## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{-5 x}}{2 \sin x-\tan x}$
## Solution

$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{-5 x}}{2 \sin x-\tan x}=\lim _{x \rightarrow 0} \frac{\left(e^{2 x}-1\right)-\left(e^{-5 x}-1\right)}{2 \sin x-\tan x}=$
$=\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{2 x}-1\right)-\left(e^{-5 x}-1\right)\right)}{\frac{1}{x}(2 \sin x-\tan x)}=$
$=\frac{\lim _{x \rightarrow 0} \frac{1}{x}\left(\left(e^{2 x}-1\right)-\left(e^{-5 x}-1\right)\right)}{\lim _{x \rightarrow 0} \frac{1}{x}(2 \sin x-\tan x)}=$
$=\left(\lim _{x \rightarrow 0} \frac{e^{2 x}-1}{x}-\lim _{x \rightarrow 0} \frac{e^{-5 x}-1}{x}\right) /\left(\lim _{x \rightarrow 0} \frac{2 \sin x}{x}-\lim _{x \rightarrow 0} \frac{\tan x}{x}\right)=$

Using the substitution of equivalent infinitesimals:

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

$e^{-5 x}-1 \sim -5 x$, as $x \rightarrow 0(-5 x \rightarrow 0)$

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

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

We get:

$=\frac{\lim _{x \rightarrow 0} \frac{2 x}{x}-\lim _{x \rightarrow 0} \frac{-5 x}{x}}{\lim _{x \rightarrow 0} \frac{2 x}{x}-\lim _{x \rightarrow 0} \frac{x}{x}}=\frac{\lim _{x \rightarrow 0} 2-\lim _{x \rightarrow 0}-5}{\lim _{x \rightarrow 0} 2-\lim _{x \rightarrow 0} 1}=\frac{2+5}{2-1}=7$

## Problem Kuznetsov Limits 15-20