Solutions To Selected Problems Solutions To Selected Problems Chapter B

Chapter A

Exercises A.0

1.

$(-\infty,\infty)$
2.

$[-7,\infty)$
3.

$(-\infty,-1]\cup[7,\infty)$
4.

$(-\infty,\infty)$
5.

$(-\infty,2)\cup(2,\infty)$
6.

$(-\infty,\infty)$
7.

$(-\infty,\infty)$
8.

$(-\infty,-2)\cup(2,\infty)$
9.

$(-\infty,0)\cup(0,\infty)$
10.

$(-\infty,\infty)$
11.
12.
13.
14.
15.
- 14
  
  11
  
  $3a^{2}-2a+6$
  
  $3(x+h)^{2}-2(x+h)+6$
  
  $\dfrac{h(3h+6x-2)}{h}$
16.
- $\sqrt{2}$
  
  undefined
  
  $\sqrt{t-2}$
  
  $\sqrt{x+h-2}$
  
  $\dfrac{\sqrt{x+h-2}-\sqrt{x-2}}{h}\\ =\dfrac{h}{h(\sqrt{x+h-2}+\sqrt{x-2})}$
17.
- $-1$
  
  $\dfrac{1}{9}$
  
  $\dfrac{1}{t+3}$
  
  $\dfrac{1}{x+h}$
  
  $\dfrac{\frac{1}{x+h}-\frac{1}{x}}{h}=-\dfrac{h}{hx(x+h)}$
18.
- $e^{-2}$
  
  $e^{3}$
  
  $e^{t+1}$
  
  $e^{x+h}$
  
  $\dfrac{e^{x+h}-e^{x}}{h}=e^{x}\dfrac{e^{h}-1}{h}$
19.

$\{2.12\}\cup[2.13,2.14)\cup(2.15,\infty)$
20.

$(-\infty,5.677]\cup[5.678,5.679)\cup(5.679,\infty)$
21.

$(0.3,0.8]$
22.

$(0.1,0.2]$

Exercises A.1

1.

Answers will vary.
2.

An indeterminate form.
3.

F
4.

The function may approach different values from the left and right, the function may grow without bound, or the function might oscillate.
5.

$1$
6.

Answers will vary.
7.

$-1$
8.

$-5$
9.

Limit does not exist
10.

$2$
11.

$1.5$
12.

Limit does not exist.
13.

Limit does not exist.
14.

$7$
15.

$1$
16.

Limit does not exist.
17.

$h$ $\frac{f(a+h)-f(a)}{h}$

$-0.1$ $-7$

$-0.01$ $-7$

$0.01$ $-7$

$0.1$ $-7$

The limit seems to be exactly $-7$ .
18.

$h$ $\frac{f(a+h)-f(a)}{h}$

$-0.1$ $9$

$-0.01$ $9$

$0.01$ $9$

$0.1$ $9$

The limit seems to be exactly 9.
19.

$h$ $\frac{f(a+h)-f(a)}{h}$

$-0.1$ $4.9$

$-0.01$ $4.99$

$0.01$ $5.01$

$0.1$ $5.1$

The limit is approx. 5.
20.

$h$ $\frac{f(a+h)-f(a)}{h}$

$-0.1$ $-0.114943$

$-0.01$ $-0.111483$

$0.01$ $-0.110742$

$0.1$ $-0.107527$

The limit is approx. $-0.11$ .
21.

$h$ $\frac{f(a+h)-f(a)}{h}$

$-0.1$ $29.4$

$-0.01$ $29.04$

$0.01$ $28.96$

$0.1$ $28.6$

The limit is approx. $29$ .
22.

$h$ $\frac{f(a+h)-f(a)}{h}$

$-0.1$ $0.202027$

$-0.01$ $0.2002$

$0.01$ $0.1998$

$0.1$ $0.198026$

The limit is approx. $0.2$ .
23.

$h$ $\frac{f(a+h)-f(a)}{h}$

$-0.1$ $-0.998334$

$-0.01$ $-0.999983$

$0.01$ $-0.999983$

$0.1$ $-0.998334$

The limit is approx. $-1$ .
24.

$h$ $\frac{f(a+h)-f(a)}{h}$

$-0.1$ $-0.0499583$

$-0.01$ $-0.00499996$

$0.01$ $0.00499996$

$0.1$ $0.0499583$

The limit is approx. $0$ .
25.

$h$ $\frac{f(a+h)-f(a)}{h}$

$-0.1$ $0.251582$

$-0.01$ $0.250156$

$0.01$ $0.249844$

$0.1$ $0.248457$

The limit is approx. $0.25$ .
26.

$h$ $\frac{f(a+h)-f(a)}{h}$

$-0.1$ $2.58$

$-0.01$ $2.70$

$0.01$ $2.73$

$0.1$ $2.85$

The limit is approx. $2.7$ .

Exercises A.2

1.

$\epsilon$ should be given first, and the restriction $\left\lvert x-a\right\rvert<\delta$ implies $\left\lvert f(x)-K\right\rvert<\epsilon$ , not the other way around.
2.

The $y$ -tolerance.
3.

T
4.

T
5.

$\delta\leq 0.45$
6.

$\delta\leq 0.71$
7.

Given $\epsilon>0$ , choose $\delta=\frac{\epsilon}{2}$ .

$\displaystyle\left\lvert x-4\right\rvert<\delta=\frac{\epsilon}{2}$

$\displaystyle\left\lvert 2x-8\right\rvert<\epsilon$

$\displaystyle\left\lvert(2x+5)-(13)\right\rvert<\epsilon.$

Thus $\displaystyle\lim_{x\to 4}(2x+5)=13$ .
8.

Given $\epsilon>0$ , choose $\delta=\epsilon$ .

$\displaystyle\left\lvert x-5\right\rvert<\delta=\epsilon$

$\displaystyle\left\lvert 5-x\right\rvert<\epsilon$

$\displaystyle\left\lvert 3-x-(-2)\right\rvert<\epsilon.$

Thus $\displaystyle\lim_{x\to 5}(3-x)=-2$ .
9.

Given $\epsilon>0$ , let $\delta=\dfrac{\epsilon}{4}$ . Then:

$\displaystyle\left\lvert x-5\right\rvert$ $\displaystyle<\delta=\frac{\epsilon}{4}$

$\displaystyle 4\left\lvert x-5\right\rvert$ $\displaystyle<\frac{\epsilon}{4}\cdot 4$

$\displaystyle\left\lvert 4x-20\right\rvert$ $\displaystyle<\epsilon$

$\displaystyle\left\lvert 4x-12-8\right\rvert$ $\displaystyle<\epsilon$

Thus $\displaystyle\lim_{x\to 5}(4x-12)=8$ .
10.

Given $\epsilon>0$ , let $\delta=\dfrac{\epsilon}{2}$ . Then:

$\displaystyle\left\lvert x-3\right\rvert$ $\displaystyle<\delta=\frac{\epsilon}{2}$

$\displaystyle 2\left\lvert x-3\right\rvert$ $\displaystyle<\frac{\epsilon}{2}\cdot 2$

$\displaystyle\left\lvert-2x+6\right\rvert$ $\displaystyle<\epsilon$

$\displaystyle\left\lvert 5-2x+1\right\rvert$ $\displaystyle<\epsilon$

Thus $\displaystyle\lim_{x\to 3}(5-2x)=-1$ .
11.

Given $\epsilon>0$ , let $\delta=\frac{\epsilon}{7}$ .

$\displaystyle\left\lvert x-3\right\rvert<\delta=\frac{\epsilon}{7}$

$\displaystyle\left\lvert x-3\right\rvert<\frac{\epsilon}{x+3}$

$\displaystyle\left\lvert x-3\right\rvert\cdot\left\lvert x+3\right\rvert<\frac% {\epsilon}{x+3}\cdot\left\lvert x+3\right\rvert$

Assuming $x$ is near 3, $x+3$ is positive and we can drop the absolute value signs on the right.

$\displaystyle\left\lvert x-3\right\rvert\cdot\left\lvert x+3\right\rvert<\frac% {\epsilon}{x+3}\cdot(x+3)$

$\displaystyle\left\lvert x^{2}-9\right\rvert<\epsilon$

$\displaystyle\left\lvert(x^{2}-3)-6\right\rvert<\epsilon.$

Thus, $\displaystyle\lim_{x\to 3}\left(x^{2}-3\right)=6$ .
12.

Given $\epsilon>0$ , let $\delta=\frac{\epsilon}{10}$ . Then:

$\displaystyle\left\lvert x-4\right\rvert<\delta=\frac{\epsilon}{10}$

$\displaystyle\left\lvert x-4\right\rvert<\frac{\epsilon}{x+5}$

$\displaystyle\left\lvert x-4\right\rvert\cdot\left\lvert x+5\right\rvert<\frac% {\epsilon}{x+5}\cdot\left\lvert x+5\right\rvert$

Assuming $x$ is near 4, $x+5$ is positive and we can drop the absolute value signs on the right.

$\displaystyle\left\lvert x-4\right\rvert\cdot\left\lvert x+5\right\rvert<\frac% {\epsilon}{x+5}\cdot(x+5)$

$\displaystyle\left\lvert x^{2}+x-20\right\rvert<\epsilon$

$\displaystyle\left\lvert(x^{2}+x-5)-15\right\rvert<\epsilon.$

Thus, $\displaystyle\lim_{x\to 4}\left(x^{2}+x-5\right)=15$ .
13.

Given $\epsilon>0$ , let $\delta=\min\{1,\frac{\epsilon}{9}\}$ . Then:

$\displaystyle\left\lvert x-1\right\rvert<\delta$

$\displaystyle\left\lvert x-1\right\rvert<\frac{\epsilon}{9}$

$\displaystyle\left\lvert x-1\right\rvert<\frac{\epsilon}{2x+5}$

$\displaystyle\left\lvert x-1\right\rvert\cdot\left\lvert 2x+5\right\rvert<% \frac{\epsilon}{2x+5}\cdot\left\lvert 2x+5\right\rvert=\epsilon$

$\displaystyle\left\lvert 2x^{2}+3x-5\right\rvert<\epsilon$

$\displaystyle\left\lvert(2x^{2}+3x+1)-6\right\rvert<\epsilon.$

Thus, $\displaystyle\lim_{x\to 1}\left(2x^{2}+3x+1\right)=6$ .
14.

Given $\epsilon>0$ , let $\delta=\frac{\epsilon}{19}$ .

$\displaystyle\left\lvert x-2\right\rvert<\delta=\frac{\epsilon}{19}$

$\displaystyle\left\lvert x-2\right\rvert<\frac{\epsilon}{x^{2}+2x+4}$

$\displaystyle\left\lvert x-2\right\rvert\cdot\left\lvert x^{2}+2x+4\right% \rvert<\frac{\epsilon}{x^{2}+2x+4}\cdot\left\lvert x^{2}+2x+4\right\rvert$

Assuming $x$ is near 2, $x^{2}+2x+4$ is positive and we can drop the absolute value signs on the right.

$\displaystyle\left\lvert x-2\right\rvert\cdot\left\lvert x^{2}+2x+4\right% \rvert<\frac{\epsilon}{x^{2}+2x+4}\cdot(x^{2}+2x+4)$

$\displaystyle\left\lvert x^{3}-8\right\rvert<\epsilon$

$\displaystyle\left\lvert(x^{3}-1)-7\right\rvert<\epsilon,$

which is what we wanted to prove.
15.

Let $\epsilon>0$ be given. We wish to find $\delta>0$ such that when $\left\lvert x-2\right\rvert<\delta$ , $\left\lvert f(x)-5\right\rvert<\epsilon$ . However, since $f(x)=5$ , a constant function, the latter inequality is simply $\left\lvert 5-5\right\rvert<\epsilon$ , which is always true. Thus we can choose any $\delta$ we like; we arbitrarily choose $\delta=\epsilon$ .
16.

Given $\epsilon>0$ , let $\delta=\frac{\ln(1+\epsilon)}{2}<\left\lvert\frac{\ln(1-\epsilon)}{2}\right\rvert$ .

$\displaystyle\left\lvert x\right\rvert<\delta$

$\displaystyle\left\lvert x\right\rvert<\frac{\ln(1+\epsilon)}{2}<\left\lvert% \frac{\ln(1-\epsilon)}{2}\right\rvert$

$\displaystyle\frac{\ln(1-\epsilon)}{2}<x<\frac{\ln(1+\epsilon)}{2}$

$\displaystyle\ln(1-\epsilon)<2x<\ln(1+\epsilon)$

$\displaystyle 1-\epsilon<e^{2x}<1+\epsilon$

$\displaystyle-\epsilon<e^{2x}-1<\epsilon$

$\displaystyle\left\lvert e^{2x}-1-(0)\right\rvert<\epsilon,$

which is what we wanted to prove.
17.

Given $\epsilon>0$ , let $\delta=\min\{\frac{1}{2},\frac{\epsilon}{2}\}$ . Then:

$\displaystyle\left\lvert x-1\right\rvert<\delta$

$\displaystyle\left\lvert x-1\right\rvert<\frac{\epsilon}{2}$

$\displaystyle\left\lvert x-1\right\rvert<\epsilon\cdot x$

$\displaystyle\left\lvert x-1\right\rvert/x<\epsilon$

$\displaystyle\left\lvert(x-1)/x\right\rvert<\epsilon$

$\displaystyle\left\lvert 1-1/x\right\rvert<\epsilon$

$\displaystyle\left\lvert(1/x)-1\right\rvert<\epsilon,$

which is what we wanted to prove.
18.

Let $\epsilon>0$ be given. We wish to find $\delta>0$ such that when $\left\lvert x-0\right\rvert<\delta$ , $\left\lvert f(x)-0\right\rvert<\epsilon$ . In simpler terms, we want to show that when $\left\lvert x\right\rvert<\delta$ , $\left\lvert\sin x\right\rvert<\epsilon$ .

Set $\delta=\epsilon$ . We start with assuming that $\left\lvert x\right\rvert<\delta$ . Using the hint, we have that $\left\lvert\sin x\right\rvert<\left\lvert x\right\rvert<\delta=\epsilon$ . Hence if $\left\lvert x\right\rvert<\delta$ , we know immediately that $\left\lvert\sin x\right\rvert<\epsilon$ .

Exercises A.3

1.

Answers will vary.
2.

Answers will vary.
3.

Answers will vary.
4.

Answers will vary.
5.

As $x$ is near 1, both $f$ and $g$ are near 0, but $f$ is approximately twice the size of $g$ . (I.e., $f(x)\approx 2g(x)$ .)
6.

T; by Theorem 1.3.4, $\lim_{x\to 1}\ln x=\ln 1=0$ .
7.

9
8.

6
9.

0
10.

Limit does not exist.
11.

3
12.

Not possible to know.
13.

3
14.

$-45$
15.

$1$
16.

$-1$
17.

$0$
18.

$\pi$
19.

$7$
20.

$-0.000000015\approx 0$
21.

$1/2$
22.

$0$
23.

Limit does not exist
24.

$64$
25.

$2$
26.

$0$
27.

$\frac{\pi^{2}+3\pi+5}{5\pi^{2}-2\pi-3}\approx 0.6064$
28.

$\frac{3\pi+1}{1-\pi}$
29.

$-8$
30.

$-1$
31.

$10$
32.

$-2$
33.

$-3/2$
34.

$-7/8$
35.

$1/6$
36.

$1/4$
37.

$-1/9$
38.

$1$
39.

$-8$
40.

$1/8$
41.

$0$
42.

$0$
43.

$1$
44.

$9$
45.

$0$
46.

$1$
47.

$3$
48.

$5/8$
49.

$1$
50.

$\pi/180$
51.

$4/3$
52.

$5/7$
53.
- Apply Part 1 of Theorem 1.3.1.
  
  Apply Theorem 1.3.7; $g(x)=\frac{x}{x}$ is the same as $g(x)=1$ everywhere except at $x=0$ . Thus $\displaystyle\lim_{x\to 0}g(x)=\lim_{x\to 0}1=1.$
  
  The function $f(x)$ is always 0, so $g\big(f(x)\big)$ is never defined as $g(x)$ is not defined at $x=0$ . Therefore the limit does not exist.
  
  The theorem requires that $\displaystyle\lim_{x\to 0}g(x)$ be equal to $g(0)$ . They are not equal, so the conditions of the theorem are not satisfied, and hence the theorem is not violated.
54.

We find $\displaystyle\lim_{x\to 0}\frac{\cos^{2}x-1}{x(\cos x+1)}=\lim_{x\to 0}\frac{% \sin^{2}x}{x(\cos x+1)}=\lim_{x\to 0}\frac{\sin x}{x}\lim_{x\to 0}\frac{\sin x% }{\cos x+1}=0$ .
55.
- $\pm\sqrt{-\frac{c}{a}}$
  
  0 and $-\frac{b}{a}$
  
  $-\frac{c}{b}$ and undefined

Exercises A.4

1.

The function approaches different values from the left and right; the function grows without bound; the function oscillates.
2.

F
3.

F
4.

T
5.
- 2
  
  2
  
  2
  
  1
  
  As $f$ is not defined for $x<0$ , this limit is not defined.
  
  1
6.
- 1
  
  2
  
  Does not exist.
  
  2
  
  0
  
  As $f$ is not defined for $x>2$ , this limit is not defined.
7.
- 2
  
  2
  
  2
  
  2
8.
- 4
  
  $-4$
  
  Does not exist.
  
  0
9.
- 2
  
  2
  
  2
  
  0
  
  2
  
  2
  
  2
  
  Not defined
10.
- $a-1$
  
  $a$
  
  Does not exist.
  
  $a$
11.

DNE
12.

0
13.

DNE
14.

DNE
15.
- 2
  
  $-4$
  
  Does not exist.
  
  2
16.
- $-1$
  
  0
  
  Does not exist.
  
  0
17.
- 0
  
  0
  
  0
  
  0
  
  2
  
  2
  
  2
  
  2
18.
- $-1$
  
  0
  
  Does not exist.
  
  0
19.
- $1-\cos^{2}a=\sin^{2}a$
  
  $\sin^{2}a$
  
  $\sin^{2}a$
  
  $\sin^{2}a$
20.
- 2
  
  0
  
  Does not exist
  
  1
21.
- 4
  
  4
  
  4
  
  3
22.
- $c$
  
  $c$
  
  $c$
  
  $c$
23.
- $-1$
  
  $1$
  
  Does not exist
  
  $0$
24.
- $-1$
  
  $1$
  
  Does not exist.
  
  Undefined
25.

Answers will vary.
26.

Answers will vary.
27.

Answers will vary.
28.

Answers will vary.
29.

$-3/5$
30.

$2/3$
31.

$\frac{1}{2\sqrt{3}}$
32.

$2$
33.

$-1.63$

Exercises A.5

1.

F
2.

T
3.

F
4.

T
5.

F
6.

Answers will vary.
7.

Answers will vary.
8.

The limit of $f$ as $x$ approaches 7 does not exist, hence $f$ is not continuous. (Note: $f$ could be defined at 7!)
9.
- $\infty$
  
  $\infty$
10.
- $-\infty$
  
  $\infty$
  
  Limit does not exist
  
  $\infty$
  
  $\infty$
  
  $\infty$
11.
- $1$
  
  $0$
  
  $1/2$
  
  $1/2$
12.
- Limit does not exist
  
  Limit does not exist
13.
- Limit does not exist
  
  Limit does not exist
14.
- $10$
  
  $\infty$
15.
Tables will vary.
- $x$ $f(x)$
  
  $2.9$ $-15.1224$
  
  $2.99$ $-159.12$
  
  $2.999$ $-1599.12$
  
  It seems $\lim_{x\to 3^{-}}f(x)=-\infty$ .
  
  $x$ $f(x)$
  
  $3.1$ $16.8824$
  
  $3.01$ $160.88$
  
  $3.001$ $1600.88$
  
  It seems $\lim_{x\to 3^{+}}f(x)=\infty$ .
  
  It seems $\lim_{x\to 3}f(x)$ does not exist.
16.
Tables will vary.
- $x$ $f(x)$
  
  $2.9$ $-335.64$
  
  $2.99$ $-30350.6$
  
  It seems $\lim_{x\to 3^{-}}f(x)=-\infty$ .
  
  $x$ $f(x)$
  
  $3.1$ $-265.61$
  
  $3.01$ $-29650.6$
  
  It seems $\lim_{x\to 3^{+}}f(x)=-\infty$ .
  
  It seems $\lim_{x\to 3}f(x)=-\infty$ .
17.
Tables will vary.
- $x$ $f(x)$
  
  $2.9$ $132.857$
  
  $2.99$ $12124.4$
  
  It seems $\lim_{x\to 3^{-}}f(x)=\infty$ .
  
  $x$ $f(x)$
  
  $3.1$ $108.039$
  
  $3.01$ $11876.4$
  
  It seems $\lim_{x\to 3^{+}}f(x)=\infty$ .
  
  It seems $\lim_{x\to 3}f(x)=\infty$ .
18.
Tables will vary.
- $x$ $f(x)$
  
  $2.9$ $-0.632$
  
  $2.99$ $-0.6032$
  
  $2.999$ $-0.60032$
  
  It seems $\lim_{x\to 3^{-}}f(x)=-0.6$ .
  
  $x$ $f(x)$
  
  $3.1$ $-0.5686$
  
  $3.01$ $-0.5968$
  
  $3.001$ $-0.59968$
  
  It seems $\lim_{x\to 3^{+}}f(x)=-0.6$ .
  
  It seems $\lim_{x\to 3}f(x)=-0.6$ .
19.

Horizontal asymptote at $y=2$ ; vertical asymptotes at $x=-5,4$ .
20.

Horizontal asymptote at $y=-3/5$ ; vertical asymptote at $x=3$ .
21.

Horizontal asymptote at $y=0$ ; vertical asymptotes at $x=-1,0$ .
22.

No horizontal asymptote; vertical asymptote at $x=1$ .
23.

No horizontal or vertical asymptotes.
24.

Horizontal asymptote at $y=-1$ ; no vertical asymptotes
25.

$y=2$
26.

$y=3$ and $y=-3$
27.

$\infty$
28.

$-\infty$
29.

$-\infty$
30.

$\infty$
31.

$-2/3$
32.

$-1$
33.

$-\sqrt{10}/2$
34.

$1/2$
35.

Let $\epsilon>0$ be given. We wish to find $\delta>0$ such that when $\left\lvert x-1\right\rvert<\delta$ , $\left\lvert f(x)-3\right\rvert<\epsilon$ .

Scratch-Work: Consider $\left\lvert f(x)-3\right\rvert<\epsilon$ , keeping in mind we want to make a statement about $\left\lvert x-1\right\rvert$ :

$\displaystyle\left\lvert f(x)-3\right\rvert$ $\displaystyle<\epsilon$

$\displaystyle\left\lvert 5x-2-3\right\rvert$ $\displaystyle<\epsilon$

$\displaystyle\left\lvert 5x-5\right\rvert$ $\displaystyle<\epsilon$

$\displaystyle 5\left\lvert x-1\right\rvert$ $\displaystyle<\epsilon$

$\displaystyle\left\lvert x-1\right\rvert$ $\displaystyle<\frac{\epsilon}{5}$

suggesting $\delta=\frac{\epsilon}{5}$ .

Proof: Given $\epsilon>0$ , let $\delta=\dfrac{\epsilon}{5}$ . Then:

$\displaystyle\left\lvert x-1\right\rvert$ $\displaystyle<\delta$

$\displaystyle\left\lvert x-1\right\rvert$ $\displaystyle<\frac{\epsilon}{5}$

$\displaystyle 5\left\lvert x-1\right\rvert$ $\displaystyle<\frac{\epsilon}{5}\cdot 5$

$\displaystyle\left\lvert 5x-5\right\rvert$ $\displaystyle<\epsilon$

$\displaystyle\left\lvert 5x-2-3\right\rvert$ $\displaystyle<\epsilon$

Thus $\displaystyle\lim_{x\to 1}(5x-2)=3$ .
36.
- 2
  
  $-3$
  
  $-3$
  
  $1/3$
37.

1

Exercises A.6

1.

Answers will vary.
2.

Answers will vary.
3.

A root of a function $f$ is a value $c$ such that $f(c)=0$ .
4.

Consider the function $h(x)=g(x)-f(x)$ , and use the Bisection Method to find a root of $h$ .
5.

F
6.

T
7.

T
8.

F
9.

F
10.

T
11.

No; $\displaystyle\lim_{x\to 1}f(x)=2$ , while $f(1)=1$ .
12.

No; $\displaystyle\lim_{x\to 1}f(x)$ does not exist.
13.

No; $f(1)$ does not exist.
14.

No
15.

Yes
16.

Yes
17.
- No; $\displaystyle\lim_{x\to-2}f(x)\neq f(-2)$
  
  Yes
  
  No; $f(2)$ is not defined.
18.

Yes; $\displaystyle\lim_{x\to 3\pi/2}\sin x+1=0$ , and $\sin(3\pi/2)+1=0$ .
19.
- Yes
  
  Yes
20.
- Yes
  
  No; the left and right hand limits at 1 are not equal.
21.
- Yes
  
  Yes
22.
- Yes
  
  No. $\lim_{x\to 8}f(x)=16/5\neq f(8)=5$ .
23.

$(-\infty,\infty)$
24.

$(-\infty,-2]\cup[2,\infty)$
25.

$[-2,2]$
26.

$[-1,1]$
27.

$(-\infty,-\sqrt{6}]\cup[\sqrt{6},\infty)$
28.

$(-1,1)$
29.

$(-\infty,\infty)$
30.

$(-\infty,\infty)$
31.

$(0,\infty)$
32.

$(-\infty,\infty)$
33.

$(-\infty,0]$
34.

$(-\infty,\infty)$
35.

$(-\infty,-4)\cup(-4,2)\cup(2,5)\cup(5,\infty)$
36.

$(-\infty,2)\cup(2,\infty)$
37.

Yes. The only “questionable” place is at $x=3$ , but the left and right limits agree.
38.

Yes. The only “questionable” place is at $x=0$ , but the Squeeze Theorem shows that the limits agree.
39.

Yes, by the Intermediate Value Theorem.
40.

Yes, by the Intermediate Value Theorem. In fact, we can be more specific and state such a value $c$ exists in $(0,2)$ , not just in $(-3,7)$ .
41.

We cannot say; the Intermediate Value Theorem only applies to function values between $-10$ and 10; as 11 is outside this range, we do not know.
42.

We cannot say; the Intermediate Value Theorem only applies to continuous functions. As we do know know if $h$ is continuous, we cannot say.
43.

$a=\frac{1}{3}$
44.

$a=-1$ and $\frac{4}{3}$
45.

$a=\frac{3}{4}$ and $b=-\frac{1}{4}$
46.

$a=-1$
47.

Answers will vary.
48.

Answers will vary.
49.

Answers will vary.
50.

Answers will vary.
51.

Use the Bisection Method with an appropriate interval.
52.

Use the Bisection Method with an appropriate interval.
53.

Use the Bisection Method with an appropriate interval.
54.

Use the Bisection Method with an appropriate interval.
55.
- 20
  
  25
  
  Limit does not exist
  
  25
56.

$x$ $f(x)$

$-0.81\phantom{0}$ $-2.34129$

$-0.801$ $-2.33413$

$-0.79\phantom{0}$ $-2.32542$

$-0.799$ $-2.33254$

The top two lines give an approximation of the limit from the left: $-2.33$ . The bottom two lines give an approximation from the right: $-2.33$ as well.
57.

Answers will vary.

$h$	$\frac{f(a+h)-f(a)}{h}$
$-0.1$	$-7$
$-0.01$	$-7$
$0.01$	$-7$
$0.1$	$-7$

$h$	$\frac{f(a+h)-f(a)}{h}$
$-0.1$	$4.9$
$-0.01$	$4.99$
$0.01$	$5.01$
$0.1$	$5.1$

$h$	$\frac{f(a+h)-f(a)}{h}$
$-0.1$	$-0.114943$
$-0.01$	$-0.111483$
$0.01$	$-0.110742$
$0.1$	$-0.107527$

$h$	$\frac{f(a+h)-f(a)}{h}$
$-0.1$	$29.4$
$-0.01$	$29.04$
$0.01$	$28.96$
$0.1$	$28.6$

$h$	$\frac{f(a+h)-f(a)}{h}$
$-0.1$	$0.202027$
$-0.01$	$0.2002$
$0.01$	$0.1998$
$0.1$	$0.198026$

$h$	$\frac{f(a+h)-f(a)}{h}$
$-0.1$	$-0.998334$
$-0.01$	$-0.999983$
$0.01$	$-0.999983$
$0.1$	$-0.998334$

$h$	$\frac{f(a+h)-f(a)}{h}$
$-0.1$	$-0.0499583$
$-0.01$	$-0.00499996$
$0.01$	$0.00499996$
$0.1$	$0.0499583$

$h$	$\frac{f(a+h)-f(a)}{h}$
$-0.1$	$0.251582$
$-0.01$	$0.250156$
$0.01$	$0.249844$
$0.1$	$0.248457$

$h$	$\frac{f(a+h)-f(a)}{h}$
$-0.1$	$2.58$
$-0.01$	$2.70$
$0.01$	$2.73$
$0.1$	$2.85$

	$\displaystyle\left\lvert x-4\right\rvert<\delta=\frac{\epsilon}{2}$
	$\displaystyle\left\lvert 2x-8\right\rvert<\epsilon$
	$\displaystyle\left\lvert(2x+5)-(13)\right\rvert<\epsilon.$

	$\displaystyle\left\lvert x-5\right\rvert<\delta=\epsilon$
	$\displaystyle\left\lvert 5-x\right\rvert<\epsilon$
	$\displaystyle\left\lvert 3-x-(-2)\right\rvert<\epsilon.$

	$\displaystyle\left\lvert x-5\right\rvert$	$\displaystyle<\delta=\frac{\epsilon}{4}$
	$\displaystyle 4\left\lvert x-5\right\rvert$	$\displaystyle<\frac{\epsilon}{4}\cdot 4$
	$\displaystyle\left\lvert 4x-20\right\rvert$	$\displaystyle<\epsilon$
	$\displaystyle\left\lvert 4x-12-8\right\rvert$	$\displaystyle<\epsilon$

	$\displaystyle\left\lvert x-3\right\rvert$	$\displaystyle<\delta=\frac{\epsilon}{2}$
	$\displaystyle 2\left\lvert x-3\right\rvert$	$\displaystyle<\frac{\epsilon}{2}\cdot 2$
	$\displaystyle\left\lvert-2x+6\right\rvert$	$\displaystyle<\epsilon$
	$\displaystyle\left\lvert 5-2x+1\right\rvert$	$\displaystyle<\epsilon$

	$\displaystyle\left\lvert x-3\right\rvert<\delta=\frac{\epsilon}{7}$
	$\displaystyle\left\lvert x-3\right\rvert<\frac{\epsilon}{x+3}$
	$\displaystyle\left\lvert x-3\right\rvert\cdot\left\lvert x+3\right\rvert<\frac% {\epsilon}{x+3}\cdot\left\lvert x+3\right\rvert$

	$\displaystyle\left\lvert x-3\right\rvert\cdot\left\lvert x+3\right\rvert<\frac% {\epsilon}{x+3}\cdot(x+3)$
	$\displaystyle\left\lvert x^{2}-9\right\rvert<\epsilon$
	$\displaystyle\left\lvert(x^{2}-3)-6\right\rvert<\epsilon.$

	$\displaystyle\left\lvert x-4\right\rvert\cdot\left\lvert x+5\right\rvert<\frac% {\epsilon}{x+5}\cdot(x+5)$
	$\displaystyle\left\lvert x^{2}+x-20\right\rvert<\epsilon$
	$\displaystyle\left\lvert(x^{2}+x-5)-15\right\rvert<\epsilon.$

	$\displaystyle\left\lvert x-1\right\rvert<\delta$
	$\displaystyle\left\lvert x-1\right\rvert<\frac{\epsilon}{9}$
	$\displaystyle\left\lvert x-1\right\rvert<\frac{\epsilon}{2x+5}$
	$\displaystyle\left\lvert x-1\right\rvert\cdot\left\lvert 2x+5\right\rvert<% \frac{\epsilon}{2x+5}\cdot\left\lvert 2x+5\right\rvert=\epsilon$
	$\displaystyle\left\lvert 2x^{2}+3x-5\right\rvert<\epsilon$
	$\displaystyle\left\lvert(2x^{2}+3x+1)-6\right\rvert<\epsilon.$

	$\displaystyle\left\lvert x-2\right\rvert<\delta=\frac{\epsilon}{19}$
	$\displaystyle\left\lvert x-2\right\rvert<\frac{\epsilon}{x^{2}+2x+4}$
	$\displaystyle\left\lvert x-2\right\rvert\cdot\left\lvert x^{2}+2x+4\right% \rvert<\frac{\epsilon}{x^{2}+2x+4}\cdot\left\lvert x^{2}+2x+4\right\rvert$

	$\displaystyle\left\lvert x-2\right\rvert\cdot\left\lvert x^{2}+2x+4\right% \rvert<\frac{\epsilon}{x^{2}+2x+4}\cdot(x^{2}+2x+4)$
	$\displaystyle\left\lvert x^{3}-8\right\rvert<\epsilon$
	$\displaystyle\left\lvert(x^{3}-1)-7\right\rvert<\epsilon,$

	$\displaystyle\left\lvert x\right\rvert<\delta$
	$\displaystyle\left\lvert x\right\rvert<\frac{\ln(1+\epsilon)}{2}<\left\lvert% \frac{\ln(1-\epsilon)}{2}\right\rvert$
	$\displaystyle\frac{\ln(1-\epsilon)}{2}<x<\frac{\ln(1+\epsilon)}{2}$
	$\displaystyle\ln(1-\epsilon)<2x<\ln(1+\epsilon)$
	$\displaystyle 1-\epsilon<e^{2x}<1+\epsilon$
	$\displaystyle-\epsilon<e^{2x}-1<\epsilon$
	$\displaystyle\left\lvert e^{2x}-1-(0)\right\rvert<\epsilon,$

	$\displaystyle\left\lvert f(x)-3\right\rvert$	$\displaystyle<\epsilon$
	$\displaystyle\left\lvert 5x-2-3\right\rvert$	$\displaystyle<\epsilon$
	$\displaystyle\left\lvert 5x-5\right\rvert$	$\displaystyle<\epsilon$
	$\displaystyle 5\left\lvert x-1\right\rvert$	$\displaystyle<\epsilon$
	$\displaystyle\left\lvert x-1\right\rvert$	$\displaystyle<\frac{\epsilon}{5}$

	$\displaystyle\left\lvert x-1\right\rvert$	$\displaystyle<\delta$
	$\displaystyle\left\lvert x-1\right\rvert$	$\displaystyle<\frac{\epsilon}{5}$
	$\displaystyle 5\left\lvert x-1\right\rvert$	$\displaystyle<\frac{\epsilon}{5}\cdot 5$
	$\displaystyle\left\lvert 5x-5\right\rvert$	$\displaystyle<\epsilon$
	$\displaystyle\left\lvert 5x-2-3\right\rvert$	$\displaystyle<\epsilon$

$x$	$f(x)$
$-0.81\phantom{0}$	$-2.34129$
$-0.801$	$-2.33413$
$-0.79\phantom{0}$	$-2.32542$
$-0.799$	$-2.33254$