# Chapter 1

## Exercises 1.0

1. 1.

$(-\infty,\infty)$

2. 2.

$[-7,\infty)$

3. 3.

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

4. 4.

$(-\infty,\infty)$

5. 5.

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

6. 6.

$(-\infty,\infty)$

7. 7.

$(-\infty,\infty)$

8. 8.

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

9. 9.

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

10. 10.

$(-\infty,\infty)$

11. 11.
12. 12.
13. 13.
14. 14.
15. 15.
(a) 14 (b) 11 (c) $3a^{2}-2a+6$ (d) $3(x+h)^{2}-2(x+h)+6$ (e) $\dfrac{h(3h+6x-2)}{h}$
16. 16.
(a) $\sqrt{2}$ (b) undefined (c) $\sqrt{t-2}$ (d) $\sqrt{x+h-2}$ (e) $\dfrac{\sqrt{x+h-2}-\sqrt{x-2}}{h}\lx@parboxnewline=\dfrac{h}{h(\sqrt{x+h-2}+% \sqrt{x-2})}$
17. 17.
(a) $-1$ (b) $\dfrac{1}{9}$ (c) $\dfrac{1}{t+3}$ (d) $\dfrac{1}{x+h}$ (e) $\dfrac{\frac{1}{x+h}-\frac{1}{x}}{h}=-\dfrac{h}{hx(x+h)}$
18. 18.
(a) $e^{-2}$ (b) $e^{3}$ (c) $e^{t+1}$ (d) $e^{x+h}$ (e) $\dfrac{e^{x+h}-e^{x}}{h}=e^{x}\dfrac{e^{h}-1}{h}$
19. 19.

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

20. 20.

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

21. 21.

$(0.3,0.8]$

22. 22.

$(0.1,0.2]$

## Exercises 1.1

1. 1.

2. 3.

F

3. 5.

$1$

4. 7.

$-1$

5. 9.

Limit does not exist

6. 11.

$1.5$

7. 13.

Limit does not exist.

8. 15.

$1$

9. 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 $-7$.
10. 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.
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. $29$.
12. 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. $-1$.
13. 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. $0.25$.

## Exercises 1.2

1. 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. 3.

T

3. 5.

$\delta\leq 0.45$

4. 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 $\displaystyle\lim_{x\to 4}(2x+5)=13$.
5. 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 $\displaystyle\lim_{x\to 5}(4x-12)=8$.
6. 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, $\displaystyle\lim_{x\to 3}\left(x^{2}-3\right)=6$.
7. 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, $\displaystyle\lim_{x\to 1}\left(2x^{2}+3x+1\right)=6$.
8. 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 $\delta=\epsilon$.

9. 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.

## Exercises 1.3

1. 1.

2. 3.

3. 5.

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

4. 7.

9

5. 9.

0

6. 11.

3

7. 13.

3

8. 15.

$1$

9. 17.

$0$

10. 19.

$7$

11. 21.

$1/2$

12. 23.

Limit does not exist

13. 25.

$2$

14. 27.

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

15. 29.

$-8$

16. 31.

$10$

17. 33.

$-3/2$

18. 35.

$1/6$

19. 37.

$-1/9$

20. 39.

$-8$

21. 41.

$0$

22. 43.

$1$

23. 45.

$0$

24. 47.

$3$

25. 49.

$1$

26. 51.

$4/3$

27. 53.
(a) Apply Part 1 of Theorem 1.3.1. (b) Apply Theorem 1.3.7; $g(x)=\frac{x}{x}$ is the same as $g(x)=1$ everywhere except at $x=0$. Thus $\displaystyle\lim_{x\to 0}g(x)=\lim_{x\to 0}1=1.$ (c) 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. (d) 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.

## Exercises 1.4

1. 1.

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

2. 3.

F

3. 5.
(a) 2 (b) 2 (c) 2 (d) 1 (e) As $f$ is not defined for $x<0$, this limit is not defined. (f) 1
4. 7.
(a) 2 (b) 2 (c) 2 (d) 2
5. 9.
(a) 2 (b) 2 (c) 2 (d) 0 (e) 2 (f) 2 (g) 2 (h) Not defined
6. 11.

DNE

7. 13.

DNE

8. 15.
(a) 2 (b) $-4$ (c) Does not exist. (d) 2
9. 17.
(a) 0 (b) 0 (c) 0 (d) 0 (e) 2 (f) 2 (g) 2 (h) 2
10. 19.
(a) $1-\cos^{2}a=\sin^{2}a$ (b) $\sin^{2}a$ (c) $\sin^{2}a$ (d) $\sin^{2}a$
11. 21.
(a) 4 (b) 4 (c) 4 (d) 3
12. 23.
(a) $-1$ (b) $1$ (c) Does not exist (d) $0$
13. 25.

14. 27.

15. 29.

$-3/5$

16. 31.

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

17. 33.

$-1.63$

## Exercises 1.5

1. 1.

F

2. 3.

F

3. 5.

F

4. 7.

5. 9.
(a) $\infty$ (b) $\infty$
6. 11.
(a) $1$ (b) $0$ (c) $1/2$ (d) $1/2$
7. 13.
(a) Limit does not exist (b) Limit does not exist
8. 15.
Tables will vary. (a) $x$ $f(x)$ $2.9$ $-15.1224$ $2.99$ $-159.12$ $2.999$ $-1599.12$ It seems $\lim_{x\to 3^{-}}f(x)=-\infty$. (b) $x$ $f(x)$ $3.1$ $16.8824$ $3.01$ $160.88$ $3.001$ $1600.88$ It seems $\lim_{x\to 3^{+}}f(x)=\infty$. (c) It seems $\lim_{x\to 3}f(x)$ does not exist.
9. 17.
Tables will vary. (a) $x$ $f(x)$ $2.9$ $132.857$ $2.99$ $12124.4$ It seems $\lim_{x\to 3^{-}}f(x)=\infty$. (b) $x$ $f(x)$ $3.1$ $108.039$ $3.01$ $11876.4$ It seems $\lim_{x\to 3^{+}}f(x)=\infty$. (c) It seems $\lim_{x\to 3}f(x)=\infty$.
10. 19.

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

11. 21.

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

12. 23.

No horizontal or vertical asymptotes.

13. 25.

$y=2$

14. 27.

$\infty$

15. 29.

$-\infty$

16. 31.

$-2/3$

17. 33.

$-\sqrt{10}/2$

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

19. 37.

1

## Exercises 1.6

1. 1.

2. 3.

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

3. 5.

F

4. 7.

T

5. 9.

F

6. 11.

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

7. 13.

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

8. 15.

Yes

9. 17.
(a) No; $\displaystyle\lim_{x\to-2}f(x)\neq f(-2)$ (b) Yes (c) No; $f(2)$ is not defined.
10. 19.
(a) Yes (b) Yes
11. 21.
(a) Yes (b) Yes
12. 23.

$(-\infty,\infty)$

13. 25.

$[-2,2]$

14. 27.

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

15. 29.

$(-\infty,\infty)$

16. 31.

$(0,\infty)$

17. 33.

$(-\infty,0]$

18. 35.

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

19. 37.

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

20. 39.

Yes, by the Intermediate Value Theorem.

21. 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.

22. 43.

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

23. 45.

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

24. 47.

25. 49.

26. 51.

Use the Bisection Method with an appropriate interval.

27. 53.

Use the Bisection Method with an appropriate interval.

28. 55.
(a) 20 (b) 25 (c) Limit does not exist (d) 25
29. 57. 