## Problem Statement

Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{2 x \sin x}{1-\cos x}$
## Solution

Let's use the substitution of equivalent infinitesimals:

$1-\cos x \sim \frac{x^{2}}{2}$, as $x \rightarrow 0$

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

We get:

$$
\lim _{x \rightarrow 0} \frac{2 x \sin x}{1-\cos x}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 0} \frac{2 x \cdot x}{\frac{x^{2}}{2}}=\lim _{x \rightarrow 0} \frac{2}{\frac{1}{2}}=4
$$

## Problem Kuznetsov Limits 12-31