MCQ
$\mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {1 - \cos \,2x} \right)}^2}}}{{2x\,\tan \,x - x\,\tan \,2x}}$ is
- A$2$
- B$ - \frac{1}{2}$
- ✓$-2$
- D$ \frac{1}{2}$
$\, = \,\mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {{{\sin }^2}x} \right)}^2}}}{{2x\left( {x + \frac{{{x^3}}}{3} + \frac{{2{x^5}}}{{15}} + ....} \right) - x\left( {2x + \frac{{{2^3}{x^3}}}{3} + \frac{{{2^5}{x^5}}}{{15}} + ....} \right)}}$
$\, = \mathop {\lim }\limits_{x \to 0} \frac{{4{{\left( {x + \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} - ....} \right)}^4}}}{{{x^4}\left( {\frac{2}{3} - \frac{8}{5}} \right) + {x^6}\left( {\frac{4}{{15}} - \frac{{64}}{{15}}} \right)}}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{{4{{\left( {1 + \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} - ....} \right)}^4}}}{{ - 2 + {x^2}\left( { - \frac{{60}}{{15}}} \right) + ......}}$
(dividing numerator & denominator by ${{x^4}}$)
$=2$
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