#### Answer

(a) False
(b) False
(c) True
(d) True
(e) True

#### Work Step by Step

(a) $\lim_{x\to2}f(x)$ does not exist.
This is false, because even though the value of $f(2)$ jumps to $0$, for values of $x$ approaching $x=2$ from the left, $f(x)$ approaches $1$, and for $x$ approaching $x=2$ from the right, $f(x)$ approaches $1$ also.
Therefore, as $x\to2$, $f(x)$ approaches $1$, or $\lim_{x\to2}f(x)=1$
(b) $\lim_{x\to2}f(x)=1$
This is also false, as proved above that $\lim_{x\to2}f(x)=1$
(c) $\lim_{x\to1}f(x)$ does not exist.
This is true, since $f(x)$ jumps at $x=1$. For values of $x$ approaching $x=1$ from the left, $f(x)$ approaches $-2$, and for $x$ approaching $x=1$ from the right, $f(x)$ approaches $0$.
Therefore, there is no single value that $f(x)$ approaches when $x\to1$.
(d) $\lim_{x\to c}f(x)$ exists at every point $c$ in $(-1,1)$
This is true. The graph as $x$ in $(-1,1)$ is continuous, and for each $x$, there is only one corresponding value of $y$, so all the limits here exist.
(e) $\lim_{x\to c}f(x)$ exists at every point $c$ in $(1,3)$
This is also true. As $x$ in $(1,2)$ and $(2,3)$, as shown in $(d)$, the graph is continuous and one-to-one, so all the limits would exist as a result.
At $x=2$, as proved in $(a)$, its limit does exist and equal $1$.
Therefore, $\lim_{x\to c}f(x)$ exists at every point $c$ in $(1,3)$