#### Answer

(a) $\lim\limits_{x \to -3^{-}} h(x) = 4$
(b) $\lim\limits_{x \to -3^{+}} h(x) = 4$
(c) $\lim\limits_{x \to -3} h(x) = 4$
(d) $h(-3)$ does not exist
(e) $\lim\limits_{x \to 0^{-}} h(x) = 1$
(f) $\lim\limits_{x \to 0^{+}} h(x) = -1$
(g) $\lim\limits_{x \to 0} h(x)$ does not exist
(h) $h(0) = 1$
(i) $\lim\limits_{x \to 2} h(x) = 2$
(j) $h(2)$ does not exist
(k) $\lim\limits_{x \to 5^{+}} h(x) = 3$
(l) $\lim\limits_{x \to 5^{-}} h(x)$ does not exist

#### Work Step by Step

(a) $h(x)$ approaches $y = 4$ as the function approaches $x = -3$ from the left.
(b) $h(x)$ approaches $y = 4$ as the function approaches $x = -3$ from the right.
(c) As $h(x)$ approaches $x = -3$ from both directions, it also approaches $y = 4$.
(d) The function $h(x)$ has no $y$ value at $x = -3$, which is why there is a hole in the function at that point.
(e) $h(x)$ approaches $y = 1$ as the function approaches $x = 0$ from the left.
(f) $h(x)$ approaches $y = -1$ as the function approaches $x = 0$ from the right.
(g) The limits from the left and from the right do not match.
(h) Although $h(x)$ approaches $y = -1$ from the right, the function is only present at $y = 1$.
(i) $h(x)$ approaches $y = 2$ as it approaches $x = 2$ from both sides.
(j) $h(x)$ has no $y$ value at $x = 2$.
(k) $h(x)$ approaches $y = 3$ as the function approaches $x = 5$ from the right.
(l) $h(x)$ oscillates between $y = 2$ and $y = 4$, meaning that it is approaching no single value as it gets closer to $x = 5$ from the left.