## Condition of the problem

Calculate the limit of the function:

$$
\lim _{x \rightarrow 0} \sqrt{4 \cos 3 x+x \cdot \operatorname{arctg}\left(\frac{1}{x}\right)}
$$
## Solution

Since $\operatorname{arctg}\left(\frac{1}{x}\right)_{\text { is bounded, then }}$

$$
x \cdot \operatorname{arctg}\left(\frac{1}{x}\right) \rightarrow 0 \underset{\text { as } x \rightarrow 0}{ }
$$

Then:

$\lim _{x \rightarrow 0} \sqrt{4 \cos 3 x+x \cdot \operatorname{arctg}\left(\frac{1}{x}\right)}=\sqrt{4 \cos (3 \cdot 0)+0}=\sqrt{4}=2$