## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0} \frac{e^{4 x}-e^{-2 x}}{2 \operatorname{arctan} x-\sin x}$
## Solution

$\lim _{x \rightarrow 0} \frac{e^{4 x}-e^{-2 x}}{2 \operatorname{arctg} x-\sin x}=\lim _{x \rightarrow 0} \frac{\left(e^{4 x}-1\right)-\left(e^{-2 x}-1\right)}{2 \operatorname{arctg} x-\sin x}=$
$=\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{4 x}-1\right)-\left(e^{-2 x}-1\right)\right)}{\frac{1}{x}(2 \operatorname{arctg} x-\sin x)}=$
$=\frac{\lim _{x \rightarrow 0} \frac{1}{x}\left(\left(e^{4 x}-1\right)-\left(e^{-2 x}-1\right)\right)}{\lim _{x \rightarrow 0} \frac{1}{x}(2 \operatorname{arctg} x-\sin x)}=$
$=\left(\lim _{x \rightarrow 0} \frac{e^{4 x}-1}{x}-\lim _{x \rightarrow 0} \frac{e^{-2 x}-1}{x}\right) /\left(\lim _{x \rightarrow 0} \frac{2 \operatorname{arctg} x}{x}-\lim _{x \rightarrow 0} \frac{\sin x}{x}\right)=$

Using the substitution of equivalent infinitesimals:

$e^{4 x}-1 \sim 4 x$, as $x \rightarrow 0(4 x \rightarrow 0)$

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

$\operatorname{arctg} x \sim x$, as $x \rightarrow 0$

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

We get:

$=\frac{\lim _{x \rightarrow 0} \frac{4 x}{x}-\lim _{x \rightarrow 0} \frac{-2 x}{x}}{\lim _{x \rightarrow 0} \frac{2 x}{x}-\lim _{x \rightarrow 0} \frac{x}{x}}=\frac{\lim _{x \rightarrow 0} 4-\lim _{x \rightarrow 0}-2}{\lim _{x \rightarrow 0} 2-\lim _{x \rightarrow 0} 1}=\frac{4+2}{2-1}=6$

## Problem Kuznetsov Limits 15-8