## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 4} \frac{\sqrt{x}-2}{\sqrt[3]{x^{2}-16}}$
## Solution

$$
\begin{aligned}
& \lim _{x \rightarrow 4} \frac{\sqrt{x}-2}{\sqrt[3]{x^{2}-16}}=\lim _{x \rightarrow 4} \frac{\sqrt{x}-2}{\sqrt[3]{(x-4)(x+4)}}= \\
& =\lim _{x \rightarrow 4} \frac{\sqrt{x}-2}{\sqrt[3]{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)}}= \\
& =\lim _{x \rightarrow 4} \frac{\sqrt{x}-2}{\sqrt[3]{(\sqrt{x}-2)} \sqrt[3]{(\sqrt{x}+2)(x+4)}}= \\
& =\lim _{x \rightarrow 4} \frac{\sqrt[3]{(\sqrt{x}-2)^{2}}}{\sqrt[3]{(\sqrt{x}+2)(x+4)}}=\frac{\sqrt[3]{(\sqrt{4}-2)^{2}}}{\sqrt[3]{(\sqrt{4}+2)(4+4)}}=
\end{aligned}
$$

$=\frac{\sqrt[3]{0^{2}}}{\sqrt[3]{(2+2) 8}}=0$

## Problem Kuznetsov Limits 11-29