Answer
$$
f(x)=\frac{x^{2}-9}{3+x}
$$
$ x$-value where the given function is discontinuous. is
$$ x=-3 $$
and
$$ f(-3)$$
do not exist, also
$$ \lim _{x \rightarrow -3} f(x)
$$
is exist and equal $-6$
Work Step by Step
$$
f(x)=\frac{x^{2}-9}{3+x}
$$
This function is discontinuous wherever the denominator is zero.
There is a discontinuity when $x=-3 $ .
so the given function is discontinuous at $x=-3 $, therefore $ f(-3)$ do not exist.
Now, we investigate $ \lim _{x \rightarrow -3} f(x) $ at $x=-3$
$$
\begin{split}
\lim _{x \rightarrow -3} f(x) &=\lim _{x \rightarrow -3}\left(\frac{x^{2}-9}{3+x} \right) \\
&=\lim _{x \rightarrow -3}\left(\frac{(x-3)(x+3)}{(3+x)} \right)\\
&=\lim _{x \rightarrow -3}\left(x-3 \right)\\
&=-3-3 \\
&=-6
\end{split}
$$
so,
$$ \lim _{x \rightarrow -3} f(x) $$
is exist and equal $-6$.
Hence, $ x$-value where the given function is discontinuous. is
$$ x=-3 $$
and
$$ f(-3)$$
do not exist, also
$$ \lim _{x \rightarrow -3} f(x)
$$
is exist and equal $-6$.
