## Problem Statement

Calculate the limit of the function:

$$
\lim _{x \rightarrow 0} \frac{1+x \sin x-\cos 2 x}{\sin ^{2} x}
$$
## Solution

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

Using the substitution of equivalent infinitesimals:

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

We get:

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