Answer
$$\frac{8}{3}$$
Work Step by Step
\begin{aligned}
\lim _{x \rightarrow 64} \frac{x^{2 / 3}-16}{\sqrt{x}-8} &=\lim _{x \rightarrow 64} \frac{\left(x^{1 / 3}-4\right)\left(x^{1 / 3}+4\right)}{\sqrt{x}-8}\\
&=\lim _{x \rightarrow 64} \frac{\left(x^{1 / 3}-4\right)\left(x^{1 / 3}+4\right)}{\sqrt{x}-8} \cdot \frac{\left(x^{2 / 3}+4 x^{1 / 3}+16\right)(\sqrt{x}+8)}{(\sqrt{x}+8)\left(x^{2 / 3}+4 x^{1 / 3}+16\right)} \\
&=\lim _{x \rightarrow 64} \frac{(x-64)\left(x^{1 / 3}+4\right)(\sqrt{x}+8)}{(x-64)\left(x^{2 / 3}+4 x^{1 / 3}+16\right)}\\
&=\lim _{x \rightarrow 64} \frac{\left(x^{1 / 3}+4\right)(\sqrt{x+8})}{x^{2 / 3}+4 x^{1 / 3}+16}\\
&=\frac{(4+4)(8+8)}{16+16+16}\\
&=\frac{8}{3}
\end{aligned}
