A) 1
B) \[e\]
C) \[{{e}^{-1}}\]
D) \[\frac{1}{2}\]
Correct Answer: A
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{{{e}^{\tan x}}-{{e}^{x}}}{\tan x-x}=\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{{{e}^{x}}[{{e}^{\tan x-x}}-1]}{\tan x-x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\,{{e}^{x}}\,.\underset{x\to 0}{\mathop{\lim }}\,\,\frac{{{e}^{\tan x-x}}-1}{\tan x-x}={{e}^{0}}\times 1=1\].You need to login to perform this action.
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