$\underset{x\to 0}{lim} \frac{2^{\left|x\right|}{e}^{\left|x\right|}-\left|x\right|-\left|x\right|{\log }_{e}2-1}{x\tan x}$ is equal to

Solution:

     $\underset{x\to 0}{lim} \frac{2^{\left|x\right|}{e}^{\left|x\right|}-\left|x\right|-\left|x\right|{\log }_{e}2-1}{x\tan x}$

$\begin{array}{l}=\underset{x\to 0}{lim} \frac{(2e{)}^{\left|x\right|}-|x|{\log }_e(2e)-1}{x^2\left(\frac{\tan x}{x}\right)}\\ =\underset{x\to 0}{lim} \frac{1+\left|x\right|{\log }_e\left(2e\right)+\frac{|x{|}^2}{2!}{\left({\log }_e\left(2e\right)\right)}^2+\dots -\left|x\right|{\log }_e\left(2e\right)-1}{x^2\left(\frac{\tan x}{x}\right)}\end{array}$

$\begin{array}{l}=\underset{x\to 0}{lim} \frac{\frac{|x{|}^2}{2!}{\left({\log }_e\left(2e\right)\right)}^2+\dots ..}{x^2\left(\frac{\tan x}{x}\right)}=\frac{1}{2!}{\left({\log }_e\left(2e\right)\right)}^2=\frac{1}{2}{\left({\log }_{e}2+1\right)}^2\\ =\frac{1}{2}(\ln 2+1{)}^2\end{array}$