## 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:

$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{1-\sqrt{\cos x}}{x \cdot \sin x}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 0} \frac{(1-\sqrt{\cos x})(1+\sqrt{\cos x})}{x \cdot x(1+\sqrt{\cos x})}= \\
& =\lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}(1+\sqrt{\cos x})}=\lim _{x \rightarrow 0} \frac{\frac{x^{2}}{2}}{x^{2}(1+\sqrt{\cos x})}= \\
& =\lim _{x \rightarrow 0} \frac{1}{2(1+\sqrt{\cos x})}=\frac{1}{2(1+\sqrt{\cos 0})}=\frac{1}{2 \cdot 2}=\frac{1}{4}
\end{aligned}
$$

## Problem Kuznetsov Limits 12-21
## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0} \frac{1-\sqrt{\cos x}}{x \cdot \sin x}$