## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\ln (\sin x)}{(2 x-\pi)^{2}}$
## Solution

Substitution:

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

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

We get:

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

Using the substitution of equivalent infinitesimals:

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

We get:

$=\lim _{y \rightarrow 0} \frac{-2 \sin ^{2} \frac{y}{2}}{4 y^{2}}=$

Using the substitution of equivalent infinitesimals:

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

We get:

$=\lim _{y \rightarrow 0} \frac{-2\left(\frac{y}{2}\right)^{2}}{4 y^{2}}=\lim _{y \rightarrow 0} \frac{-\left(\frac{1}{2}\right)}{4}=\lim _{y \rightarrow 0}-\frac{1}{8}=-\frac{1}{8}$

## Problem Kuznetsov Limits 14-26