## Problem Statement

Calculate the limit of the function:

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

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

Using the substitution of equivalent infinitesimals:

$$
\begin{aligned}
& 1-\cos x \sim \frac{x^{2}}{2}, \text { as } x \rightarrow 0 \\
& 1-\cos \sqrt{x} \sim \frac{(\sqrt{x})^{2}}{2}, \text { as } x \rightarrow 0
\end{aligned}
$$

We get:

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

## Problem Kuznetsov Limits 16-22