## Problem Statement

Calculate the limit of the function:

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

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

Using the substitution of equivalent infinitesimals:

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

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

$\tan x^{2} \sim x^{2}$, as $x \rightarrow 0\left(x^{2} \rightarrow 0\right)$

We get:

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

## Problem Kuznetsov Limits 15-29