## Problem Statement

Calculate the limit of the function:

$$
\lim _{x \rightarrow 2}\left(\frac{\sin (3 \pi x)}{\sin (\pi x)}\right)^{\sin ^{2}(x-2)}
$$
## Solution

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

Problem Kuznetsov Limits 20-5