Answer
The function $f(x)$ is discontinuous at $x=-2$ and $x=3$
(a)
$$
f(-2) =1,\quad f(3) =1
$$ .
(b)
$$
\lim _{x \rightarrow-2^{-}} f(x)=-1, \quad \lim _{x \rightarrow 3^{-}} f(x)=-1
$$
(c)
$$
\lim _{x \rightarrow-2^{+}} f(x)=-1, \quad \lim _{x \rightarrow 3^{+}} f(x)=-1
$$
(d)
Since
$$
\lim _{x \rightarrow-2^{-}} f(x)=\lim _{x \rightarrow-2^{+}} f(x)=-1
$$
and
$$
\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{+}} f(x)=-1
$$
then,
$$
\lim _{x \rightarrow-2} f(x)=-1, \quad \lim _{x \rightarrow 3} f(x)=-1
$$
(e)
$$
\lim _{x \rightarrow -2} f(x) \ne f(-2), \quad \lim _{x \rightarrow 3} f(x) \ne f(3).
$$
Work Step by Step
The function $f(x)$ is discontinuous at $x=-2$ and $x=3$
(a)
$$
f(-2) =1,\quad f(3) =1
$$ .
(b)
$$
\lim _{x \rightarrow-2^{-}} f(x)=-1, \quad \lim _{x \rightarrow 3^{-}} f(x)=-1
$$
(c)
$$
\lim _{x \rightarrow-2^{+}} f(x)=-1, \quad \lim _{x \rightarrow 3^{+}} f(x)=-1
$$
(d)
Since
$$
\lim _{x \rightarrow-2^{-}} f(x)=\lim _{x \rightarrow-2^{+}} f(x)=-1
$$
and
$$
\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{+}} f(x)=-1
$$
then,
$$
\lim _{x \rightarrow-2} f(x)=-1, \quad \lim _{x \rightarrow 3} f(x)=-1
$$
(e)
$$
\lim _{x \rightarrow -2} f(x) \ne f(-2), \quad \lim _{x \rightarrow 3} f(x) \ne f(3).
$$
