## Calculus (3rd Edition)

$f (x)$ has a removable discontinuity at $x = 4$
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 $f (4) = 8$