Answer
$f (x)$ has a removable discontinuity at $x = 4$
Work Step by Step
We find the limit of the given function at $x=4$:
\begin{align*}
\lim _{x \rightarrow 4} \frac{x^{2}-16}{x-4}&=\lim _{x \rightarrow 4}(x+4)\\
&=8
\end{align*}
We see that the function value does not equal the limit at $x=4$:
$$ f(4)=10\neq\lim _{x \rightarrow 4} \frac{x^{2}-16}{x-4} =8$$
Thus, $f (x)$ has a removable discontinuity at $x = 4$. To remove the discontinuity, we must redefine $f (4) = 8$
