Answer
$$56$$
Work Step by Step
We have
$$
\lim _{x \rightarrow 8} \frac{x^{5/3}-2x-16}{x^{1/3}-2}=\frac{32-32}{2-2}=\frac{0}{0}
$$
is an intermediate form, then we can apply L’Hôpital’s Rule as follows
$$
\lim _{x \rightarrow 8} \frac{x^{5/3}-2x-16}{x^{1/3}-2}=\lim _{x \rightarrow 8} \frac{\frac{5}{3}x^{2/3}-2}{\frac{1}{3}x^{-2/3}}=\frac{\frac{20}{3}-2}{\frac{1}{12}}=80-24=56.
$$
