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

limx→0 2|x|e|x|−|x|−|x|loge⁡2−1xtan⁡x=limx→0 (2e)|x|−|x|loge⁡(2e)−1x2tan⁡xx=limx→0 1+|x|loge⁡(2e)+|x|22!loge⁡(2e)2+…−|x|loge⁡(2e)−1x2tan⁡xx=limx→0 |x|22!loge⁡(2e)2+…..x2tan⁡xx=12!loge⁡(2e)2=12loge⁡2+12=12(ln⁡2+1)2

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

limx→0 2|x|e|x|−|x|−|x|loge⁡2−1xtan⁡x=limx→0 (2e)|x|−|x|loge⁡(2e)−1x2tan⁡xx=limx→0 1+|x|loge⁡(2e)+|x|22!loge⁡(2e)2+…−|x|loge⁡(2e)−1x2tan⁡xx=limx→0 |x|22!loge⁡(2e)2+…..x2tan⁡xx=12!loge⁡(2e)2=12loge⁡2+12=12(ln⁡2+1)2

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$lim_{x \to 0} \frac{2^{| x |} e^{| x |} - | x | - | x | \log_{e} 2 - 1}{x \tan x}$ is equal to

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

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

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

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

answer is D.

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Detailed Solution

$lim_{x \to 0} \frac{2^{| x |} e^{| x |} - | x | - | x | \log_{e} 2 - 1}{x \tan x}$

$\begin{array}{l} = lim_{x \to 0} \frac{(2 e)^{| x |} - | x | \log_{e} (2 e) - 1}{x^{2} (\frac{\tan x}{x})} \\ = lim_{x \to 0} \frac{1 + | x | \log_{e} (2 e) + \frac{| x |^{2}}{2!} {(\log_{e} (2 e))}^{2} + \dots - | x | \log_{e} (2 e) - 1}{x^{2} (\frac{\tan x}{x})} \end{array}$

$\begin{array}{l} = lim_{x \to 0} \frac{\frac{| x |^{2}}{2!} {(\log_{e} (2 e))}^{2} + \dots . .}{x^{2} (\frac{\tan x}{x})} = \frac{1}{2!} {(\log_{e} (2 e))}^{2} = \frac{1}{2} {(\log_{e} 2 + 1)}^{2} \\ = \frac{1}{2} (\ln 2 + 1)^{2} \end{array}$