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

The given limit,

$ = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {1 + \sin x + \frac{{{{\sin }^2}x}}{{2!}} + \frac{{{{\sin }^3}x}}{{3!}} + ..........} \right) - \left( {1 - \sin x + \frac{{{{\sin }^2}x}}{{2!}} - \frac{{{{\sin }^3}x}}{{3!}} + ..........} \right) - 2\left( {x + \frac{{{x^3}}}{3} + \frac{2}{{15}}{x^5} + ..........} \right)}}{{\left( {x + \frac{{{x^3}}}{3} + \frac{2}{{15}}{x^5} + ..........} \right) - x}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{2\left( {\sin x + \frac{{{{\sin }^3}x}}{{3!}} + \frac{{{{\sin }^5}x}}{{5!}} + ..........} \right) - 2\left( {x + \frac{{{x^3}}}{3} + \frac{2}{{15}}{x^5} + ..........} \right)}}{{\frac{{{x^3}}}{3} + \frac{2}{{15}}{x^5} + ..........}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{2\left( {\sin x - x + \frac{{{{\sin }^3}x}}{{3!}} - \frac{{{x^3}}}{3} + \frac{{{{\sin }^5}x}}{{5!}} - \frac{2}{{15}}{x^5} + ..........} \right)}}{{\frac{{{x^3}}}{3} + \frac{2}{{15}}{x^5} + ..........}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{2\left( {x - \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} - ........... - x + \frac{{{{\sin }^3}x}}{{3!}} - \frac{{{x^3}}}{3} + \frac{{{{\sin }^5}x}}{{5!}} - \frac{2}{{15}}{x^5} + ..........} \right)}}{{\frac{{{x^3}}}{3} + \frac{2}{{15}}{x^5} + ..........}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{2\left( { - \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} - ........... + \frac{{{{\sin }^3}x}}{{3!}} - \frac{{{x^3}}}{3} + \frac{{{{\sin }^5}x}}{{5!}} - \frac{2}{{15}}{x^5} + ..........} \right)}}{{\frac{{{x^3}}}{3} + \frac{2}{{15}}{x^5} + ..........}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{2\left( { - \frac{1}{{3!}} + \frac{{{x^2}}}{{5!}} - ........... + \frac{{{{\sin }^3}x}}{{{x^3}3!}} - \frac{1}{3} + \frac{{{{\sin }^5}x}}{{{x^3}5!}} - \frac{2}{{15}}{x^2} + ..........} \right)}}{{\frac{1}{3} + \frac{2}{{15}}{x^2} + ..........}}$

$ = \frac{{2\left( { - \frac{1}{{3!}} + \frac{1}{{3!}} - \frac{1}{3}} \right)}}{{\frac{1}{3}}} =  - 2$

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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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