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

$ \Rightarrow \;{\log _e}A\; = \;\mathop {\lim }\limits_{x \to 0} \frac{1}{{{x^2}}}\log \;\left( {1 + \frac{{2{x^2}}}{{1 + 3{x^2}}}} \right)$

$ = \mathop {\lim }\limits_{x \to 0} \frac{1}{{{x^2}}}\left[ {\frac{{2{x^2}}}{{1 + 3{x^2}}} - \frac{1}{2}\left( {\frac{{2{x^2}}}{{1 + 3{x^2}}}} \right)\;\; + \;\;...} \right]$

$ = \;\mathop {\lim }\limits_{x \to 0} \left[ {\frac{2}{{1 + 3{x^2}}} - \frac{1}{2}\;\;\frac{{2{x^2}}}{{{{(1 + 3{x^2})}^2}}}\;\; + \;\;...} \right]\;\; = \;\;2$

$\therefore \;A\; = \;{e^2}$