## Problem Statement

Calculate the limit of the function:

$$
\lim _{x \rightarrow 0} \ln \left(\left(e^{x^{2}}-\cos x\right) \cos \left(\frac{1}{x}\right)+\tan\left(x+\frac{\pi}{3}\right)\right)
$$
## Solution

Since $\cos \left(\frac{1}{x}\right)_{\text {- is bounded, and }}$

$$
\begin{aligned}
& \lim _{x \rightarrow 0} e^{x^{2}}-\cos x=e^{0^{2}}-\cos 0=e^{0}-1=1-1=0 \\
& \left(e^{x^{2}}-\cos x\right) \cos \left(\frac{1}{x}\right) \rightarrow 0, \text { then } \\
& \text {, as } x \rightarrow 0
\end{aligned}
$$

$$
\begin{aligned}
& \lim _{x \rightarrow 0} \ln \left(\left(e^{x^{2}}-\cos x\right) \cos \left(\frac{1}{x}\right)+\operatorname{tg}\left(x+\frac{\pi}{3}\right)\right)=\ln \left(0+\operatorname{tg}\left(0+\frac{\pi}{3}\right)\right)= \\
& =\ln \left(\operatorname{tg}\left(\frac{\pi}{3}\right)\right)=\ln \sqrt{3}
\end{aligned}
$$