## Problem Statement

Calculate the limit of the function:

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

Substitution:

$x=y+1 \Rightarrow y=x-1$

$x \rightarrow 1 \Rightarrow y \rightarrow 0$

We get:

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

$$
\begin{aligned}
& =\lim _{y \rightarrow 0} \frac{\cos \left(\frac{\pi}{2} y+\frac{\pi}{2}\right)}{1-\sqrt{y+1}}=\lim _{y \rightarrow 0} \frac{-\sin \left(\frac{\pi}{2} y\right)(1+\sqrt{y+1})}{(1-\sqrt{y+1})(1+\sqrt{y+1})}= \\
& =\lim _{y \rightarrow 0} \frac{-\sin \left(\frac{\pi}{2} y\right)(1+\sqrt{y+1})}{1-(y+1)}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$\sin \left(\frac{\pi}{2} y\right) \sim \frac{\pi}{2} y_{, \text {as }} y \rightarrow 0\left(\frac{\pi}{2} y \rightarrow 0\right)$

We get:

$$
\begin{aligned}
& =\lim _{y \rightarrow 0} \frac{-\frac{\pi}{2} y(1+\sqrt{y+1})}{-y}=\lim _{y \rightarrow 0} \frac{\pi}{2}(1+\sqrt{y+1})= \\
& =\frac{\pi}{2}(1+\sqrt{0+1})=\frac{\pi}{2}(1+1)=\pi
\end{aligned}
$$

## Problem Kuznetsov Limits 13-28