Answer
The function $f(x)$ is discontinuous at $x=-5$ and $x=0$
(a)
$f(-5)$ does not exist.
$f(0) $ does not exist.
(b)
$$
\lim _{x \rightarrow-5^{-}} f(x)=\infty \quad \quad \text {(limit does not exist) }
$$
and
$$
\lim _{x \rightarrow 0^{-}} f(x)=0
$$
(c)
$$
\lim _{x \rightarrow-5^{+}} f(x)=-\infty \quad \quad \text {(limit does not exist) }
$$
and
$$
\lim _{x \rightarrow 0^{+}} f(x)=0
$$
(d)
Since
$$
\lim _{x \rightarrow-5^{-}} f(x) \ne \lim _{x \rightarrow-5^{+}} f(x)
$$
then,
$$
\lim _{x \rightarrow-5} f(x) \quad \quad \text {(does not exist) }
$$
and
$$
\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=0
$$
then,
$$
\lim _{x \rightarrow0} f(x)=0.
$$
(e)
$f(-5)$ does not exist.
$$
\lim _{x \rightarrow-5} f(x) \quad \quad \text {(does not exist) }
$$
$f(0) $ does not exist.
Work Step by Step
The function $f(x)$ is discontinuous at $x=-5$ and $x=0$
(a)
$f(-5)$ does not exist.
$f(0) $ does not exist.
(b)
$$
\lim _{x \rightarrow-5^{-}} f(x)=\infty \quad \quad \text {(limit does not exist) }
$$
and
$$
\lim _{x \rightarrow 0^{-}} f(x)=0
$$
(c)
$$
\lim _{x \rightarrow-5^{+}} f(x)=-\infty \quad \quad \text {(limit does not exist) }
$$
and
$$
\lim _{x \rightarrow 0^{+}} f(x)=0
$$
(d)
Since
$$
\lim _{x \rightarrow-5^{-}} f(x) \ne \lim _{x \rightarrow-5^{+}} f(x)
$$
then,
$$
\lim _{x \rightarrow-5} f(x) \quad \quad \text {(does not exist) }
$$
and
$$
\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=0
$$
then,
$$
\lim _{x \rightarrow0} f(x)=0.
$$
(e)
$f(-5)$ does not exist.
$$
\lim _{x \rightarrow-5} f(x) \quad \quad \text {(does not exist) }
$$
$f(0) $ does not exist.
