## Problem Statement

Calculate the limit of the function:

$$
\lim _{x \rightarrow 0}\left(\frac{\sin 4 x}{x}\right)^{\frac{2}{x+2}}
$$
## Solution

$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(\frac{\sin 4 x}{x}\right)^{\frac{2}{x+2}}=\left(\lim _{x \rightarrow 0} \frac{\sin 4 x}{x}\right)^{\lim _{x \rightarrow 0} \frac{2}{x+2}}= \\
& =\left(\lim _{x \rightarrow 0} \frac{\sin 4 x}{x}\right)^{\frac{2}{0+2}}=\left(\lim _{x \rightarrow 0} \frac{\sin 4 x}{x}\right)^{1}= \\
& =\lim _{x \rightarrow 0} \frac{\sin 4 x}{x}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$\sin 4 x \sim 4 x$, as $x \rightarrow 0(4 x \rightarrow 0)$

We get:

$$
=\lim _{x \rightarrow 0} \frac{4 x}{x}=\lim _{x \rightarrow 0} \frac{4}{1}=4
$$

## Problem Kuznetsov Problem Kuznetsov Limits 18-3