Question

Let $m$ and $n$ be two positive integers greater than $1$. If
$\underset{\alpha \to 0}{lim}\frac{e^{\cos \left({\alpha }^n\right)}-e}{{\alpha }^m}=-\frac{e}{2}$, then the value of $\frac{m}{n}$ is

Answer 1

$\underset{\alpha \to 0}{lim}\frac{e^{\cos \left({\alpha }^n\right)}-e}{{\alpha }^m}=e\underset{\alpha \to 0}{lim}\frac{e^{\cos \left({\alpha }^n\right)-1}-1}{{\alpha }^m}=e\underset{\alpha \to 0}{lim}\frac{e^{\cos \left({\alpha }^n\right)-1}-1}{\cos \left({\alpha }^n\right)-1}\times \frac{\cos \left({\alpha }^n\right)-1}{{\alpha }^m}=e\underset{\alpha \to 0}{lim}\frac{\cos \left({\alpha }^n\right)-1}{{\alpha }^m}=-e\underset{\alpha \to 0}{lim}\frac{2{\sin }^2\frac{{\alpha }^n}{2}}{{\alpha }^m}=-e\underset{\alpha \to 0}{lim}\frac{2{\sin }^2\frac{{\alpha }^n}{2}}{4{\left(\frac{{\alpha }^n}{2}\right)}^2}\times \frac{{\alpha }^{2n}}{{\alpha }^m}=-\frac{e}{2}{\alpha }^{2n-m}=-\frac{e}{2}\Rightarrow 2n-m=0\therefore \frac{m}{n}=2$