If $\lim_{\alpha \to 0} \frac{e^{\cos(\alpha^n)} - e}{\alpha^m} = \frac{-e}{2}$ where $m$ and $n$ are positive integers greater than 1, then the value of $\frac{m}{n}$ is:
Step-by-Step Solution
Key Concept: Use the expansion near \(0\):
\[
\cos(\alpha^n)=1-\frac{\alpha^{2n}}{2}+O(\alpha^{4n}).
\]
Then
\[
e^{\cos(\alpha^n)}
=e^{1-\alpha^{2n}/2+O(\alpha^{4n})}
=e\left(1-\frac{\alpha^{2n}}{2}+O(\alpha^{4n})\right).
\]
So the numerator behaves like
\[
-\frac{e}{2}\alpha^{2n}.
\]
For the given limit to equal \(-e/2\), we need
\[
m=2n.
\]
We are given
\[
\lim_{\alpha\to 0}
\frac{e^{\cos(\alpha^n)}-e}{\alpha^m}
=-\frac{e}{2}.
\]
For small \(\alpha\),
\[
\cos(\alpha^n)
=1-\frac{\alpha^{2n}}{2}+O(\alpha^{4n}).
\]
Therefore
\[
e^{\cos(\alpha^n)}
=e^{1-\alpha^{2n}/2+O(\alpha^{4n})}.
\]
So
\[
e^{\cos(\alpha^n)}
=e\cdot e^{-\alpha^{2n}/2+O(\alpha^{4n})}.
\]
Using \(e^t=1+t+O(t^2)\),
\[
e^{-\alpha^{2n}/2+O(\alpha^{4n})}
=1-\frac{\alpha^{2n}}{2}+O(\alpha^{4n}).
\]
Hence
\[
e^{\cos(\alpha^n)}-e
=e\left(1-\frac{\alpha^{2n}}{2}+O(\alpha^{4n})\right)-e.
\]
Thus
\[
e^{\cos(\alpha^n)}-e
=-\frac{e}{2}\alpha^{2n}+O(\alpha^{4n}).
\]
So
\[
\frac{e^{\cos(\alpha^n)}-e}{\alpha^m}
=-\frac{e}{2}\alpha^{2n-m}+O(\alpha^{4n-m}).
\]
For the limit to be the non-zero finite value \(-e/2\), we must have
\[
2n-m=0.
\]
Therefore
\[
m=2n.
\]
Hence
\[
\frac{m}{n}=2.
\]
\[
\boxed{2}
\]
Correct Answer: 2