## Problem Statement

Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{\sin 2 x}-e^{\sin x}}{\tan x}$
## Solution

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

$=\lim _{x \rightarrow 0} \frac{e^{\sin 2 x}-1}{\tan x}-\lim _{x \rightarrow 0} \frac{e^{\sin x}-1}{\tan x}=$

Using the substitution of equivalent infinitesimals:

$e^{\sin 2 x}-1 \sim \sin 2 x$, as $x \rightarrow 0(\sin 2 x \rightarrow 0)$

$e^{\sin x}-1 \sim \sin x$, as $x \rightarrow 0(\sin x \rightarrow 0)$

$\tan x \sim x$, as $x \rightarrow 0$

We get:

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

Using the substitution of equivalent infinitesimals:

$\sin x \sim x$, as $x \rightarrow 0$

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

We get:

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

## Problem Kuznetsov Limits 16-17