Ratio of Exponents in Exponential Limit Evaluation

To find the value of $\frac{m}{n}$, we begin with the given limit:

$\lim_{\alpha \to 0} \left( \frac{e^{\cos(\alpha^n)} - e}{\alpha^m} \right) = -\frac{e}{2}$

Factoring out $e$ from the numerator, we get:

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

Multiply and divide the expression inside the limit by $(\cos(\alpha^n) - 1)$:

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

Since $\lim_{x \to 0} \frac{e^x - 1}{x} = 1$ (where $x = \cos(\alpha^n) - 1 \to 0$ as $\alpha \to 0$), the first limit evaluates to $1$:

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

Simplifying by cancelling $e$ on both sides:

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

Using the Taylor series expansion for $\cos(x) = 1 - \frac{x^2}{2!} + O(x^4)$, we substitute $x = \alpha^n$:

$\cos(\alpha^n) - 1 = -\frac{\alpha^{2n}}{2} + O(\alpha^{4n})$

Substitute this back into the limit:

$\lim_{\alpha \to 0} \left( \frac{-\frac{\alpha^{2n}}{2} + O(\alpha^{4n})}{\alpha^m} \right) = -\frac{1}{2}$

$-\frac{1}{2} \cdot \lim_{\alpha \to 0} \alpha^{2n - m} = -\frac{1}{2}$

For this limit to evaluate to a non-zero finite value (specifically $-\frac{1}{2}$), the exponent of $\alpha$ must be $0$:

$2n - m = 0 \implies m = 2n$

Therefore, the ratio $\frac{m}{n}$ is:

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