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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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${\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$

The range of the function $f(x) = {\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$ is: (A) $[ - 2,2]$ (B) $\left[ {\frac{1}{{\sqrt 5 }},\sqrt 5 } \right]$ (C) $(0,\sqrt 5 )$ (D) $[ 0,2]$ Solution We have, $f(x) = {\log _{\sqrt 5 }}\left( {3 - 2\sin \frac{{3\pi }}{4}\sin x + 2\cos \frac{\pi }{4}\cos x} \right)$ $ \Rightarrow f(x) = {\log _{\sqrt 5 }}\left[ {3 + \sqrt 2 (\cos x - \sin x)} \right]$ Now, $ - \sqrt 2  \le \cos x - \sin x \le \sqrt 2 $ $\therefore - 2 \le \sqrt 2 (\cos x - \sin x) \le 2$ $\therefore 1 \le 3 + \sqrt 2 (\cos x - \sin x) \le 5$ $\therefore{\log _{\sqrt 5 }}1 \le {\log _{\sqrt 5 }}[3 + \sqrt 2 (\cos x - \sin x)] \le {\log _{\sqrt 5 }}5$ $ \Rightarrow 0 \le f(x) \le 2$ Answer: (D)
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