STD 11 - 12. limits — JEE कक्षा 11 साइन्स — Question

$ = \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}$.