## Problem Statement

Calculate the limit of the function:

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

$$
\begin{aligned}
& \lim _{x \rightarrow 16} \frac{\sqrt[4]{x}-2}{\sqrt[3]{(\sqrt{x}-4)^{2}}}=\lim _{x \rightarrow 16} \frac{\sqrt[4]{x}-2}{\sqrt[3]{(\sqrt[4]{x}-2)^{2}(\sqrt[4]{x}+2)^{2}}}= \\
& =\lim _{x \rightarrow 16} \frac{\sqrt[4]{x}-2}{(\sqrt[4]{x}-2)^{\frac{2}{3}} \sqrt[3]{(\sqrt[4]{x}+2)^{2}}}=\lim _{x \rightarrow 16} \frac{(\sqrt[4]{x}-2)^{\frac{1}{3}}}{\sqrt[3]{(\sqrt[4]{x}+2)^{2}}}= \\
& =\frac{(\sqrt[4]{16}-2)^{\frac{1}{3}}}{\sqrt[3]{(\sqrt[4]{16}+2)^{2}}}=\frac{(2-2)^{\frac{1}{3}}}{\sqrt[3]{(2+2)^{2}}}=\frac{0}{\sqrt[3]{4^{2}}}=0
\end{aligned}
$$

## Problem Kuznetsov Limits 11-27