## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0} \sqrt[x^{2}]{2-\cos x}$
## Solution

$\lim _{x \rightarrow 0} \sqrt[x^{2}]{2-\cos x}=\lim _{x \rightarrow 0}(2-\cos x)^{\frac{1}{x^{2}}}=$

$=\lim _{x \rightarrow 0}\left(e^{\ln (2-\cos x)}\right)^{\frac{1}{x^{2}}}=\lim _{x \rightarrow 0} e^{\frac{\ln (2-\cos x)}{x^{2}}}=$

$=\exp \left\{\lim _{x \rightarrow 0} \frac{\ln (1+(1-\cos x))}{x^{2}}\right\}=$

Using the substitution of equivalent infinitesimals:

$\ln (1+(1-\cos x)) \sim 1-\cos x$, as $x \rightarrow 0(1-\cos x \rightarrow 0)$

We get:

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

Using the substitution of equivalent infinitesimals:
$1-\cos x \sim \frac{x^{2}}{2}$, as $x \rightarrow 0$

We get:

$=\exp \left\{\lim _{x \rightarrow 0} \frac{\frac{x^{2}}{2}}{x^{2}}\right\}=\exp \left\{\lim _{x \rightarrow 0} \frac{1}{2}\right\}=e^{\frac{1}{2}}=\sqrt{e}$

## Problem Kuznetsov Limits 17-20