## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\ln 2 x-\ln \pi}{\sin \left(\frac{5 x}{2}\right) \cos x}$
## Solution

$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\ln 2 x-\ln \pi}{\sin \left(\frac{5 x}{2}\right) \cos x}=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\ln \frac{2 x}{\pi}}{\sin \left(\frac{5 x}{2}\right) \cos x}=$

Substitution:

$x=y+\frac{\pi}{2} \Rightarrow y=x-\frac{\pi}{2}$

$x \rightarrow \frac{\pi}{2} \Rightarrow y \rightarrow 0$

We get:

$$
\begin{aligned}
& =\lim _{y \rightarrow 0} \frac{\ln \frac{2\left(y+\frac{\pi}{2}\right)}{\pi}}{\sin \left(\frac{5}{2}\left(y+\frac{\pi}{2}\right)\right) \cos \left(y+\frac{\pi}{2}\right)}=\lim _{y \rightarrow 0} \frac{\ln \left(\frac{2 y}{\pi}+1\right)}{\sin \left(\frac{5}{2} y+\frac{5 \pi}{4}\right)(-\sin y)}= \\
& =\lim _{y \rightarrow 0} \frac{\ln \left(\frac{2 y}{\pi}+1\right)}{-\sin \left(\frac{5}{2} y+\frac{\pi}{4}\right)(-\sin y)}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$\sin y \sim y$, as $y \rightarrow 0$

$\ln \left(1+\frac{2 y}{\pi}\right) \sim \frac{2 y}{\pi}$, as $y \rightarrow 0\left(\frac{2 y}{\pi} \rightarrow 0\right)$

We get:

$=\lim _{y \rightarrow 0} \frac{\frac{2 y}{\pi}}{-\sin \left(\frac{5}{2} y+\frac{\pi}{4}\right)(-y)}=\lim _{y \rightarrow 0} \frac{2}{\pi \sin \left(\frac{5}{2} y+\frac{\pi}{4}\right)}=$

$$
=\frac{2}{\pi \sin \left(\frac{5}{2} \cdot 0+\frac{\pi}{4}\right)}=\frac{2}{\pi \sin \frac{\pi}{4}}=\frac{2}{\pi \frac{\sqrt{2}}{2}}=\frac{2 \sqrt{2}}{\pi}
$$

## Problem Kuznetsov Limits 13-17