## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0} \frac{x^{2}\left(e^{x}-e^{-x}\right)}{e^{x^{3}+1}-e}$
## Solution

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

$$
\begin{aligned}
& e^{2 x}-1 \sim 2 x, \text { as } x \rightarrow 0(2 x \rightarrow 0) \\
& e^{x^{3}}-1 \sim x^{3}, \text { as } x \rightarrow 0\left(x^{3} \rightarrow 0\right)
\end{aligned}
$$

We get:

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

## Problem Kuznetsov Limits 16-8