Question

Let $m$ and $n$ be two positive integers greater than 1. If

$\underset{\alpha \to 0}{lim}\left(\frac{e^{cos\left({\alpha }^n\right)}-e}{{\alpha }^m}\right)=-\left(\frac{e}{2}\right)$The value of $\frac{m}{n}$ is

Answer 1

$e\underset{\alpha \to 0}{lim}(\frac{e^{\cos {\alpha }^n-1}-1}{\cos {\alpha }^n-1}\cdot \frac{\cos {\alpha }^n-1}{{\alpha }^m})=\frac{-e}{2}$

$\Rightarrow \underset{\alpha \to 0}{lim}\left(\frac{1-\cos {\alpha }^n}{{\alpha }^m}\right)=\frac{1}{2}$

$\Rightarrow \underset{\alpha \to 0}{lim}\left(\frac{1-\cos {\alpha }^n}{{\left({\alpha }^n\right)}^2}\right)\cdot {\alpha }^{2n-m}=\frac{1}{2}$

$\Rightarrow 2n-m=0$

$\therefore \frac{m}{n}=2$