## Problem Statement

Calculate the limit of the function:

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

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