## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0}\left(\frac{\sin 2 x}{x}\right)^{1+x}$
## Solution

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

$=\left(\lim _{x \rightarrow 0}\left(\frac{\sin 2 x}{x}\right)\right)^{1}=\lim _{x \rightarrow 0} \frac{\sin 2 x}{x}=$

Using the substitution of equivalent infinitesimals:

$\sin 2 x \sim 2 x$, as $x \rightarrow 0(2 x \rightarrow 0)$

We get:

$$
=\lim _{x \rightarrow 0} \frac{2 x}{x}=\lim _{x \rightarrow 0} \frac{2}{1}=2
$$

## Problem Kuznetsov Limits 18-1