Solutions to Odd-Numbered Exercises

1. The value of the function, the output, at $x=a$ is $f\left(a\right)$. When the $\underset{x\to a}{\mathrm{lim}}f\left(x\right)$ is taken, the values of $x$ get infinitely close to $a$ but never equal $a$. As the values of $x$ approach $a$ from the left and right, the limit is the value that the function is approaching.

3. –4

5. –4

7. 2

9. does not exist

11. 4

13. does not exist

15. Answers will vary.

17. Answers will vary.

19. Answers will vary.

21. Answers will vary.

23. 7.38906

25. 54.59815

27. ${e}^{6}\approx 403.428794$, ${e}^{7}\approx 1096.633158$, ${e}^{n}$

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

31. $\underset{x\to 3}{\mathrm{lim}}\left(\frac{{x}^{2}-x - 6}{{x}^{2}-9}\right)=\frac{5}{6}\approx 0.83$

33. $\underset{x\to 1}{\mathrm{lim}}\left(\frac{{x}^{2}-1}{{x}^{2}-3x+2}\right)=-2.00$

35. $\underset{x\to 1}{\mathrm{lim}}\left(\frac{10 - 10{x}^{2}}{{x}^{2}-3x+2}\right)=20.00$

37. $\underset{x\to \frac{-1}{2}}{\mathrm{lim}}\left(\frac{x}{4{x}^{2}+4x+1}\right)$ does not exist. Function values decrease without bound as $x$ approaches –0.5 from either left or right.

39. $\underset{x\to 0}{\mathrm{lim}}\frac{7\tan x}{3x}=\frac{7}{3}$

41. $\underset{x\to 0}{\mathrm{lim}}\frac{2\sin x}{4\tan x}=\frac{1}{2}$

43. $\underset{x\to 0}{\mathrm{lim}}{e}^{{e}^{-\text{ }\frac{1}{{x}^{2}}}}=1.0$

45. $\underset{x\to -{1}^{-}}{\mathrm{lim}}\frac{|x+1|}{x+1}=\frac{-\left(x+1\right)}{\left(x+1\right)}=-1$ and $\underset{x\to -{1}^{+}}{\mathrm{lim}}\frac{|x+1|}{x+1}=\frac{\left(x+1\right)}{\left(x+1\right)}=1$; since the right-hand limit does not equal the left-hand limit, $\underset{x\to -1}{\mathrm{lim}}\frac{|x+1|}{x+1}$ does not exist.

47. $\underset{x\to -1}{\mathrm{lim}}\frac{1}{{\left(x+1\right)}^{2}}$ does not exist. The function increases without bound as $x$ approaches $-1$ from either side.

49. $\underset{x\to 0}{\mathrm{lim}}\frac{5}{1-{e}^{\frac{2}{x}}}$ does not exist. Function values approach 5 from the left and approach 0 from the right.

51. Through examination of the postulates and an understanding of relativistic physics, as $v\to c$, $m\to \infty$. Take this one step further to the solution,

$\underset{v\to {c}^{-}}{\mathrm{lim}}m=\underset{v\to {c}^{-}}{\mathrm{lim}}\frac{{m}_{o}}{\sqrt{1-\left({v}^{2}/{c}^{2}\right)}}=\infty$