Search any question & find its solution
Question:
Answered & Verified by Expert
The value of $\operatorname{Lt}_{x \rightarrow 0}\left(\frac{1+5 x^2}{1+3 x^2}\right)^{\frac{1}{x^2}}$ is
Options:
Solution:
1167 Upvotes
Verified Answer
The correct answer is:
$\mathrm{e}^2$
Hints: $\operatorname{Lim}_{x \rightarrow 0}\left(\frac{1+5 x^2}{1+3 x^2}\right)^{\frac{1}{x^2}}=\mathrm{e}^{\operatorname{Lim}_{x \rightarrow 0} \frac{1}{x^2}\left(\frac{1+5 x^2}{1+3 x^2}-1\right)}=\mathrm{e}^{\operatorname{Lim}_{x \rightarrow 0} \frac{2 x^2}{x^2\left(1+3 x^2\right)}}=\mathrm{e}^2$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.