Review Exercises for the Math 151/161 Final Exam

Checkpoint

2.4

$\underset{\mathrm{ten}\to \mathrm{ane}}{\text{lim}}\frac{\frac{1}{x}-\mathrm{i}}{x-1}=\mathrm{-1}$

2.five

$\underset{x\to 2}{\text{lim}}h\left(x\right)=\mathrm{-i}.$

2.half dozen

$\underset{x\to 2}{\text{lim}}\frac{\left|x^2-\mathrm{four}\right|}{x-2}$ does not exist.

2.7

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

2.eight

a. $\underset{\mathrm{ten}\to 0^{-}}{\text{lim}}\frac{\mathrm{ane}}{x^2}=\text{+}\infty ;$ b. $\underset{\mathrm{10}\to 0^{+}}{\text{lim}}\frac{1}{x^2}=\text{+}\infty ;$ c. $\underset{\mathrm{10}\to 0}{\text{lim}}\frac{1}{x^2}=\text{+}\infty$

2.ix

a. $\underset{x\to {\mathrm{ii}}^{-}}{\text{lim}}\frac{1}{{\left(\mathrm{10}-\mathrm{ii}\right)}^3}=\text{−}\infty ;$ b. $\underset{\mathrm{ten}\to {\mathrm{ii}}^{+}}{\text{lim}}\frac{\mathrm{ane}}{{\left(x-\mathrm{two}\right)}^3}=\text{+}\infty ;$ c. $\underset{x\to \mathrm{ii}}{\text{lim}}\frac{1}{{\left(x-2\right)}^3}$ DNE. The line $\mathrm{10}=2$ is the vertical asymptote of $f\left(x\right)=\mathrm{one}\text{/}{\left(x-2\right)}^3.$

ii.17

$\underset{\mathrm{ten}\to {\mathrm{-1}}^{-}}{\text{lim}}f\left(x\right)=\mathrm{-one}$

2.21

f is non continuous at 1 because $f\left(1\right)=2\ne 3=\underset{x\to 1}{\text{lim}}f\left(\mathrm{ten}\right).$

2.22

$f\left(x\right)$ is continuous at every existent number.

2.23

Discontinuous at 1; removable

2.24

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

2.26

$f\left(0\right)=1>0,f\left(1\right)=\mathrm{-2}<0;f\left(x\right)$ is continuous over $\left[0,\mathrm{one}\right].$ It must have a goose egg on this interval.

two.27

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

Thus, $\left|\left(3\mathrm{10}-2\right)-4\right|=\left|\mathrm{three}x-6\right|=\left|3\right|·\left|x-2\right|<3·\delta =3·\left(\epsilon \text{/}3\right)=\epsilon .$

Therefore, $\underset{x\to \mathrm{two}}{\text{lim}}3\mathrm{10}-\mathrm{two}=4.$

ii.28

Choose $\delta =\text{min}\left\{\mathrm{nine}-{\left(3-\epsilon \right)}^2,{\left(\mathrm{three}+\epsilon \right)}^{\mathrm{ii}}-\mathrm{ix}\right\}.$

two.29

$\left|{\mathrm{ten}}^{\mathrm{ii}}-1\right|=\left|x-\mathrm{i}\right|·\left|\mathrm{ten}+\mathrm{ane}\right|<\epsilon \text{/}3·3=\epsilon$

Section two.1 Exercises

one .

a. 2.2100000; b. two.0201000; c. 2.0020010; d. 2.0002000; e. (ane.meg, 2.2100000); f. (1.0100000, two.0201000); g. (i.0010000, 2.0020010); h. (1.0001000, 2.0002000); i. 2.meg; j. two.0100000; k. 2.0010000; l. 2.0001000

7 .

a. 2.0248457; b. ii.0024984; c. two.0002500; d. 2.0000250; e. (4.1000000,ii.0248457); f. (4.0100000,2.0024984); k. (4.0010000,two.0002500); h. (iv.00010000,2.0000250); i. 0.24845673; j. 0.24984395; thousand. 0.24998438; l. 0.24999844

nine .

$y=\frac{x}{\mathrm{four}}+\mathrm{one}$

thirteen .

a. −0.95238095; b. −0.99009901; c. −0.99502488; d. −0.99900100; due east. (−1;.0500000,−0;.95238095); f. (−1;.0100000,−0;.9909901); g. (−1;.0050000,−0;.99502488); h. (ane.0010000,−0;.99900100); i. −0.95238095; j. −0.99009901; yard. −0.99502488; l. −0.99900100

15 .

$y=\text{−}x-2$

17 .

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

25 .

Under, ane unit²; over: 4 unit². The verbal area of the ii triangles is $\frac{\mathrm{i}}{2}\left(\mathrm{ane}\right)\left(\mathrm{i}\right)+\frac{\mathrm{one}}{2}\left(2\right)\left(2\right)=\mathrm{2.v}{\text{units}}^{\mathrm{two}}.$

27 .

Under, 0.96 unit²; over, ane.92 unit². The verbal area of the semicircle with radius one is $\frac{\pi {\left(\mathrm{i}\right)}^2}{2}=\frac{\pi }{2}$ unitⁱⁱ.

29 .

Approximately 1.3333333 unit of measurement²

Section 2.ii Exercises

31 .

$\underset{\mathrm{10}\to \mathrm{ane}}{\text{lim}}f\left(\mathrm{10}\right)$ does non exist because $\underset{x\to {\mathrm{i}}^{-}}{\text{lim}}f\left(x\right)=\mathrm{-2}\ne \underset{x\to {\mathrm{i}}^{+}}{\text{lim}}f\left(\mathrm{10}\right)=2.$

33 .

$\underset{\mathrm{ten}\to 0}{\text{lim}}{\left(1+x\right)}^{1\text{/}x}=\mathrm{ii.7183}$

35 .

a. 1.98669331; b. 1.99986667; c. i.99999867; d. 1.99999999; e. 1.98669331; f. ane.99986667; g. one.99999867; h. 1.99999999; $\underset{x\to 0}{\text{lim}}\frac{\text{sin}2x}{\mathrm{10}}=2$

37 .

$\underset{\mathrm{ten}\to 0}{\text{lim}}\frac{\text{sin}ax}{x}=a$

39 .

a. −0.80000000; b. −0.98000000; c. −0.99800000; d. −0.99980000; e. −1.2000000; f. −1.0200000; yard. −i.0020000; h. −one.0002000; $\underset{\mathrm{10}\to 1}{\text{lim}}\left(\mathrm{i}-\mathrm{ii}x\right)=\mathrm{-1}$

41 .

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

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; $\therefore \underset{x\to 2}{\text{lim}}\frac{1-\frac{2}{\mathrm{ten}}}{x^2-4}=0.1250=\frac{1}{8}$

45 .

a. 10.00000; b. 100.00000; c. yard.0000; d. ten,000.000; Guess: $\underset{\alpha \to 0^{+}}{\text{lim}}\frac{1}{\alpha }\text{cos}\left(\frac{\pi }{\alpha }\right)=\infty ,$ actual: DNE

47 .

False; $\underset{x\to {\mathrm{-2}}^{+}}{\text{lim}}f\left(x\right)=\text{+}\infty$

49 .

False; $\underset{\mathrm{ten}\to \mathrm{half\; dozen}}{\text{lim}}f\left(x\right)$ DNE since $\underset{x\to {\mathrm{six}}^{-}}{\text{lim}}f\left(x\right)=\mathrm{ii}$ and $\underset{\mathrm{ten}\to 6^{+}}{\text{lim}}f\left(x\right)=5.$

77 .

Answers may vary.

79 .

Answers may vary.

81 .

a. ${\rho }_2$ b. ${\rho }_1$ c. DNE unless ${\rho }_1={\rho }_2.$ Every bit you approach $x_{\text{SF}}$ from the left, you are in the loftier-density area of the shock. When you approach from the right, you have not experienced the "stupor" nevertheless and are at a lower density.

Section two.3 Exercises

83 .

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

85 .

Use root law: $\underset{x\to \mathrm{-two}}{\text{lim}}\sqrt{x^2-6x+3}=\sqrt{\underset{\mathrm{10}\to \mathrm{-ii}}{\text{lim}}\left(x^2-6\mathrm{ten}+3\right)}=\sqrt{\mathrm{nineteen}}$

93 .

$\underset{x\to 4}{\text{lim}}\frac{x^2-\mathrm{xvi}}{x-4}=\frac{16-16}{4-4}=\frac{0}{0};$ and so, $\underset{\mathrm{10}\to 4}{\text{lim}}\frac{x^2-16}{\mathrm{ten}-\mathrm{iv}}=\underset{x\to \mathrm{iv}}{\text{lim}}\frac{\left(x+4\right)\left(x-4\right)}{x-4}=8$

95 .

$\underset{\mathrm{10}\to 6}{\text{lim}}\frac{3x-18}{2x-12}=\frac{18-18}{12-12}=\frac{0}{0};$ so, $\underset{x\to 6}{\text{lim}}\frac{3\mathrm{10}-18}{2\mathrm{ten}-12}=\underset{x\to 6}{\text{lim}}\frac{3\left(x-\mathrm{vi}\right)}{2\left(\mathrm{ten}-\mathrm{six}\right)}=\frac{3}{\mathrm{two}}$

97 .

$\underset{\mathrm{ten}\to 9}{\text{lim}}\frac{t-\mathrm{nine}}{\sqrt{t}-3}=\frac{9-9}{\mathrm{three}-\mathrm{three}}=\frac{0}{0};$ then, $\underset{t\to 9}{\text{lim}}\frac{t-9}{\sqrt{t}-3}=\underset{t\to 9}{\text{lim}}\frac{t-9}{\sqrt{t}-3}\frac{\sqrt{t}+3}{\sqrt{t}+3}=\underset{t\to \mathrm{ix}}{\text{lim}}\left(\sqrt{t}+3\right)=6$

99 .

$\underset{\theta \to \pi }{\text{lim}}\frac{\text{sin}\theta }{\text{tan}\theta }=\frac{\text{sin}\pi }{\text{tan}\pi }=\frac{0}{0};$ then, $\underset{\theta \to \pi }{\text{lim}}\frac{\text{sin}\theta }{\text{tan}\theta }=\underset{\theta \to \pi }{\text{lim}}\frac{\text{sin}\theta }{\frac{\text{sin}\theta }{\text{cos}\theta }}=\underset{\theta \to \pi }{\text{lim}}\text{cos}\theta =\mathrm{-one}$

101 .

$\underset{x\to 1\text{/}2}{\text{lim}}\frac{2{\mathrm{ten}}^2+\mathrm{iii}x-\mathrm{two}}{2\mathrm{10}-\mathrm{one}}=\frac{\frac{1}{\mathrm{ii}}+\frac{3}{2}-2}{1-1}=\frac{0}{0};$ and then, $\underset{x\to 1\text{/}2}{\text{lim}}\frac{\mathrm{ii}{x}^2+3x-2}{2x-\mathrm{one}}=\underset{x\to 1\text{/}2}{\text{lim}}\frac{\left(2x-1\right)\left(x+\mathrm{two}\right)}{2\mathrm{10}-\mathrm{i}}=\frac{5}{2}$

107 .

$\underset{x\to 6}{\text{lim}}2f\left(\mathrm{10}\right)m\left(x\right)=2\underset{\mathrm{10}\to 6}{\text{lim}}f\left(x\right)\underset{x\to 6}{\text{lim}}\mathrm{1000}\left(x\right)=72$

109 .

$\underset{\mathrm{ten}\to 6}{\text{lim}}\left(f\left(x\right)+\frac{\mathrm{one}}{\mathrm{iii}}\mathrm{thousand}\left(x\right)\right)=\underset{x\to 6}{\text{lim}}f\left(x\right)+\frac{1}{3}\underset{\mathrm{ten}\to 6}{\text{lim}}g\left(x\right)=7$

111 .

$\underset{x\to 6}{\text{lim}}\sqrt{\mathrm{yard}\left(\mathrm{10}\right)-f\left(\mathrm{ten}\right)}=\sqrt{\underset{x\to 6}{\text{lim}}g\left(\mathrm{ten}\right)-\underset{\mathrm{10}\to 6}{\text{lim}}f\left(x\right)}=\sqrt{5}$

113 .

$\underset{x\to 6}{\text{lim}}\left[\left(\mathrm{10}+1\right)f\left(\mathrm{ten}\right)\right]=\left(\underset{x\to \mathrm{half-dozen}}{\text{lim}}\left(\mathrm{10}+\mathrm{i}\right)\right)\left(\underset{\mathrm{10}\to \mathrm{half\; dozen}}{\text{lim}}f\left(x\right)\right)=28$

115 .

a. 9; b. vii

117 .

a. 1; b. ane

119 .

$\underset{x\to {\mathrm{-iii}}^{-}}{\text{lim}}\left(f\left(x\right)-3g\left(x\right)\right)=\underset{x\to {\mathrm{-3}}^{-}}{\text{lim}}f\left(\mathrm{ten}\right)-3\underset{\mathrm{10}\to {\mathrm{-iii}}^{-}}{\text{lim}}\mathrm{grand}\left(x\right)=0+\mathrm{six}=\mathrm{vi}$

121 .

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

123 .

$\underset{x\to 1}{\text{lim}}\sqrt[3]{f\left(\mathrm{10}\right)-\mathrm{thou}\left(x\right)}=\sqrt[\mathrm{three}]{\underset{\mathrm{10}\to \mathrm{i}}{\text{lim}}f\left(\mathrm{ten}\right)-\underset{\mathrm{ten}\to 1}{\text{lim}}\mathrm{thousand}\left(x\right)}=\sqrt[3]{\mathrm{ii}+5}=\sqrt[\mathrm{three}]{7}$

125 .

$\underset{x\to \mathrm{-9}}{\text{lim}}\left(xf\left(x\right)+2k\left(\mathrm{ten}\right)\right)=\left(\underset{x\to \mathrm{-9}}{\text{lim}}x\right)\left(\underset{x\to \mathrm{-9}}{\text{lim}}f\left(x\right)\right)+2\underset{x\to \mathrm{-nine}}{\text{lim}}\left(m\left(x\right)\right)=\left(\mathrm{-ix}\right)\left(6\right)+\mathrm{two}\left(4\right)=\mathrm{-46}$

127 .

The limit is nil.

129 .

a.

b. ∞. The magnitude of the electrical field equally yous approach the particle q becomes infinite. Information technology does not make physical sense to evaluate negative distance.

Department ii.4 Exercises

131 .

The office is divers for all x in the interval $\left(0,\infty \right).$

133 .

Removable discontinuity at $\mathrm{10}=0;$ infinite aperture at $\mathrm{10}=\mathrm{one}$

135 .

Infinite discontinuity at $\mathrm{10}=\text{ln}\mathrm{two}$

137 .

Space discontinuities at $\mathrm{ten}=\frac{\left(\mathrm{two}k+1\right)\pi }{4},$ for $k=0,\pm \mathrm{i},\pm 2,\pm 3\text{,…}$

139 .

No. Information technology is a removable discontinuity.

141 .

Yes. Information technology is continuous.

143 .

Yes. It is continuous.

151 .

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

153 .

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

155 .

a.

b. It is not possible to redefine $f\left(\mathrm{one}\right)$ since the aperture is a jump aperture.

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 .

Fake. Consider $f\left(x\right)=\{\begin{array}{l}x\text{if}x\ne 0\\ 4\text{if}x=0\end{array}.$

165 .

False. IVT only says that there is at least one solution; it does not guarantee that there is exactly one. Consider $f\left(\mathrm{ten}\right)=\text{cos}\left(\mathrm{ten}\right)$ on $\left[-\pi ,\mathrm{ii}\pi \right].$

167 .

Faux. The IVT does not work in contrary! Consider ${\left(x-1\right)}^2$ over the interval $\left[\mathrm{-2},\mathrm{ii}\right].$

169 .

$R=0.0001519\text{m}$

171 .

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

173 .

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

Department 2.5 Exercises

177 .

For every $\epsilon >0,$ in that location exists a $\delta >0,$ then that if $0<\left|t-b\right|<\delta ,$ then $\left|\mathrm{one\; thousand}\left(t\right)-M\right|<\epsilon$

179 .

For every $\epsilon >0,$ there exists a $\delta >0,$ so that if $0<\left|\mathrm{10}-a\right|<\delta ,$ then $\left|\phi \left(x\right)-A\right|<\epsilon$

187 .

$\delta <0.3900$

189 .

Permit $\delta =\epsilon .$ If $0<\left|x-3\right|<\epsilon ,$ so $\left|x+3-6\right|=\left|x-3\right|<\epsilon .$

191 .

Allow $\delta =\sqrt[4]{\epsilon }.$ If $0<\left|\mathrm{10}\right|<\sqrt[\mathrm{four}]{\epsilon },$ so $\left|x^{\mathrm{four}}\right|=x^{\mathrm{four}}<\epsilon .$

193 .

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

195 .

Allow $\delta =\epsilon \text{/}\mathrm{v}.$ If $1-\epsilon \text{/}\mathrm{v}<\mathrm{10}<\mathrm{one},$ so $\left|f\left(x\right)-3\right|=5x-\mathrm{v}<\epsilon .$

197 .

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

199 .

0.328 cm, $\epsilon =8,\delta =0.33,a=12,L=144$

205 .

$f\left(x\right)-g\left(x\right)=f\left(x\right)+\left(\mathrm{-1}\right)g\left(x\right)$

Review Exercises

211 .

Faux. A removable discontinuity is possible.

223 .

Since $\mathrm{-i}\le \text{cos}(2\pi \mathrm{ten})\le 1,$ then $-{\mathrm{10}}^2\le {\mathrm{10}}^{\mathrm{ii}}\text{cos}(\mathrm{ii}\pi x)\le x^2.$ Since $\underset{x\to 0}{\text{lim}}{\mathrm{ten}}^{\mathrm{ii}}=0=\underset{\mathrm{ten}\to 0}{\text{lim}}-{\mathrm{ten}}^2,$ it follows that $\underset{x\to 0}{\text{lim}}{x}^2\text{cos}\left(\mathrm{two}\pi \mathrm{ten}\right)=0.$

225 .

$[2,\infty ]$

231 .

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

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Source: https://openstax.org/books/calculus-volume-1/pages/chapter-2

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