## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 1} \frac{\sin x+\sin (\pi x) \cdot \operatorname{arctg} \frac{1+x}{1-x}}{1+\cos x}$
## Solution

Since $\operatorname{arctg} \frac{1+x}{1-x}_{\text {- is bounded, and }}$

$$
\begin{aligned}
& \lim _{x \rightarrow 1} \sin (\pi x)=\sin (\pi \cdot 1)=\sin \pi=0 \\
& \sin (\pi x) \cdot \operatorname{arctg} \frac{1+x}{1-x} \rightarrow 0, \text { then } \\
& \text {, as } x \rightarrow 0
\end{aligned}
$$

Then:

$\lim _{x \rightarrow 1} \frac{\sin x+\sin (\pi x) \cdot \operatorname{arctg} \frac{1+x}{1-x}}{1+\cos x}=\frac{\sin 1+0}{1+\cos 1}=\frac{\sin 1}{1+\cos 1}$