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

The given limit,

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

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

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

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

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

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

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