limx→0 2|x|e|x|−|x|−|x|loge⁡2−1xtan⁡x is equal to

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

1. A

$\frac{1}{2}\left(\mathrm{ln}2{\right)}^{2}+\frac{1}{2}\left(\mathrm{ln}2\right)+1$

2. B

$\left(\mathrm{ln}2{\right)}^{2}+\frac{1}{2}\left(\mathrm{ln}2\right)+1$

3. C

$\left(\mathrm{ln}2{\right)}^{2}+\left(\mathrm{ln}2\right)+\frac{1}{2}$

4. D

$\frac{1}{2}\left(\mathrm{ln}2{\right)}^{2}+\left(\mathrm{ln}2\right)+\frac{1}{2}$

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### Solution:

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

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

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

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