Chapter 2

2.25

12.006001

17 unit²

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

$\underset{x\to 2}{\text{lim}}h\left(x\right)=\mathrm{-1}.$

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

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

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

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

Does not exist.

$11\sqrt{10}$

−13;

$\frac{1}{3}$

$\frac{1}{4}$

−1;

$\frac{1}{4}$

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

+∞

0

0

f is not continuous at 1 because $f\left(1\right)=2\ne 3=\underset{x\to 1}{\text{lim}}f\left(x\right).$

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

Discontinuous at 1; removable

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

0

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

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

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

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

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

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

$\delta ={\epsilon }^2$

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

$y=2x$

3

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

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

π

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

$y=\text{−}x-2$

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

5.2 m/sec

−9.8 m/sec

6 m/sec

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

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

Approximately 1.3333333 unit²

$\underset{x\to 1}{\text{lim}}f\left(x\right)$ does not exist because $\underset{x\to 1^{-}}{\text{lim}}f\left(x\right)=\mathrm{-2}\ne \underset{x\to 1^{+}}{\text{lim}}f\left(x\right)=2.$

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

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

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

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

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 $\underset{x\to 0}{\text{lim}}\frac{z-1}{z^2\left(z+3\right)}=\text{−}\infty$

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

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

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

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

2

1

1

DNE

0

DNE

2

3

DNE

0

−2

DNE

0

Answers may vary.

Answers may vary.

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.

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

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

49

1

$-\frac{5}{7}$

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

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

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

$\underset{\theta \to \pi }{\text{lim}}\frac{\text{sin}\theta }{\text{tan}\theta }=\frac{\text{sin}\pi }{\text{tan}\pi }=\frac{0}{0};$ then, $\underset{\theta \to \pi }{\text{lim}}\frac{\text{sin}\theta }{\text{tan}\theta }=\underset{\theta \to \pi }{\text{lim}}\frac{\text{sin}\theta }{\frac{\text{sin}\theta }{\text{cos}\theta }}=\underset{\theta \to \pi }{\text{lim}}\text{cos}\theta =\mathrm{-1}$

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

−∞

−∞

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

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

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

$\underset{x\to 6}{\text{lim}}\left[\left(x+1\right)f\left(x\right)\right]=\left(\underset{x\to 6}{\text{lim}}\left(x+1\right)\right)\left(\underset{x\to 6}{\text{lim}}f\left(x\right)\right)=28$

a. 9; b. 7

a. 1; b. 1

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

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

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

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

The limit is zero.

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.

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

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

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

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

No. It is a removable discontinuity.

Yes. It is continuous.

Yes. It is continuous.

$k=\mathrm{-5}$

$k=\mathrm{-1}$

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

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, $h\left(2\right)=3>0$ and $h\left(5\right)=\mathrm{-3}<0.$ Therefore, by the IVT, there is a value $x=c$ such that $h\left(c\right)=0.$

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

a.

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

Answers may vary; see the following example:

Answers may vary; see the following example:

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

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

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

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

$R=0.0001519\text{m}$

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

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.

Nowhere

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

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

$\delta \le 0.25$

$\delta \le 2$

$\delta \le 1$

$\delta <0.3900$

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

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

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

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

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

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

Answers may vary.

0

$\begin{array}{l}\underset{x\to a}{\lim }\left(f\left(x\right)\right)+\underset{x\to a}{\lim }\left(g\left(x\right)\right)\\ =L+M\end{array}$

Answers may vary.

False

False. A removable discontinuity is possible.

5

$8\text{/}7$

DNE

$2\text{/}3$

−4;

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

$[2,\infty ]$

$c=\mathrm{-1}$

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

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