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$\mathop {\lim }\limits_{x \to 0} {e^{\frac{{\tan x - \sin x}}{{{{\sin }^3}x}}}} = ?$

The given limit = ${e^{\mathop {\lim }\limits_{x \to 0} \frac{{\tan x - \sin x}}{{{{\sin }^3}x}}}}$

$ = {e^{\mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{\cos x}} - 1}}{{{{\sin }^2}x}}}}$

$ = {e^{\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x}}{{\cos x.{{\sin }^2}x}}}}$

$ = {e^{\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x}}{{\cos x.(1 - {{\cos }^2}x)}}}}$

$ = {e^{\mathop {\lim }\limits_{x \to 0} \frac{1}{{\cos x.(1 + \cos x)}}}}$

$ = {e^{1/2}} = \sqrt e $

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