Condition of the problem

Calculate the limit of the function:

$\lim _{x \rightarrow \frac{\pi}{4}} \frac{1-\sin 2 x}{(\pi-4 x)^{2}}$
## Solution

Substitution:

$x=y+\frac{\pi}{4} \Rightarrow y=x-\frac{\pi}{4}$

$x \rightarrow \frac{\pi}{4} \Rightarrow y \rightarrow 0$

We get:

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

Using the substitution of equivalent infinitesimals:

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

We get:

$$
=\lim _{y \rightarrow 0} \frac{\frac{(2 y)^{2}}{2}}{16 y^{2}}=\lim _{y \rightarrow 0} \frac{2 y^{2}}{16 y^{2}}=\lim _{y \rightarrow 0} \frac{1}{8}=\frac{1}{8}
$$

## Problem Kuznetsov Limits 13-4