Answer:
Option A
Explanation:
Given $\lim_{\alpha \rightarrow0}\left(\frac{e^{\cos(\alpha^{n})}-e}{\alpha^{m}}\right)=-\left(\frac{e}{2}\right)$
$\Rightarrow$ $\lim_{\alpha \rightarrow0}\frac{e[e^{\cos(\alpha^{n})-1}-1].\cos (\alpha^{n})-1}{\cos(\alpha^{n})-1 \alpha^{m}}=\frac{-e}{2}$
$\Rightarrow$ $\lim_{\alpha \rightarrow 0}e\left\{\frac{e^{\cos(\alpha^{n})-1}-1}{\cos(\alpha^{n})-1}\right\}$ .
$\lim_{\alpha \rightarrow 0}\frac{-2\sin^{2}\frac{\alpha^{n}}{2}}{\alpha^{m}}$
=-e/2
$\Rightarrow$ $e\times1\times(-2)\lim_{\alpha \rightarrow 0}\frac{\sin^{2}\left(\frac{\alpha^{n}}{2}\right)}{\frac{\alpha^{2n}}{4}}.\frac{\alpha^{2n}}{4\alpha^{m}}=\frac{-e}{2}$
$\Rightarrow$ $e\times1\times(-2)\times 1\lim_{\alpha \rightarrow 0}\frac{\alpha^{2n-m}}{4}=\frac{-e}{2}$
For this to exist.
2n-m=0 $\Rightarrow$ $\frac{m}{n}=2$
