Answer
$$6$$
Work Step by Step
Since we have
$$
\lim _{x \rightarrow 4} \frac{ x^{3}-64 }{x^2-16}= \frac{0}{0}.
$$
Then we can apply L’Hôpital’s Rule as follows
$$
\lim _{x \rightarrow 4} \frac{ x^{3}-64 }{x^2-16}= \lim _{x \rightarrow 4} \frac{ 3x^{2} }{2x}=\frac{48}{8}=6.
$$
