STD 11 - 12. limits — JEE STD 11 Science — Question

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

             $\, = \mathop {\lim }\limits_{x \to 0} \frac{{4{{\left( {x + \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} - ....} \right)}^4}}}{{{x^4}\left( {\frac{2}{3} - \frac{8}{5}} \right) + {x^6}\left( {\frac{4}{{15}} - \frac{{64}}{{15}}} \right)}}$

             $ = \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 ${{x^4}}$)

          $=2$