight into the formal definition of the limit that this method provides is invaluable. However, we may also approach limit proofs from a purely algebraic point of view. In many cases, an algebraic approach may not only provide us with additional insight into the definition, it may prove to be simpler as well. Furthermore, an algebraic approach is the primary tool used in proofs of statements about limits. For Example 2.42, we take on a purely algebraic approach.

Proving a Statement about the Limit of a Specific Function (Algebraic Approach) 证明关于特定函数极限的命题（代数方法）

Prove that $\underset{x\rightarrow-1}{\text{lim}}\left( {x^{2} - 2x + 3} \right) = 6.$

Solution 解答

Let’s use our outline from the Problem-Solving Strategy:

1. Let $\varepsilon > 0.$

2. Choose $\delta = \text{min}\left\{ {1,{\varepsilon\text{/}5}} \right\}.$ This choice of $\delta$ may appear odd at first glance, but it was obtained by taking a look at our ultimate desired inequality: $\left| {\left( {x^{2} - 2x + 3} \right) - 6} \right| < \varepsilon.$ This inequality is equivalent to $\left| {x + 1} \right| \cdot \left| {x - 3} \right| < \varepsilon.$ At this point, the temptation simply to choose $\delta = \frac{\varepsilon}{x - 3}$ is very strong. Unfortunately, our choice of $\delta$ must depend on ε only and no other variable. If we can replace $\left| {x - 3} \right|$ by a numerical value, our problem can be resolved. This is the place where assuming $\delta \leq 1$ comes into play. The choice of $\delta \leq 1$ here is arbitrary. We could have just as easily used any other positive number. In some proofs, greater care in this choice may be necessary. Now, since $\delta \leq 1$ and $\left| {x + 1} \right| < \delta \leq 1,$ we are able to show that $\left| {x - 3} \right| < 5.$ Consequently, $\left| {x + 1} \right| \cdot \left| {x - 3} \right| < \left| {x + 1} \right| \cdot 5.$ At this point we realize that we also need $\delta \leq {\varepsilon\text{/}5}.$ Thus, we choose $\delta = \text{min}\left\{ {1,{\varepsilon\text{/}5}} \right\}.$

3. Assume $0 < \left| {x + 1} \right| < \delta.$ Thus,

$$\left| {x + 1} \right| < 1\ \text{and}\ {\left| {x + 1} \right| < \frac{\varepsilon}{5}.}$$

Since $\left| {x + 1} \right| < 1,$ we may conclude that $-1 < x + 1 < 1.$ Thus, by subtracting 4 from all parts of the inequality, we obtain $-5 < x - 3 < \text{−}3.$ Consequently, $\left| {x - 3} \right| < 5.$ This gives us

$$\left| {\left( {x^{2} - 2x + 3} \right) - 6} \right| = \left| {x + 1} \right| \cdot \left| {x - 3} \right| < \frac{\varepsilon}{5} \cdot 5 = \varepsilon.$$

Therefore,

$$\underset{x\rightarrow-1}{\text{lim}}\left( {x^{2} - 2x + 3} \right) = 6.$$

Complete the proof that $\underset{x\rightarrow 1}{\text{lim}}x^{2} = 1.$

Let $\varepsilon > 0;$ choose $\delta = \text{min}\left\{ {1,{\varepsilon\text{/}3}} \right\};$ assume $0 < \left| {x - 1} \right| < \delta.$

Since $\left| {x - 1} \right| < 1,$ we may conclude that $-1 < x - 1 < 1.$ Thus, $1 < x + 1 < 3.$ Hence, $\left| {x + 1} \right| < 3.$

You will find that, in general, the more complex a function, the more likely it is that the algebraic approach is the easiest to apply. The algebraic approach is also more useful in proving statements about limits.

Proving Limit Laws 证明极限法则

We now demonstrate how to use the epsilon-delta definition of a limit to construct a rigorous proof of one of the limit laws. The triangle inequality is used at a key point of the proof, so we first review this key property of absolute value.

The triangle inequality states that if a and b are any real numbers, then $\left| {a + b} \right| \leq |a| + |b|.$

Proof 证明

We prove the following limit law: If $\underset{x\rightarrow a}{\text{lim}}f(x) = L$ and $\underset{x\rightarrow a}{\text{lim}}g(x) = M,$ then $\underset{x\rightarrow a}{\text{lim}}\left( {f(x) + g(x)} \right) = L + M.$

Let $\varepsilon > 0.$

Choose $\delta_{1} > 0$ so that if $0 < \left| {x - a} \right| < \delta_{1},$ then $\left| {f(x) - L} \right| < {\varepsilon\text{/}2}.$

Choose $\delta_{2} > 0$ so that if $0 < \left| {x - a} \right| < \delta_{2},$ then $\left| {g(x) - M} \right| < {\varepsilon\text{/}2}.$

Choose $\delta = \text{min}\left\{ {\delta_{1},\delta_{2}} \right\}.$

Assume $0 < \left| {x - a} \right| < \delta.$

Thus,

$$0 < \left| {x - a} \right| < \delta_{1}\ \text{and}\ {0 < \left| {x - a} \right| < \delta_{2}.}$$

Hence,

$$\begin{array}{cl} \left| {\left( {f(x) + g(x)} \right) - \left( {L + M} \right)} \right| & {= \left| {\left( {f(x) - L} \right) + \left( {g(x) - M} \right)} \right|} \\ & {\leq \left| {f(x) - L} \right| + \left| {g(x) - M} \right|} \\ & {< \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.} \end{array}$$

□

We now explore what it means for a limit not to exist. The limit $\underset{x\rightarrow a}{\text{lim}}f(x)$ does not exist if there is no real number L for which $\underset{x\rightarrow a}{\text{lim}}f(x) = L.$ Thus, for all real numbers L, $\underset{x\rightarrow a}{\text{lim}}f(x) \neq L.$ To understand what this means, we look at each part of the definition of $\underset{x\rightarrow a}{\text{lim}}f(x) = L$ together with its opposite. A translation of the definition is given in Table 2.10.

Table 2.10 Translation of the Definition of $\underset{x\rightarrow a}{\text{lim}}f(x) = L$ and its Opposite

Finally, we may state what it means for a limit not to exist. The limit $\underset{x\rightarrow a}{\text{lim}}f(x)$ does not exist if for every real number L, there exists a real number $\varepsilon > 0$ so that for all $\delta > 0,$ there is an x satisfying $0 < \left| {x - a} \right| < \delta,$ so that $\left| {f(x) - L} \right| \geq \varepsilon.$ Let’s apply this in Example 2.43 to show that a limit does not exist.

Showing That a Limit Does Not Exist 证明某个极限不存在

Show that $\underset{x\rightarrow 0}{\text{lim}}\frac{|x|}{x}$ does not exist. The graph of $f(x) = {|x|\text{/}x}$ is shown here:

Solution 解答

Suppose that L is a candidate for a limit. Choose $\varepsilon = {1\text{/}{2.}}$

Let $\delta > 0.$ Either $L \geq 0$ or $L < 0.$ If $L \geq 0,$ then let $x = - {\delta\text{/}2}.$ Thus,

$$\left| {x - 0} \right| = \left| {- \frac{\delta}{2} - 0} \right| = \frac{\delta}{2} < \delta$$

and

$$\left| {\frac{\left| {- \frac{\delta}{2}} \right|}{- \frac{\delta}{2}} - L} \right| = \left| {-1 - L} \right| = L + 1 \geq 1 > \frac{1}{2} = \varepsilon.$$

On the other hand, if $L < 0,$ then let $x = {\delta\text{/}2}.$ Thus,

$$\left| {x - 0} \right| = \left| {\frac{\delta}{2} - 0} \right| = \frac{\delta}{2} < \delta$$

and

$$\left| {\frac{\left| \frac{\delta}{2} \right|}{\frac{\delta}{2}} - L} \right| = \left| {1 - L} \right| = |L| + 1 \geq 1 > \frac{1}{2} = \varepsilon.$$

Thus, for any value of L, $\underset{x\rightarrow 0}{\text{lim}}\frac{|x|}{x} \neq L.$

One-Sided and Infinite Limits 单侧极限与无穷极限

Just as we first gained an intuitive understanding of limits and then moved on to a more rigorous definition of a limit, we now revisit one-sided limits. To do this, we modify the epsilon-delta definition of a limit to give formal epsilon-delta definitions for limits from the right and left at a point. These definitions only require slight modifications from the definition of the limit. In the definition of the limit from the right, the inequality $0 < x - a < \delta$ replaces $0 < \left| {x - a} \right| < \delta,$ which ensures that we only consider values of x that are greater than (to the right of) a. Similarly, in the definition of the limit from the left, the inequality $- \delta < x - a < 0$ replaces $0 < \left| {x - a} \right| < \delta,$ which ensures that we only consider values of x that are less than (to the left of) a.

Limit from the Right: Let $f(x)$ be defined over an open interval of the form $\left( {a,b} \right)$ where $a < b.$ Then,

$$\underset{x\rightarrow a^{+}}{\text{lim}}f(x) = L$$

if for every $\varepsilon > 0,$ there exists a $\delta > 0$ such that if $0 < x - a < \delta,$ then $\left| {f(x) - L} \right| < \varepsilon.$

Limit from the Left: Let $f(x)$ be defined over an open interval of the form $\left( {b,a} \right)$ where $b < a.$ Then,

$$\underset{x\rightarrow a^{-}}{\text{lim}}f(x) = L$$

if for every $\varepsilon > 0,$ there exists a $\delta > 0$ such that if $- \delta < x - a < 0,$ then $\left| {f(x) - L} \right| < \varepsilon.$

Proving a Statement about a Limit From the Right 证明关于右极限的命题

Prove that $\underset{x\rightarrow 4^{+}}{\text{lim}}\sqrt{x - 4} = 0.$

Solution 解答

Let $\varepsilon > 0.$

Choose $\delta = \varepsilon^{2}.$ Since we ultimately want $\left| {\sqrt{x - 4} - 0} \right| < \varepsilon,$ we manipulate this inequality to get $\sqrt{x - 4} < \varepsilon$ or, equivalently, $0 < x - 4 < \varepsilon^{2},$ making $\delta = \varepsilon^{2}$ a clear choice. We may also determine $\delta$ geometrically, as shown in Figure 2.42.

Assume $0 < x - 4 < \delta.$ Thus, $0 < x - 4 < \varepsilon^{2}.$ Hence, $0 < \sqrt{x - 4} < \varepsilon.$ Finally, $\left| {\sqrt{x - 4} - 0} \right| < \varepsilon.$

Therefore, $\underset{x\rightarrow 4^{+}}{\text{lim}}\sqrt{x - 4} = 0.$

Find $\delta$ corresponding to ε for a proof that $\underset{x\rightarrow 1^{-}}{\text{lim}}\sqrt{1 - x} = 0.$

We conclude the process of converting our intuitive ideas of various types of limits to rigorous formal definitions by pursuing a formal definition of infinite limits. To have $\underset{x\rightarrow a}{\text{lim}}f(x) = \text{+}\infty,$ we want the values of the function $f(x)$ to get larger and larger as x approaches a. Instead of the requirement that $\left| {f(x) - L} \right| < \varepsilon$ for arbitrarily small ε when $0 < \left| {x - a} \right| < \delta$ for small enough $\delta,$ we want $f(x) > M$ for arbitrarily large positive M when $0 < \left| {x - a} \right| < \delta$ for small enough $\delta.$ Figure 2.43 illustrates this idea by showing the value of $\delta$ for successively larger values of M.

Let $f(x)$ be defined for all $x \neq a$ in an open interval containing a. Then, we have an infinite limit

$$\underset{x\rightarrow a}{\text{lim}}f(x) = \text{+}\infty$$

if for every $M > 0,$ there exists $\delta > 0$ such that if $0 < \left| {x - a} \right| < \delta,$ then $f(x) > M.$

Let $f(x)$ be defined for all $x \neq a$ in an open interval containing a. Then, we have a negative infinite limit

$$\underset{x\rightarrow a}{\text{lim}}f(x) = \text{−}\infty$$

if for every $M > 0,$ there exists $\delta > 0$ such that if $0 < \left| {x - a} \right| < \delta,$ then $f(x) < \text{−}M.$

Section 2.5 Exercises 2.5 节习题

In the following exercises, write the appropriate $\varepsilon$-$\delta$ definition for each of the given statements.

176.

$\underset{x\rightarrow a}{\text{lim}}f(x) = N$

177.

$\underset{t\rightarrow b}{\text{lim}}g(t) = M$

178.

$\underset{x\rightarrow c}{\text{lim}}h(x) = L$

179.

$\underset{x\rightarrow a}{\text{lim}}\varphi(x) = A$

The following graph of the function f satisfies $\underset{x\rightarrow 2}{\text{lim}}f(x) = 2.$ In the following exercises, determine a value of $\delta > 0$ that satisfies each statement.

180.

If $0 < \left| {x - 2} \right| < \delta,$ then $\left| {f(x) - 2} \right| < 1.$

181.

If $0 < \left| {x - 2} \right| < \delta,$ then $\left| {f(x) - 2} \right| < 0.5.$

The following graph of the function f satisfies $\underset{x\rightarrow 3}{\text{lim}}f(x) = -1.$ In the following exercises, determine a value of $\delta > 0$ that satisfies each statement.

182.

If $0 < \left| {x - 3} \right| < \delta,$ then $\left| {f(x) + 1} \right| < 1.$

183.

If $0 < \left| {x - 3} \right| < \delta,$ then $\left| {f(x) + 1} \right| < 2.$

The following graph of the function f satisfies $\underset{x\rightarrow 3}{\text{lim}}f(x) = 2.$ In the following exercises, for each value of ε, find a value of $\delta > 0$ such that the precise definition of limit holds true.

184.

$\varepsilon = 1.5$

185.

$\varepsilon = 3$

\[T\] In the following exercises, use a graphing calculator to find a number $\delta$ such that the statements hold true.

186.

$\left| {\text{sin}\mspace{2mu}\left( {2x} \right) - \frac{1}{2}} \right| < 0.1,$ whenever $\left| {x - \frac{\pi}{12}} \right| < \delta$

187.

$\left| {\sqrt{x - 4} - 2} \right| < 0.1,\ \text{whenever}\ \left| {x - 8} \right| < \delta$

In the following exercises, use the precise definition of limit to prove the given limits.

188.

$\underset{x\rightarrow 2}{\text{lim}}\left( {5x + 8} \right) = 18$

189.

$\underset{x\rightarrow 3}{\text{lim}}\frac{x^{2} - 9}{x - 3} = 6$

190.

$\underset{x\rightarrow 2}{\text{lim}}\frac{2x^{2} - 3x - 2}{x - 2} = 5$

191.

$\underset{x\rightarrow 0}{\text{lim}}x^{4} = 0$

192.

$\underset{x\rightarrow 2}{\text{lim}}(x^{2} + 2x) = 8$

In the following exercises, use the precise definition of limit to prove the given one-sided limits.

193.

$\underset{x\rightarrow 5^{-}}{\text{lim}}\sqrt{5 - x} = 0$

194.

$\underset{x\rightarrow 0^{+}}{\text{lim}}f(x) = -2,\ \text{where}\ f(x) = \left\{ \begin{matrix} {8x - 3,\ \text{if}\ x < 0} \\ {4x - 2,\ \text{if}\ x \geq 0} \end{matrix} \right..$

195.

$\underset{x\rightarrow 1^{-}}{\text{lim}}f(x) = 3,\ \text{where}\ f(x) = \left\{ \begin{matrix} {5x - 2,\ \text{if}\ x < 1} \\ {7x - 1,\ \text{if}\ x \geq 1} \end{matrix} \right..$

In the following exercises, use the precise definition of limit to prove the given infinite limits.

196.

$\underset{x\rightarrow 0}{\text{lim}}\frac{1}{x^{2}} = \infty$

197.

$\underset{x\rightarrow-1}{\text{lim}}\frac{3}{\left( {x + 1} \right)^{2}} = \infty$

198.

$\underset{x\rightarrow 2}{\text{lim}} - \frac{1}{\left( {x - 2} \right)^{2}} = \text{−}\infty$

199.

An engineer is using a machine to cut a flat square of Aerogel of area 144 cm². If there is a maximum error tolerance in the area of 8 cm², how accurately must the engineer cut on the side, assuming all sides have the same length? How do these numbers relate to $\delta,$ ε, a, and L?

200.

Use the precise definition of limit to prove that the following limit does not exist: $\underset{x\rightarrow 1}{\text{lim}}\frac{\left| {x - 1} \right|}{x - 1}.$

201.

Using precise definitions of limits, prove that $\underset{x\rightarrow 0}{\text{lim}}f(x)$ does not exist, given that $f(x)$ is the ceiling function. (*Hint*: Try any $\delta < 1\text{.)}$

202.

Using precise definitions of limits, prove that $\underset{x\rightarrow 0}{\text{lim}}f(x)$ does not exist: $f(x) = \left\{ \begin{array}{l} {1\ \text{if}\ x\ \text{is rational}} \\ {0\ \text{if}\ x\ \text{is irrational}} \end{array} \right..$ (*Hint*: Think about how you can always choose a rational number $0 < r < d,$ but $\left| {f(r) - 0} \right| = 1\text{.)}$

203.

Using precise definitions of limits, determine $\underset{x\rightarrow 0}{\text{lim}}f(x)$ for $f(x) = \left\{ \begin{array}{l} {x\ \text{if}\ x\ \text{is rational}} \\ {0\ \text{if}\ x\ \text{is irrational}} \end{array} \right..$ (*Hint*: Break into two cases, x rational and x irrational.)

204.

Using the function from the previous exercise, use the precise definition of limits to show that $\underset{x\rightarrow a}{\text{lim}}f(x)$ does not exist for $a \neq 0.$

For the following exercises, suppose that $\underset{x\rightarrow a}{\text{lim}}f(x) = L$ and $\underset{x\rightarrow a}{\text{lim}}g(x) = M$ both exist. Use the precise definition of limits to prove the following limit laws:

205.

$\underset{x\rightarrow a}{\text{lim}}\left( {f(x) + g(x)} \right) = L + M$

206.

$\underset{x\rightarrow a}{\text{lim}}\left\lbrack {cf(x)} \right\rbrack = cL$ for any real constant c (*Hint*: Consider two cases: $c = 0$ and $c \neq 0\text{.)}$

207.

$\underset{x\rightarrow a}{\text{lim}}\left\lbrack {f(x)g(x)} \right\rbrack = LM.$ (*Hint*: $\left| {f(x)g(x) - LM} \right| =$ $\left| {f(x)g(x) - f(x)M + f(x)M - LM} \right| \leq \left| {f(x)} \right|\left| {g(x) - M} \right| + |M|{\left| {f(x) - L} \right|\text{.)}}$

Key Terms 关键术语

average velocity

the change in an object’s position divided by the length of a time period; the average velocity of an object over a time interval $\left\lbrack {t,a} \right\rbrack$ (if $t < a$ or $\left\lbrack {a,t} \right\rbrack$ if $\left. t > a \right)$, with a position given by $s(t),$ that is $v_{\text{ave}} = \frac{s(t) - s(a)}{t - a}$

constant multiple law for limits

the limit law $\underset{x\rightarrow a}{\text{lim}}cf(x) = c \cdot \underset{x\rightarrow a}{\text{lim}}f(x) = cL$

continuity at a point

A function $f(x)$ is continuous at a point *a* if and only if the following three conditions are satisfied: (1) $f(a)$ is defined, (2) $\underset{x\rightarrow a}{\text{lim}}f(x)$ exists, and (3) $\underset{x\rightarrow a}{\text{lim}}f(x) = f(a)$

continuity from the left

A function is continuous from the left at *b* if $\underset{x\rightarrow b^{-}}{\text{lim}}f(x) = f(b)$

continuity from the right

A function is continuous from the right at *a* if $\underset{x\rightarrow a^{+}}{\text{lim}}f(x) = f(a)$

continuity over an interval

a function that can be traced with a pencil without lifting the pencil; a function is continuous over an open interval if it is continuous at every point in the interval; a function $f(x)$ is continuous over a closed interval of the form $\left\lbrack {a,b} \right\rbrack$ if it is continuous at every point in $\left( {a,b} \right),$ and it is continuous from the right at *a* and from the left at *b*

difference law for limits

the limit law $\underset{x\rightarrow a}{\text{lim}}\left( {f(x) - g(x)} \right) = \underset{x\rightarrow a}{\text{lim}}f(x) - \underset{x\rightarrow a}{\text{lim}}g(x) = L - M$

differential calculus

the field of calculus concerned with the study of derivatives and their applications

discontinuity at a point

A function is discontinuous at a point or has a discontinuity at a point if it is not continuous at the point

epsilon-delta definition of the limit

$\underset{x\rightarrow a}{\text{lim}}f(x) = L$ if for every $\varepsilon > 0,$ there exists a $\delta > 0$ such that if $0 < \left| {x - a} \right| < \delta,$ then $\left| {f(x) - L} \right| < \varepsilon$

infinite discontinuity

An infinite discontinuity occurs at a point *a* if $\underset{x\rightarrow a^{-}}{\text{lim}}f(x) = \text{±}\infty$ or $\underset{x\rightarrow a^{+}}{\text{lim}}f(x) = \text{±}\infty$

infinite limit

A function has an infinite limit at a point *a* if it either increases or decreases without bound as it approaches *a*

instantaneous velocity

The instantaneous velocity of an object with a position function that is given by $s(t)$ is the value that the average velocities on intervals of the form $\left\lbrack {t,a} \right\rbrack$ and $\left\lbrack {a,t} \right\rbrack$ approach as the values of *t* move closer to $a,$ provided such a value exists

integral calculus

the study of integrals and their applications

Intermediate Value Theorem

Let *f* be continuous over a closed bounded interval $\left\lbrack {\text{a},\text{b}} \right\rbrack;$ if *z* is any real number between $f(a)$ and $f(b),$ then there is a number *c* in $\left\lbrack {a,b} \right\rbrack$ satisfying $f(c) = z$

intuitive definition of the limit

If all values of the function $f(x)$ approach the real number *L* as the values of $x\left( {\neq a} \right)$ approach *a*, $f(x)$ approaches *L*

jump discontinuity

A jump discontinuity occurs at a point *a* if $\underset{x\rightarrow a^{-}}{\text{lim}}f(x)$ and $\underset{x\rightarrow a^{+}}{\text{lim}}f(x)$ both exist, but $\underset{x\rightarrow a^{-}}{\text{lim}}f(x) \neq \underset{x\rightarrow a^{+}}{\text{lim}}f(x)$

limit

the process of letting *x* or *t* approach *a* in an expression; the limit of a function $f(x)$ as *x* approaches *a* is the value that $f(x)$ approaches as *x* approaches *a*

limit laws

the individual properties of limits; for each of the individual laws, let $f(x)$ and $g(x)$ be defined for all $x \neq a$ over some open interval containing *a*; assume that *L* and *M* are real numbers so that $\underset{x\rightarrow a}{\text{lim}}f(x) = L$ and $\underset{x\rightarrow a}{\text{lim}}g(x) = M;$ let *c* be a constant

multivariable calculus

the study of the calculus of functions of two or more variables

one-sided limit

A one-sided limit of a function is a limit taken from either the left or the right

power law for limits

the limit law $\underset{x\rightarrow a}{\text{lim}}\left( {f(x)} \right)^{n} = \left( {\underset{x\rightarrow a}{\text{lim}}f(x)} \right)^{n} = L^{n}$ for every positive integer *n*

product law for limits

the limit law $\underset{x\rightarrow a}{\text{lim}}\left( {f(x) \cdot g(x)} \right) = \underset{x\rightarrow a}{\text{lim}}f(x) \cdot \underset{x\rightarrow a}{\text{lim}}g(x) = L \cdot M$

quotient law for limits

the limit law $\underset{x\rightarrow a}{\text{lim}}\frac{f(x)}{g(x)} = \frac{\underset{x\rightarrow a}{\text{lim}}f(x)}{\underset{x\rightarrow a}{\text{lim}}g(x)} = \frac{L}{M}$ for $M \neq 0$

removable discontinuity

A removable discontinuity occurs at a point *a* if $f(x)$ is discontinuous at *a*, but $\underset{x\rightarrow a}{\text{lim}}f(x)$ exists

root law for limits

the limit law $\underset{x\rightarrow a}{\text{lim}}\sqrt[n]{f(x)} = \sqrt[n]{\underset{x\rightarrow a}{\text{lim}}f(x)} = \sqrt[n]{L}$ for all *L* if *n* is odd and for $L \geq 0$ if *n* is even

secant

A secant line to a function $f(x)$ at *a* is a line through the point $\left( {a,f(a)} \right)$ and another point on the function; the slope of the secant line is given by $m_{\text{sec}} = \frac{f(x) - f(a)}{x - a}$

squeeze theorem

states that if $f(x) \leq g(x) \leq h(x)$ for all $x \neq a$ over an open interval containing *a* and $\underset{x\rightarrow a}{\text{lim}}f(x) = L = \underset{x\rightarrow a}{\text{lim}}h(x)$ where *L* is a real number, then $\underset{x\rightarrow a}{\text{lim}}g(x) = L$

sum law for limits

The limit law $\underset{x\rightarrow a}{\text{lim}}\left( {f(x) + g(x)} \right) = \underset{x\rightarrow a}{\text{lim}}f(x) + \underset{x\rightarrow a}{\text{lim}}g(x) = L + M$

tangent

A tangent line to the graph of a function at a point $\left( {a,f(a)} \right)$ is the line that secant lines through $\left( {a,f(a)} \right)$ approach as they are taken through points on the function with *x*-values that approach *a*; the slope of the tangent line to a graph at *a* measures the rate of change of the function at *a*

triangle inequality

If *a* and *b* are any real numbers, then $\left| {a + b} \right| \leq |a| + |b|$

vertical asymptote

A function has a vertical asymptote at $x = a$ if the limit as *x* approaches *a* from the right or left is infinite

Key Equations 公式表

Key Concepts 核心概念

2.1 A Preview of Calculus 2.1 微积分学预览

• Differential calculus arose from trying to solve the problem of determining the slope of a line tangent to a curve at a point. The slope of the tangent line indicates the rate of change of the function, also called the *derivative*. Calculating a derivative requires finding a limit.
• Integral calculus arose from trying to solve the problem of finding the area of a region between the graph of a function and the *x*-axis. We can approximate the area by dividing it into thin rectangles and summing the areas of these rectangles. This summation leads to the value of a function called the *integral*. The integral is also calculated by finding a limit and, in fact, is related to the derivative of a function.
• Multivariable calculus enables us to solve problems in three-dimensional space, including determining motion in space and finding volumes of solids.

• 微分学源于试图解决「求曲线在某点处切线斜率」这一问题。切线的斜率表示函数的变化率，也称为导数。求导数需要计算一个极限。
• 积分学源于试图解决「求函数图像与 x 轴之间区域的面积」这一问题。我们可以把该区域划分为若干细矩形，再把这些矩形的面积相加来近似面积。这种求和得到了一个称为积分的函数值。积分同样通过求极限来计算，并且实际上与函数的导数相关。
• 多元微积分使我们能够解决三维空间中的问题，包括确定空间中的运动以及求立体的体积。

2.2 The Limit of a Function 2.2 函数的极限

• A table of values or graph may be used to estimate a limit.
• If the limit of a function at a point does not exist, it is still possible that the limits from the left and right at that point may exist.
• If the limits of a function from the left and right exist and are equal, then the limit of the function is that common value.
• We may use limits to describe infinite behavior of a function at a point.

• 数值表或图像可用于估计一个极限。
• 即使函数在某点的极限不存在，该点处的左极限与右极限仍可能存在。
• 若函数的左极限与右极限都存在且相等，则函数的极限即为该公共值。
• 我们可以用极限来描述函数在某点处的无穷性态。

2.3 The Limit Laws 2.3 极限法则

• The limit laws allow us to evaluate limits of functions without having to go through step-by-step processes each time.
• For polynomials and rational functions, $\underset{x\rightarrow a}{\text{lim}}f(x) = f(a).$
• You can evaluate the limit of a function by factoring and canceling, by multiplying by a conjugate, or by simplifying a complex fraction.
• The squeeze theorem allows you to find the limit of a function if the function is always greater than one function and less than another function with limits that are known.

• 极限法则使我们能够直接计算函数的极限，而不必每次都逐步推导。
• 对于多项式与有理函数，$\underset{x\rightarrow a}{\text{lim}}f(x) = f(a).$
• 可以通过因式分解并约分、乘以共轭式，或化简繁分式来求函数极限。
• 夹逼定理可用于求函数极限：若该函数恒大于一个函数且小于另一个函数，且后两者的极限已知，则可求得该函数的极限。

2.4 Continuity 2.4 连续性

• For a function to be continuous at a point, it must be defined at that point, its limit must exist at the point, and the value of the function at that point must equal the value of the limit at that point.
• Discontinuities may be classified as removable, jump, or infinite.
• A function is continuous over an open interval if it is continuous at every point in the interval. It is continuous over a closed interval if it is continuous at every point in its interior and is continuous at its endpoints.
• The composite function theorem states: If $f(x)$ is continuous at *L* and $\underset{x\rightarrow a}{\text{lim}}g(x) = L,$ then $\underset{x\rightarrow a}{\text{lim}}f\left( {g(x)} \right) = f\left( {\underset{x\rightarrow a}{\text{lim}}g(x)} \right) = f(L).$
• The Intermediate Value Theorem guarantees that if a function is continuous over a closed interval, then the function takes on every value between the values at its endpoints.

• 函数在某点连续，必须在该点有定义、在该点的极限存在，且函数在该点的值必须等于该点极限的值。
• 间断可分为可去间断、跳跃间断或无穷间断。
• 若函数在开区间内每一点都连续，则它在开区间上连续；若函数在闭区间内部每一点连续且在端点处连续，则它在闭区间上连续。
• 复合函数定理：若 $f(x)$ 在 L 处连续且 $\underset{x\rightarrow a}{\text{lim}}g(x) = L,$ 则 $\underset{x\rightarrow a}{\text{lim}}f\left( {g(x)} \right) = f\left( {\underset{x\rightarrow a}{\text{lim}}g(x)} \right) = f(L).$
• 介值定理保证：若函数在闭区间上连续，则它取遍其端点函数值之间的每一个值。

2.5 The Precise Definition of a Limit 2.5 极限的精确定义

• The intuitive notion of a limit may be converted into a rigorous mathematical definition known as the *epsilon-delta definition of the limit*.
• The epsilon-delta definition may be used to prove statements about limits.
• The epsilon-delta definition of a limit may be modified to define one-sided limits.

• 极限的直观概念可以转化为一个严格的数学定义，即极限的 ε-δ 定义。
• ε-δ 定义可用于证明有关极限的命题。
• 极限的 ε-δ 定义可经修改用于定义单侧极限。

Review Exercises 复习题

True or False. In the following exercises, justify your answer with a proof or a counterexample.

208.

A function has to be continuous at $x = a$ if the $\underset{x\rightarrow a}{\text{lim}}f(x)$ exists.

209.

You can use the quotient rule to evaluate $\underset{x\rightarrow 0}{\text{lim}}\frac{\text{sin}\mspace{2mu} x}{x}.$

210.

If there is a vertical asymptote at $x = a$ for the function $f(x),$ then *f* is undefined at the point $x = a.$

211.

If $\underset{x\rightarrow a}{\text{lim}}f(x)$ does not exist, then *f* is undefined at the point $x = a.$

212.

Using the graph, find each limit or explain why the limit does not exist.

1. $\underset{x\rightarrow-1}{\text{lim}}f(x)$

2. $\underset{x\rightarrow 1}{\text{lim}}f(x)$

3. $\underset{x\rightarrow 0^{+}}{\text{lim}}f(x)$

4. $\underset{x\rightarrow 2}{\text{lim}}f(x)$

In the following exercises, evaluate the limit algebraically or explain why the limit does not exist.

213.

$\underset{x\rightarrow 2}{\text{lim}}\frac{2x^{2} - 3x - 2}{x - 2}$

214.

$\underset{x\rightarrow 0}{\text{lim}}3x^{2} - 2x + 4$

215.

$\underset{x\rightarrow 3}{\text{lim}}\frac{x^{3} - 2x^{2} - 1}{3x - 2}$

216.

$\underset{x\rightarrow\pi\text{/}2}{\text{lim}}\frac{\text{cot}\mspace{2mu} x}{\text{cos}\mspace{2mu} x}$

217.

$\underset{x\rightarrow-5}{\text{lim}}\frac{x^{2} + 25}{x + 5}$

218.

$\underset{x\rightarrow 2}{\text{lim}}\frac{3x^{2} - 2x - 8}{x^{2} - 4}$

219.

$\underset{x\rightarrow 1}{\text{lim}}\frac{x^{2} - 1}{x^{3} - 1}$

220.

$\underset{x\rightarrow 1}{\text{lim}}\frac{x^{2} - 1}{\sqrt{x} - 1}$

221.

$\underset{x\rightarrow 4}{\text{lim}}\frac{4 - x}{\sqrt{x} - 2}$

222.

$\underset{x\rightarrow 4}{\text{lim}}\frac{1}{\sqrt{x} - 2}$

In the following exercises, use the squeeze theorem to prove the limit.

223.

$\underset{x\rightarrow 0}{\text{lim}}x^{2}\text{cos}\mspace{2mu}\left( {2\pi x} \right) = 0$

224.

$\underset{x\rightarrow 0}{\text{lim}}x^{3}\text{sin}\mspace{2mu}\left( \frac{\pi}{x} \right) = 0$

225.

Determine the domain such that the function $f(x) = \sqrt{x - 2} + xe^{x}$ is continuous over its domain.

In the following exercises, determine the value of *c* such that the function remains continuous. Draw your resulting function to ensure it is continuous.

226.

$f(x) = \left\{ \begin{array}{l} {x^{2} + 1,x > c} \\ {2x,x \leq c} \end{array} \right.$

227.

$f(x) = \left\{ \begin{array}{l} {\sqrt{x + 1},x > \text{−}1} \\ {x^{2} + c,x \leq - 1} \end{array} \right.$

In the following exercises, use the precise definition of limit to prove the limit.

228.

$\underset{x\rightarrow 1}{\text{lim}}(8x + 16) = 24$

229.

$\underset{x\rightarrow 0}{\text{lim}}x^{3} = 0$

230.

A ball is thrown into the air and the vertical position is given by $x(t) = -4.9t^{2} + 25t + 5.$ Use the Intermediate Value Theorem to show that the ball must land on the ground sometime between 5 sec and 6 sec after the throw.

231.

A particle moving along a line has a displacement according to the function $x(t) = t^{2} - 2t + 4,$ where *x* is measured in meters and *t* is measured in seconds. Find the average velocity over the time period $t = \left\lbrack {0,2} \right\rbrack.$

232.

From the previous exercises, estimate the instantaneous velocity at $t = 2$ by checking the average velocity within $t = 0.01\ \text{sec}\text{.}$