## Problem Statement

Calculate the limit of the function:

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

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

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

Using the substitution of equivalent infinitesimals:

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

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

We get:

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

## Problem Kuznetsov Limits 16-1