## Solutions to Try Its

1. a. removable discontinuity at $x=6$;
b. jump discontinuity at $x=4$

2. yes

3. No, the function is not continuous at $x=3$. There exists a removable discontinuity at $x=3$.

4. $x=6$

## Solutions to Odd-Numbered Exercises

1. Informally, if a function is continuous at $x=c$, then there is no break in the graph of the function at $f\left(c\right)$, and $f\left(c\right)$ is defined.

3. discontinuous at $a=-3$ ; $f\left(-3\right)$ does not exist

5. removable discontinuity at $a=-4$ ; $f\left(-4\right)$ is not defined

7. discontinuous at $a=3$ ; $\underset{x\to 3}{\mathrm{lim}}f\left(x\right)=3$, but $f\left(3\right)=6$, which is not equal to the limit.

9. $\underset{x\to 2}{\mathrm{lim}}f\left(x\right)$ does not exist.

11. $\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=4;\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=1$ . Therefore, $\underset{x\to 1}{\mathrm{lim}}f\left(x\right)$ does not exist.

13. $\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=5\ne \underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=-1$ . Thus $\underset{x\to 1}{\mathrm{lim}}f\left(x\right)$ does not exist.

15. $\underset{x\to -{3}^{-}}{\mathrm{lim}}f\left(x\right)=-6$ , $\underset{x\to -{3}^{+}}{\mathrm{lim}}f\left(x\right)=-\frac{1}{3}$

Therefore, $\underset{x\to -3}{\mathrm{lim}}f\left(x\right)$ does not exist.

17. $f\left(2\right)$ is not defined.

19. $f\left(-3\right)$ is not defined.

21. $f\left(0\right)$ is not defined.

23. Continuous on $\left(-\infty ,\infty \right)$

25. Continuous on $\left(-\infty ,\infty \right)$

27. Discontinuous at $x=0$ and $x=2$

29. Discontinuous at $x=0$

31. Continuous on $\left(0,\infty \right)$

33. Continuous on $\left[4,\infty \right)$

35. Continuous on $\left(-\infty ,\infty \right)$ .

37. 1, but not 2 or 3

39. 1 and 2, but not 3

41. $f\left(0\right)$ is undefined.

43. $\left(-\infty ,0\right)\cup \left(0,\infty \right)$

45. At $x=-1$, the limit does not exist. At $x=1$, $f\left(1\right)$ does not exist.
At $x=2$, there appears to be a vertical asymptote, and the limit does not exist.

47. $\frac{{x}^{3}+6{x}^{2}-7x}{\left(x+7\right)\left(x - 1\right)}$

49. $fx=\begin{cases}x^{2}+4 \hfill& x\neq 1 \\ 2 \hfill& x=1\end{cases}$