## Problem Statement

Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\frac{2+x}{3-x}\right)^{x}$
## Solution

$\lim _{x \rightarrow 0}\left(\frac{2+x}{3-x}\right)^{x}=\left(\frac{2+0}{3-0}\right)^{0}=\left(\frac{2}{3}\right)^{0}=1$

## Problem Kuznetsov Limits 18-2