## Problem Statement

Calculate the limit of the function:

$$
\lim _{x \rightarrow 1} \tan\left(\cos x+\sin \frac{x-1}{x+1} \cdot \cos \frac{x+1}{x-1}\right)
$$
## Solution

Since $\cos \frac{x+1}{x-1}$ is bounded, and
$\lim _{x \rightarrow 1} \sin \frac{x-1}{x+1}=\sin \frac{1-1}{1+1}=\sin 0=0$, then

$\sin \frac{x-1}{x+1} \cdot \cos \frac{x+1}{x-1} \rightarrow 0 \quad$, as $x \rightarrow 1$

Then:

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