## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{3 x}}{\operatorname{arctg} x-x^{2}}$
## Solution

$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{3 x}}{\operatorname{arctg} x-x^{2}}=\lim _{x \rightarrow 0} \frac{\left(e^{2 x}-1\right)-\left(e^{3 x}-1\right)}{\operatorname{arctg} x-x^{2}}=$

$=\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{2 x}-1\right)-\left(e^{3 x}-1\right)\right)}{\frac{1}{x}\left(\operatorname{arctg} x-x^{2}\right)}=$

$=\frac{\lim _{x \rightarrow 0} \frac{1}{x}\left(\left(e^{2 x}-1\right)-\left(e^{3 x}-1\right)\right)}{\lim _{x \rightarrow 0} \frac{1}{x}\left(\operatorname{arctg} x-x^{2}\right)}=$

$=\left(\lim _{x \rightarrow 0} \frac{e^{2 x}-1}{x}-\lim _{x \rightarrow 0} \frac{e^{3 x}-1}{x}\right) /\left(\lim _{x \rightarrow 0} \frac{\operatorname{arctg} x}{x}-\lim _{x \rightarrow 0} \frac{x^{2}}{x}\right)=$

Using the substitution of equivalent infinitesimals:

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

$e^{3 x}-1 \sim 3 x$, as $x \rightarrow 0(3 x \rightarrow 0)$

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

We get:

$$
=\frac{\lim _{x \rightarrow 0} \frac{2 x}{x}-\lim _{x \rightarrow 0} \frac{3 x}{x}}{\lim _{x \rightarrow 0} \frac{x}{x}-\lim _{x \rightarrow 0} \frac{x^{2}}{x}}=\frac{\lim _{x \rightarrow 0} 2-\lim _{x \rightarrow 0} 3}{\lim _{x \rightarrow 0} 1-\lim _{x \rightarrow 0} x}=\frac{2-3}{1-0}=-1
$$

## Problem Kuznetsov Limits 15-6