## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0}(\cos \sqrt{x})^{\frac{1}{x}}$
## Solution

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

Using the substitution of equivalent infinitesimals:

$$
\ln \left(1-2 \sin ^{2} \frac{\sqrt{x}}{2}\right) \sim-2 \sin ^{2} \frac{\sqrt{x}}{2}, \quad x \rightarrow 0\left(-2 \sin ^{2} \frac{\sqrt{x}}{2} \rightarrow 0\right)
$$

We get:

$=\exp \left\{\lim _{x \rightarrow 0} \frac{-2 \sin ^{2} \frac{\sqrt{x}}{2}}{x}\right\}=$

Using the substitution of equivalent infinitesimals:

$$
\sin \frac{\sqrt{x}}{2} \sim \frac{\sqrt{x}}{2}, \quad x \rightarrow 0\left(\frac{\sqrt{x}}{2} \rightarrow 0\right)
$$

We get:

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

## Problem Kuznetsov Limits 17-2