## Problem Statement

Calculate the limit of the function:

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

Substitution:

$$
\begin{aligned}
& x=y+\pi \Rightarrow y=x-\pi \\
& x \rightarrow \pi \Rightarrow y \rightarrow 0
\end{aligned}
$$

We get:

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

Using the substitution of equivalent infinitesimals:

$1-\cos y \sim \frac{y^{2}}{2}$, as $y \rightarrow 0$

We get:

$=\lim _{y \rightarrow 0} \frac{\frac{y^{2}}{2}}{-y}=\lim _{y \rightarrow 0}-\frac{y}{2}=-\frac{0}{2}=0$

## Problem Kuznetsov Limits 13-24