## Problem Statement

Calculate the limit of the function:

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

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

Using the substitution of equivalent infinitesimals:

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

$\sin 2 x \sim 2 x$, as $x \rightarrow 0(2 x \rightarrow 0)$

We get:

$$
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{-\frac{x^{2}}{2}}{(2 x)^{2}(1+\sqrt{\cos x})}=\lim _{x \rightarrow 0} \frac{-x^{2}}{8 x^{2}(1+\sqrt{\cos x})}= \\
& =\lim _{x \rightarrow 0} \frac{-1}{8(1+\sqrt{\cos x})}=\frac{-1}{8(1+\sqrt{\cos 0})}=
\end{aligned}
$$

$$
=\frac{-1}{8(1+1)}=-\frac{1}{16}
$$

## Problem Kuznetsov Limits 16-27