## Problem Statement

$\lim _{x \rightarrow 0} \frac{\tan x \cdot \cos \left(\frac{1}{x}\right)+\log (2+x)}{\log (4+x)}$
## Solution

Since $\cos \left(\frac{1}{x}\right)_{\text {- is bounded, and }} \operatorname{tg} x \rightarrow 0$, as $x \rightarrow 0$, then

$$
\operatorname{tg} x \cdot \cos \left(\frac{1}{x}\right) \rightarrow 0, \text { as } x \rightarrow 0
$$

Then:

$\lim _{x \rightarrow 0} \frac{\operatorname{tg} x \cdot \cos \left(\frac{1}{x}\right)+\lg (2+x)}{\lg (4+x)}=\frac{0+\lg (2+0)}{\lg (4+0)}=\frac{\lg 2}{\lg 2^{2}}=\frac{\lg 2}{2 \lg 2}=\frac{1}{2}$