JEE Advanced 2015 Paper 2Question 1
Let $m$ and $n$ be two positive integers greater than 1. If $$\lim_{\alpha\to0}\left(\frac{e^{\cos(\alpha^n)}-e}{\alpha^m}\right)=-\left(\frac{e}{2}\right)$$ then the value of $\frac{m}{n}$ is
Limits & Continuity
Answer$2$
Solution
$e^{\cos\alpha^n}-e=e(e^{\cos\alpha^n-1}-1)\approx-\frac e2\alpha^{2n}$. The limit is $-\frac e2$ iff $m=2n$, so $\frac mn=2$.