Answer
$$
f(x)=\frac{x-6}{x+5}
$$
$ x$-value where the given function is discontinuous. is
$$ x=-3 $$
and
$$f(-3)$$
does not exist, also
$$ \lim _{x \rightarrow -5} f(x)
$$
is not exist
Work Step by Step
$$
f(x)=\frac{x-6}{x+5}
$$
This function is discontinuous wherever the denominator is zero.
There is a discontinuity when $x=-5 $ .
so the given function is discontinuous at $x=-5 $ , therefore $f(-5)$ does not exist.
We investigate $ \lim _{x \rightarrow -5} f(x) $ at $x=-5$ first. From the left, where $x$-values are less than $-5$,
$$
\lim _{x \rightarrow -5^{-}} f(x)=\lim _{x \rightarrow -5^{-}}\left(\frac{x-6}{x+5} \right)=\infty
$$
From the right, where $x$-values are greater than -5,
$$
\lim _{x \rightarrow -5^{+}} f(x)=\lim _{x \rightarrow-5^{+}}\left(\frac{x-6}{x+5} \right)=-\infty
$$
so, $ \lim _{x \rightarrow -5} f(x) $ does not exist since
$$
\lim _{x \rightarrow -5^{+}} f(x) \ne \lim _{x \rightarrow -5^{-}}f(x)
$$
Hence, $ x$-value where the function is discontinuous. is
$x=-5 $ .
$ f(-5)$ does not exist, also
$ \lim _{x \rightarrow -5} f(x) $ does not exist.
