MCQ
$\mathop {\lim }\limits_{x \to 0} \frac{{x\tan 2x - 2x\tan x}}{{{{(1 - \cos 2x)}^2}}} = . . .$
- A$2$
- B$-2$
- ✓$\frac{1}{2}$
- D$ - \frac{1}{2}$
$ = \mathop {\lim }\limits_{x \to 0} \,\frac{{x(\tan \,\,2x - 2\tan x)}}{{{{(2\,{{\sin }^2}x)}^2}}} = \mathop {\lim }\limits_{x \to 0} \,\,\frac{1}{4}\,\frac{{x\,(\tan 2x - 2\tan x)}}{{{{\sin }^4}x}}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{1}{4}\frac{{x\left\{ {\left( {2x + \frac{1}{3}{{(2x)}^3} + \frac{2}{{15}}\,{{(2x)}^5} + ...} \right) - 2\left( {x + \frac{{{x^3}}}{3} + \frac{2}{{15}}{x^5} + ...} \right)} \right\}}}{{{x^4}\,{{\left( {1 - \frac{{{x^2}}}{{3\,\,!}} + \frac{{{x^4}}}{{5\,\,!}} + ....} \right)}^4}}}$
$ = \frac{1}{4}\,.\,\left( {\frac{8}{3} - \frac{2}{3}} \right) = \frac{2}{4} = \frac{1}{2}$.
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