Answer
$$
f(x)=\frac{7-3x}{(1-x)(3+x)}
$$
$ x$-values where the given function is discontinuous. are
$$x=1 , \quad x=-3 $$
and
$$f(0), \quad f(-3)$$
do not exist, also
$$ \lim _{x \rightarrow -3} f(x) ,\quad \lim _{x \rightarrow 1} f(x)
$$
are not exist
Work Step by Step
$$
f(x)=\frac{7-3x}{(1-x)(3+x)}
$$
This rational function is discontinuous wherever the denominator is zero.
There is a discontinuity when $x=1 $ and $x=-3 $ .
so the given function is discontinuous at $x=1 $ and $x=-3 $, therefore $f(1), f(-3)$ do not exist.
We investigate $ \lim _{x \rightarrow 1} f(x) $ at $x=1$ first. From the left, where $x$-values are less than $1$,
$$
\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}\left(\frac{7-3x}{(1-x)(3+x)} \right)=\infty
$$
From the right, where $x$-values are greater than 1,
$$
\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}\left(\frac{7-3x}{(1-x)(3+x)} \right)=-\infty
$$
so, $ \lim _{x \rightarrow 1} f(x) $ does not exist since
$$
\lim _{x \rightarrow 1^{+}} f(x) \ne \lim _{x \rightarrow 1^{-}}f(x)
$$
Now, we investigate $ \lim _{x \rightarrow -3} f(x) $ at $x=-3$ first. From the left, where $x$-values are less than $-3$,
$$
\lim _{x \rightarrow -3^{-}} f(x)=\lim _{x \rightarrow -3^{-}}\left(\frac{7-3x}{(1-x)(3+x)} \right)=-\infty
$$
From the right, where $x$-values are greater than $ -3$,
$$
\lim _{x \rightarrow -3^{+}} f(x)=\lim _{x \rightarrow -3^{+}}\left(\frac{7-3x}{(1-x)(3+x)} \right)=\infty
$$
so, $ \lim _{x \rightarrow -3} f(x) $ does not exist since
$$
\lim _{x \rightarrow -3^{+}} f(x) \ne \lim _{x \rightarrow -3^{-}}f(x)
$$
Hence, $ x$-values where the function is discontinuous. are
$x=1$ and $x=-3 $ .
$f(1), f(-3)$ do not exist, also
$ \lim _{x \rightarrow -3} f(x) $, $ \lim _{x \rightarrow 1} f(x) $ do not exist.
