## Problem Statement

Calculate the limit of the function:

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

$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{e^{x}-e^{-2 x}}{x+\sin x^{2}}=\lim _{x \rightarrow 0} \frac{\left(e^{x}-1\right)-\left(e^{-2 x}-1\right)}{x+\sin x^{2}}= \\
& =\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{x}-1\right)-\left(e^{-2 x}-1\right)\right)}{\frac{1}{x}\left(x+\sin x^{2}\right)}=
\end{aligned}
$$

$$
\begin{aligned}
& =\frac{\lim _{x \rightarrow 0} \frac{1}{x}\left(\left(e^{x}-1\right)-\left(e^{-2 x}-1\right)\right)}{\lim _{x \rightarrow 0} \frac{1}{x}\left(x+\sin x^{2}\right)}= \\
& =\left(\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}-\lim _{x \rightarrow 0} \frac{e^{-2 x}-1}{x}\right) /\left(\lim _{x \rightarrow 0} \frac{x}{x}+\lim _{x \rightarrow 0} \frac{\sin x^{2}}{x}\right)=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

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

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

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

We get:

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

## Problem Kuznetsov Limits 15-26