Chapter 2

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Chapter 2

Calculus Volume 1Chapter 2

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Chapter 2

Checkpoint

2.1

2.25

2.2

12.006001

2.3

17 unit²

2.4

$\underset{x\rightarrow 1}{\text{lim}}\frac{\frac{1}{x} - 1}{x - 1} = -1$

2.5

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

2.6

$\underset{x\rightarrow 2}{\text{lim}}\frac{\left| {x^{2} - 4} \right|}{x - 2}$ does not exist.

2.7

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

2.8

a\. $\underset{x\rightarrow 0^{-}}{\text{lim}}\frac{1}{x^{2}} = \text{+}\infty;$ b. $\underset{x\rightarrow 0^{+}}{\text{lim}}\frac{1}{x^{2}} = \text{+}\infty;$ c. $\underset{x\rightarrow 0}{\text{lim}}\frac{1}{x^{2}} = \text{+}\infty$

2.9

a\. $\underset{x\rightarrow 2^{-}}{\text{lim}}\frac{1}{\left( {x - 2} \right)^{3}} = \text{−}\infty;$ b. $\underset{x\rightarrow 2^{+}}{\text{lim}}\frac{1}{\left( {x - 2} \right)^{3}} = \text{+}\infty;$ c. $\underset{x\rightarrow 2}{\text{lim}}\frac{1}{\left( {x - 2} \right)^{3}}$ DNE. The line $x = 2$ is the vertical asymptote of $f(x) = {1\text{/}\left( {x - 2} \right)^{3}}.$

2.10

Does not exist.

2.11

$11\sqrt{10}$

2.12

−13;

2.13

$\frac{1}{3}$

2.14

$\frac{1}{4}$

2.15

−1;

2.16

$\frac{1}{4}$

2.17

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

2.18

+∞

2.19

0

2.20

0

2.21

*f* is not continuous at 1 because $f(1) = 2 \neq 3 = \underset{x\rightarrow 1}{\text{lim}}f(x).$

2.22

$f(x)$ is continuous at every real number.

2.23

Discontinuous at 1; removable

2.24

$\left\lbrack {-3,\text{+}\infty} \right)$

2.25

0

2.26

$f(0) = 1 > 0,f(1) = -2 < 0;f(x)$ is continuous over $\left\lbrack {0,1} \right\rbrack.$ It must have a zero on this interval.

2.27

Let $\varepsilon > 0;$ choose $\delta = \frac{\varepsilon}{3};$ assume $0 < \left| {x - 2} \right| < \delta.$

Thus, $\left| {\left( {3x - 2} \right) - 4} \right| = \left| {3x - 6} \right| = |3| \cdot \left| {x - 2} \right| < 3 \cdot \delta = 3 \cdot \left( {\varepsilon\text{/}3} \right) = \varepsilon.$

Therefore, $\underset{x\rightarrow 2}{\text{lim}}3x - 2 = 4.$

2.28

Choose $\delta = \text{min}\left\{ {9 - \left( {3 - \varepsilon} \right)^{2},\left( {3 + \varepsilon} \right)^{2} - 9} \right\}.$

2.29

$\left| {x^{2} - 1} \right| = \left| {x - 1} \right| \cdot \left| {x + 1} \right| < {\varepsilon\text{/}3} \cdot 3 = \varepsilon$

2.30

$\delta = \varepsilon^{2}$

Section 2.1 Exercises

1.

a\. 2.2100000; b. 2.0201000; c. 2.0020010; d. 2.0002000; e. (1.1000000, 2.2100000); f. (1.0100000, 2.0201000); g. (1.0010000, 2.0020010); h. (1.0001000, 2.0002000); i. 2.1000000; j. 2.0100000; k. 2.0010000; l. 2.0001000

3.

$y = 2x$

5.

3

7.

a\. 2.0248457; b. 2.0024984; c. 2.0002500; d. 2.0000250; e. (4.1000000,2.0248457); f. (4.0100000,2.0024984); g. (4.0010000,2.0002500); h. (4.00010000,2.0000250); i. 0.24845673; j. 0.24984395; k. 0.24998438; l. 0.24999844

9.

$y = \frac{x}{4} + 1$

11.

*π*

13.

a\. −0.95238095; b. −0.99009901; c. −0.99502488; d. −0.99900100; e. (−1;.0500000,−0;.95238095); f. (−1;.0100000,−0;.9909901); g. (−1;.0050000,−0;.99502488); h. (1.0010000,−0;.99900100); i. −0.95238095; j. −0.99009901; k. −0.99502488; l. −0.99900100

15.

$y = \text{−}x - 2$

17.

−49 m/sec (velocity of the ball is 49 m/sec downward)

19.

5.2 m/sec

21.

−9.8 m/sec

23.

6 m/sec

25.

Under, 1 unit²; over: 4 unit². The exact area of the two triangles is $\frac{1}{2}(1)(1) + \frac{1}{2}(2)(2) = 2.5{\ \text{units}}^{2}.$

27.

Under, 0.96 unit²; over, 1.92 unit². The exact area of the semicircle with radius 1 is $\frac{\pi(1)^{2}}{2} = \frac{\pi}{2}$ unit².

29.

Approximately 1.3333333 unit²

Section 2.2 Exercises

31.

$\underset{x\rightarrow 1}{\text{lim}}f(x)$ does not exist because $\underset{x\rightarrow 1^{-}}{\text{lim}}f(x) = -2 \neq \underset{x\rightarrow 1^{+}}{\text{lim}}f(x) = 2.$

33.

$\underset{x\rightarrow 0}{\text{lim}}\left( {1 + x} \right)^{1\text{/}x} = 2.7183$

35.

a\. 1.98669331; b. 1.99986667; c. 1.99999867; d. 1.99999999; e. 1.98669331; f. 1.99986667; g. 1.99999867; h. 1.99999999; $\underset{x\rightarrow 0}{\text{lim}}\frac{\text{sin}\mspace{2mu} 2x}{x} = 2$

37.

$\underset{x\rightarrow 0}{\text{lim}}\frac{\text{sin}\mspace{2mu} ax}{x} = a$

39.

a\. −0.80000000; b. −0.98000000; c. −0.99800000; d. −0.99980000; e. −1.2000000; f. −1.0200000; g. −1.0020000; h. −1.0002000; $\underset{x\rightarrow 1}{\text{lim}}\left( {1 - 2x} \right) = -1$

41.

a\. −37.931934; b. −3377.9264; c. −333,777.93; d. −33,337,778; e. −29.032258; f. −3289.0365; g. −332,889.04; h. −33,328,889 $\underset{x\rightarrow 0}{\text{lim}}\frac{z - 1}{z^{2}\left( {z + 3} \right)} = \text{−}\infty$

43.

a\. 0.13495277; b. 0.12594300; c. 0.12509381; d. 0.12500938; e. 0.11614402; f. 0.12406794; g. 0.12490631; h. 0.12499063; $\therefore\underset{x\rightarrow 2}{\text{lim}}\frac{1 - \frac{2}{x}}{x^{2} - 4} = 0.1250 = \frac{1}{8}$

45.

a\. 10.00000; b. 100.00000; c. 1000.0000; d. 10,000.000; Guess: $\underset{\alpha\rightarrow 0^{+}}{\text{lim}}\frac{1}{\alpha}\mspace{2mu}\text{cos}\mspace{2mu}\left( \frac{\pi}{\alpha} \right) = \infty,$ actual: DNE

47.

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

49.

False; $\underset{x\rightarrow 6}{\text{lim}}f(x)$ DNE since $\underset{x\rightarrow 6^{-}}{\text{lim}}f(x) = 2$ and $\underset{x\rightarrow 6^{+}}{\text{lim}}f(x) = 5.$

51.

2

53.

1

55.

1

57.

DNE

59.

0

61.

DNE

63.

2

65.

3

67.

DNE

69.

0

71.

−2

73.

DNE

75.

0

77.

Answers may vary.

79.

Answers may vary.

81.

a\. $\rho_{2}$ b. $\rho_{1}$ c. DNE unless $\rho_{1} = \rho_{2}.$ As you approach $x_{\text{SF}}$ from the left, you are in the high-density area of the shock. When you approach from the right, you have not experienced the “shock” yet and are at a lower density.

Section 2.3 Exercises

83.

Use constant multiple law and difference law: $\underset{x\rightarrow 0}{\text{lim}}\left( {4x^{2} - 2x + 3} \right) = 4\underset{x\rightarrow 0}{\text{lim}}x^{2} - 2\underset{x\rightarrow 0}{\text{lim}}x + \underset{x\rightarrow 0}{\text{lim}}3 = 3$

85.

Use root law: $\underset{x\rightarrow-2}{\text{lim}}\sqrt{x^{2} - 6x + 3} = \sqrt{\underset{x\rightarrow-2}{\text{lim}}\left( {x^{2} - 6x + 3} \right)} = \sqrt{19}$

87.

49

89.

1

91.

$- \frac{5}{7}$

93.

$\underset{x\rightarrow 4}{\text{lim}}\frac{x^{2} - 16}{x - 4} = \frac{16 - 16}{4 - 4} = \frac{0}{0};$ then, $\underset{x\rightarrow 4}{\text{lim}}\frac{x^{2} - 16}{x - 4} = \underset{x\rightarrow 4}{\text{lim}}\frac{\left( {x + 4} \right)\left( {x - 4} \right)}{x - 4} = 8$

95.

$\underset{x\rightarrow 6}{\text{lim}}\frac{3x - 18}{2x - 12} = \frac{18 - 18}{12 - 12} = \frac{0}{0};$ then, $\underset{x\rightarrow 6}{\text{lim}}\frac{3x - 18}{2x - 12} = \underset{x\rightarrow 6}{\text{lim}}\frac{3\left( {x - 6} \right)}{2\left( {x - 6} \right)} = \frac{3}{2}$

97.

$\underset{x\rightarrow 9}{\text{lim}}\frac{t - 9}{\sqrt{t} - 3} = \frac{9 - 9}{3 - 3} = \frac{0}{0};$ then, $\underset{t\rightarrow 9}{\text{lim}}\frac{t - 9}{\sqrt{t} - 3} = \underset{t\rightarrow 9}{\text{lim}}\frac{t - 9}{\sqrt{t} - 3}\frac{\sqrt{t} + 3}{\sqrt{t} + 3} = \underset{t\rightarrow 9}{\text{lim}}\left( {\sqrt{t} + 3} \right) = 6$

99.

$\underset{\theta\rightarrow\pi}{\text{lim}}\frac{\text{sin}\mspace{2mu}\theta}{\text{tan}\mspace{2mu}\theta} = \frac{\text{sin}\mspace{2mu}\pi}{\text{tan}\mspace{2mu}\pi} = \frac{0}{0};$ then, $\underset{\theta\rightarrow\pi}{\text{lim}}\frac{\text{sin}\mspace{2mu}\theta}{\text{tan}\mspace{2mu}\theta} = \underset{\theta\rightarrow\pi}{\text{lim}}\frac{\text{sin}\mspace{2mu}\theta}{\frac{\text{sin}\mspace{2mu}\theta}{\text{cos}\mspace{2mu}\theta}} = \underset{\theta\rightarrow\pi}{\text{lim}}\text{cos}\mspace{2mu}\theta = -1$

101.

$\underset{x\rightarrow 1\text{/}2}{\text{lim}}\frac{2x^{2} + 3x - 2}{2x - 1} = \frac{\frac{1}{2} + \frac{3}{2} - 2}{1 - 1} = \frac{0}{0};$ then, $\underset{x\rightarrow 1\text{/}2}{\text{lim}}\frac{2x^{2} + 3x - 2}{2x - 1} = \underset{x\rightarrow 1\text{/}2}{\text{lim}}\frac{\left( {2x - 1} \right)\left( {x + 2} \right)}{2x - 1} = \frac{5}{2}$

103.

−∞

105.

−∞

107.

$\underset{x\rightarrow 6}{\text{lim}}2f(x)g(x) = 2\underset{x\rightarrow 6}{\text{lim}}f(x)\underset{x\rightarrow 6}{\text{lim}}g(x) = 72$

109.

$\underset{x\rightarrow 6}{\text{lim}}\left( {f(x) + \frac{1}{3}g(x)} \right) = \underset{x\rightarrow 6}{\text{lim}}f(x) + \frac{1}{3}\underset{x\rightarrow 6}{\text{lim}}g(x) = 7$

111.

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

113.

$\underset{x\rightarrow 6}{\text{lim}}\left\lbrack {\left( {x + 1} \right)f(x)} \right\rbrack = \left( {\underset{x\rightarrow 6}{\text{lim}}\left( {x + 1} \right)} \right)\left( {\underset{x\rightarrow 6}{\text{lim}}f(x)} \right) = 28$

115.

a. 9; b. 7

117.

a. 1; b. 1

119.

$\underset{x\rightarrow-3^{-}}{\text{lim}}\left( {f(x) - 3g(x)} \right) = \underset{x\rightarrow-3^{-}}{\text{lim}}f(x) - 3\underset{x\rightarrow-3^{-}}{\text{lim}}g(x) = 0 + 6 = 6$

121.

$\underset{x\rightarrow-5}{\text{lim}}\frac{2 + g(x)}{f(x)} = \frac{2 + \left( {\underset{x\rightarrow-5}{\text{lim}}g(x)} \right)}{\underset{x\rightarrow-5}{\text{lim}}f(x)} = \frac{2 + 0}{2} = 1$

123.

$\underset{x\rightarrow 1}{\text{lim}}\sqrt[3]{f(x) - g(x)} = \sqrt[3]{\underset{x\rightarrow 1}{\text{lim}}f(x) - \underset{x\rightarrow 1}{\text{lim}}g(x)} = \sqrt[3]{2 + 5} = \sqrt[3]{7}$

125.

$\underset{x\rightarrow-9}{\text{lim}}\left( {xf(x) + 2g(x)} \right) = \left( {\underset{x\rightarrow-9}{\text{lim}}x} \right)\left( {\underset{x\rightarrow-9}{\text{lim}}f(x)} \right) + 2\underset{x\rightarrow-9}{\text{lim}}\left( {g(x)} \right) = (-9)(6) + 2(4) = -46$

127.

The limit is zero.

129.

a.

b. ∞. The magnitude of the electric field as you approach the particle *q* becomes infinite. It does not make physical sense to evaluate negative distance.

Section 2.4 Exercises

131.

The function is defined for all *x* in the interval $\left( {0,\infty} \right).$

133.

Removable discontinuity at $x = 0;$ infinite discontinuity at $x = 1$

135.

Infinite discontinuity at $x = \text{ln}\mspace{2mu} 2$

137.

Infinite discontinuities at $x = \frac{\left( {2k + 1} \right)\pi}{4},$ for $k = 0, \pm 1, \pm 2, \pm 3\text{,…}$

139.

No. It is a removable discontinuity.

141.

Yes. It is continuous.

143.

Yes. It is continuous.

145.

$k = -5$

147.

$k = -1$

149.

$k = \frac{16}{3}$

151.

Since both *s* and $y = t$ are continuous everywhere, then $h(t) = s(t) - t$ is continuous everywhere and, in particular, it is continuous over the closed interval $\left\lbrack {2,5} \right\rbrack.$ Also, $h(2) = 3 > 0$ and $h(5) = -3 < 0.$ Therefore, by the IVT, there is a value $x = c$ such that $h(c) = 0.$

153.

The function $f(x) = 2^{x} - x^{3}$ is continuous over the interval $\left\lbrack {1.25,1.375} \right\rbrack$ and has opposite signs at the endpoints.

155.

a.

b. It is not possible to redefine $f(1)$ since the discontinuity is a jump discontinuity.

157.

Answers may vary; see the following example:

159.

Answers may vary; see the following example:

161.

False. It is continuous over $\left( {\text{−}\infty,0} \right) \cup \left( {0,\infty} \right).$

163.

False. Consider $f(x) = \left\{ \begin{array}{l}

{x\ \text{if}\ x \neq 0} \\

{4\ \text{if}\ x = 0}

\end{array} \right..$

165.

False. IVT only says that there is at least one solution; it does not guarantee that there is exactly one. Consider $f(x) = \text{cos}\mspace{2mu}(x)$ on $\left\lbrack {- \pi,2\pi} \right\rbrack.$

167.

False. The IVT does *not* work in reverse! Consider $\left( {x - 1} \right)^{2}$ over the interval $\left\lbrack {-2,2} \right\rbrack.$

169.

$R = 0.0001519\ \text{m}$

171.

$D = 345,826\ \text{km}$

173.

For all values of $a,{f(a)}$ is defined, $\underset{\theta\rightarrow a}{\text{lim}}f(\theta)$ exists, and $\underset{\theta\rightarrow a}{\text{lim}}f(\theta) = f(a).$ Therefore, $f(\theta)$ is continuous everywhere.

175.

Nowhere

Section 2.5 Exercises

177.

For every $\varepsilon > 0,$ there exists a $\delta > 0,$ so that if $0 < \left| {t - b} \right| < \delta,$ then $\left| {g(t) - M} \right| < \varepsilon$

179.

For every $\varepsilon > 0,$ there exists a $\delta > 0,$ so that if $0 < \left| {x - a} \right| < \delta,$ then $\left| {\varphi(x) - A} \right| < \varepsilon$

181.

$\delta \leq 0.25$

183.

$\delta \leq 2$

185.

$\delta \leq 1$

187.

$\delta < 0.3900$

189.

Let $\delta = \varepsilon.$ If $0 < \left| {x - 3} \right| < \varepsilon,$ then $\left| {x + 3 - 6} \right| = \left| {x - 3} \right| < \varepsilon.$

191.

Let $\delta = \sqrt[4]{\varepsilon}.$ If $0 < |x| < \sqrt[4]{\varepsilon},$ then $\left| x^{4} \right| = x^{4} < \varepsilon.$

193.

Let $\delta = \varepsilon^{2}.$ If $5 - \varepsilon^{2} < x < 5,$ then $\left| \sqrt{5 - x} \right| = \sqrt{5 - x} < \varepsilon.$

195.

Let $\delta = {\varepsilon\text{/}5}.$ If $1 - {\varepsilon\text{/}5} < x < 1,$ then $\left| {f(x) - 3} \right| = 5x - 5 < \varepsilon.$

197.

Let $\delta = \sqrt{\frac{3}{M}}.$ If $0 < \left| {x + 1} \right| < \sqrt{\frac{3}{M}},$ then $f(x) = \frac{3}{\left( {x + 1} \right)^{2}} > M.$

199.

The engineer must cut within 0.328 cm of 12 cm on each side; $\varepsilon = 8,\delta = 0.328,a = 12,L = 144$

201.

Answers may vary.

203.

0

205.

$\begin{array}{l}

{\lim\limits_{x\rightarrow a}\left( {f(x)} \right) + \lim\limits_{x\rightarrow a}\left( {g(x)} \right)} \\

{= L + M}

\end{array}$

207.

Answers may vary.

Review Exercises

209.

False

211.

False. A removable discontinuity is possible.

213.

5

215.

$8\text{/}7$

217.

DNE

219.

$2\text{/}3$

221.

−4;

223.

Since $-1 \leq \text{cos}(2\pi x) \leq 1,$ then $- x^{2} \leq x^{2}\text{cos}(2\pi x) \leq x^{2}.$ Since $\underset{x\rightarrow 0}{\text{lim}}x^{2} = 0 = \underset{x\rightarrow 0}{\text{lim}} - x^{2},$ it follows that $\underset{x\rightarrow 0}{\text{lim}}x^{2}\text{cos}\mspace{2mu}\left( {2\pi x} \right) = 0.$

225.

$\lbrack 2,\infty\rbrack$

227.

$c = -1$

229.

$\delta = \sqrt[3]{\varepsilon}$

231.

$0\ {\text{m}\text{/}\text{sec}}$

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• Book title: Calculus Volume 1
• Publication date: Mar 30, 2016
• Location: Houston, Texas
• Book URL: https://openstax.org/books/calculus-volume-1/pages/1-introduction
• Section URL: https://openstax.org/books/calculus-volume-1/pages/chapter-2

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