Condition of the problem

Calculate the limit of the function:
$\lim _{x \rightarrow 0}(\cos \pi x)^{\frac{1}{x \cdot \sin \pi x}}$
Solution

$\lim _{x \rightarrow 0}(\cos \pi x)^{\frac{1}{x \cdot \sin \pi x}}=$

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

$=\lim _{x \rightarrow 0} e^{\ln (\cos \pi x) \cdot \frac{1}{x \cdot \sin \pi x}}=$

$=\exp \left\{\lim _{x \rightarrow 0} \ln (\cos \pi x) \cdot \frac{1}{x \cdot \sin \pi x}\right\}=$

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

Using the substitution of equivalent infinitesimals:

$\ln \left(1-2 \sin ^{2}\left(\frac{\pi x}{2}\right)\right) \sim-2 \sin ^{2}\left(\frac{\pi x}{2}\right)_{, \text {as }} x \rightarrow 0\left(-2 \sin ^{2}\left(\frac{\pi x}{2}\right) \rightarrow 0\right)$ $\sin \pi x \sim \pi x$, as $x \rightarrow 0(\pi x \rightarrow 0)$

We get:

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

Using the substitution of equivalent infinitesimals:

![](https://cdn.mathpix.com/cropped/2024_05_22_26d7e3a2c4258d4fcc5ag-12.jpg?height=136&width=851&top_left_y=1748&top_left_x=152)

We get:

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

## Problem Kuznetsov Limits 17-9