## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow \pi} \frac{1+\cos 3 x}{\sin ^{2} 7 x}$
## Solution

Substitution:

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

$x \rightarrow \pi \Rightarrow y \rightarrow 0$

We get:

$$
\begin{aligned}
& \lim _{x \rightarrow \pi} \frac{1+\cos 3 x}{\sin ^{2} 7 x}=\lim _{y \rightarrow 0} \frac{1+\cos 3(y+\pi)}{\sin ^{2} 7(y+\pi)}= \\
& =\lim _{y \rightarrow 0} \frac{1+\cos (3 y+3 \pi)}{\sin ^{2}(7 y+7 \pi)}=\lim _{y \rightarrow 0} \frac{1+\cos (3 y+\pi)}{\sin ^{2}(7 y+\pi)}= \\
& =\lim _{y \rightarrow 0} \frac{1-\cos 3 y}{(-\sin 7 y)^{2}}=\lim _{y \rightarrow 0} \frac{1-\cos 3 y}{\sin ^{2} 7 y}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$\sin 7 y \sim 7 y$, as $y \rightarrow 0(7 y \rightarrow 0)$

$1-\cos 3 y \sim \frac{(3 y)^{2}}{2}$, as $y \rightarrow 0$

We get:

$=\lim _{y \rightarrow 0} \frac{\frac{(3 y)^{2}}{2}}{(7 y)^{2}}=\lim _{y \rightarrow 0} \frac{9 y^{2}}{2 \cdot 49 y^{2}}=\lim _{y \rightarrow 0} \frac{9}{2 \cdot 49}=\frac{9}{98}$

## Problem Kuznetsov Limits 13-3