Answer
We can define $g(4)=\frac{8}{5}$ that extends $g(x)$ to be continuous at $x=4$.
Work Step by Step
$$g(x)=\frac{x^2-16}{x^2-3x-4}=\frac{x^2-16}{(x-4)(x+1)}$$
Currently, $g(x)$ is not defined at $x=4$, so $g(x)$ is not continuous at $x=4$ as well.
For $g(x)$ to be continuous at $x=4$, we need to extend $g(x)$ to include a value of $g(4)$ so that $g(4)=\lim_{x\to4}g(x)$
So we need to calculate $\lim_{x\to4}g(x)$ first:
$$\lim_{x\to4}g(x)=\lim_{x\to4}\frac{x^2-16}{(x-4)(x+1)}=\lim_{x\to4}\frac{(x-4)(x+4)}{(x-4)(x+1)}=\lim_{x\to4}\frac{x+4}{x+1}$$
$$\lim_{x\to4}g(x)=\frac{4+4}{4+1}=\frac{8}{5}$$
So we can define $g(4)=\frac{8}{5}$ that extends $g(x)$ to be continuous at $x=4$.
