#### Answer

$(a)$ $f(-2)$ does not exist.
$(b)$ $\lim_{x\to-2}f(x)$ does not exist.
$(c)$ $f(0)=4$
$(d)$ $\lim_{x\to0}f(x)$ does not exist.
$(e)$ $f(2)$ does not exist.
$(f)$ $\lim_{x\to2}f(x)=\dfrac{1}{2}$
$(g)$ $f(4)=2$
$(h)$ $\lim_{x\to4}f(x)=\infty$

#### Work Step by Step

The graph of the function $f$ is shown below:
$(a)$ $f(-2)$
From the graph, it can be seen that $f$ is not defined for $-2$ because $x=-2$ represents a vertical asymptote. $f(-2)$ does not exist.
$(b)$ $\lim_{x\to-2}f(x)$
From the graph, it can be seen that $f$ goes to $-\infty$ going to $-2$ from the left and to $\infty$ going to $-2$ from the right. Since the limits from the left and right are not the same, $\lim_{x\to-2}f(x)$ does not exist.
$(c)$ $f(0)$
From the graph, it can be easily seen that $f(0)=4$, because of the two dots present at $x=0$, the one that is filled falls on $4$.
$(d)$ $\lim_{x\to0}f(x)$
From the graph, it can be seen that $f$ goes to $\dfrac{1}{2}$ going to $0$ from the left and to $4$ going to $0$ from the right. Since the limits from the left and right are not the same, $\lim_{x\to0}f(x)$ does not exist.
$(e)$ $f(2)$
From the graph, it can be seen that $f$ is not defined for $2$ because the dot is not filled at that point. $f(2)$ does not exist.
$(f)$ $\lim_{x\to2}f(x)$
From the graph, it can be seen that $f$ goes to $\dfrac{1}{2}$ going to $2$ from the left and to $\dfrac{1}{2}$ going to $2$ from the right. Since the limits from the left and right are the same, $\lim_{x\to2}f(x)=\dfrac{1}{2}$.
$(g)$ $f(4)$
From the graph, it can be seen that $f(4)=2$, because the dot is filled at that point.
$(h)$ $\lim_{x\to4}f(x)$
From the graph, it can be seen that $f$ goes to $\infty$ going to $4$ from the left and to $\infty$ going to $4$ from the right. Since the limits from the left and right are the same, $\lim_{x\to4}f(x)=\infty$