MCQ
$\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{1 + 5{x^2}}}{{1 + 3{x^2}}}} \right)^{1/{x^2}}} = $
- ✓${e^2}$
- B$e$
- C${e^{ - 2}}$
- D${e^{ - 1}}$
✓
Answer
Correct option: A.
${e^2}$
a
(a) $\mathop {\lim }\limits_{x \to 0} \,{\left( {\frac{{1 + 5{x^2}}}{{1 + 3{x^2}}}} \right)^{1/{x^2}}} = A$
(a) $\mathop {\lim }\limits_{x \to 0} \,{\left( {\frac{{1 + 5{x^2}}}{{1 + 3{x^2}}}} \right)^{1/{x^2}}} = A$
$ \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$
$\therefore \;A\; = \;{e^2}$
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