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

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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