## Problem Statement

Calculate the limit of the function:

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

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

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

$$
\begin{aligned}
& =\frac{\lim _{x \rightarrow 0} \frac{1}{x}\left(\left(e^{3 x}-1\right)-\left(e^{-2 x}-1\right)\right)}{\lim _{x \rightarrow 0} \frac{1}{x}(2 \arcsin x-\sin x)}= \\
& =\left(\lim _{x \rightarrow 0} \frac{e^{3 x}-1}{x}-\lim _{x \rightarrow 0} \frac{e^{-2 x}-1}{x}\right) /\left(\lim _{x \rightarrow 0} \frac{2 \arcsin x}{x}-\lim _{x \rightarrow 0} \frac{x}{x}\right)=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

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

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

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

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

We get:

$=\frac{\lim _{x \rightarrow 0} \frac{3 x}{x}-\lim _{x \rightarrow 0} \frac{-2 x}{x}}{\lim _{x \rightarrow 0} \frac{2 x}{x}-\lim _{x \rightarrow 0} \frac{x}{x}}=\frac{\lim _{x \rightarrow 0} 3-\lim _{x \rightarrow 0}-2}{\lim _{x \rightarrow 0} 2-\lim _{x \rightarrow 0} 1}=\frac{3+2}{2-1}=5$

## Problem Kuznetsov Limits 15-2