## Problem Statement

Calculate the limit of the function:

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

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

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

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

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

Using the substitution of equivalent infinitesimals:

$\ln \left(1-\left(e^{x^{2}}-1\right)\right) \sim-\left(e^{x^{2}}-1\right)$, as $x \rightarrow 0\left(-\left(e^{x^{2}}-1\right) \rightarrow 0\right)$

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

We get:

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

Using the substitution of equivalent infinitesimals:

$e^{x^{2}}-1 \sim x^{2}$, as $x \rightarrow 0\left(x^{2} \rightarrow 0\right)$

We get:

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

## Problem Kuznetsov Limits 17-24