## Problem Statement

Calculate the limit of the function:

$$
\lim _{x \rightarrow 2}(\sin x)^{\frac{3}{1+x}}
$$
## Solution

$\lim _{x \rightarrow 2}(\sin x)^{\frac{3}{1+x}}=(\sin 2)^{\frac{3}{1+2}}=(\sin 2)^{1}=\sin 2$

## Problem Kuznetsov Limits 20-4