## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 1} \frac{\cos (2 \pi x)}{2+\left(e^{\sqrt{x-1}}-1\right) \operatorname{arctg} \frac{x+2}{x-1}}$
Solution

Since $\operatorname{arctg} \frac{x+2}{x-1}$ is bounded, and $e^{\sqrt{x-1}}-1 \rightarrow 0$, as $x \rightarrow 1$, then

$\left(e^{\sqrt{x-1}}-1\right) \operatorname{arctg} \frac{x+2}{x-1} \rightarrow 0 \quad$, as $x \rightarrow 1$

Then:

$\lim _{x \rightarrow 1} \frac{\cos (2 \pi x)}{2+\left(e^{\sqrt{x-1}}-1\right) \operatorname{arctg} \frac{x+2}{x-1}}=\frac{\cos (2 \pi \cdot 0)}{2+0}=\frac{\cos 0}{2}=\frac{1}{2}$