Question:

Step 1: Set up the limit expression and factor out the constant. We are given: $$\lim_{\alpha \to 0} \frac{e^{\cos(\alpha^n)} - e}{\alpha^m} = \frac{-e}{2}$$ Factor out $e$ from the numerator: $$\lim_{\alpha \to 0} \frac{e\left(e^{\cos(\alpha^n)-1} - 1\right)}{\alpha^m}$$ Step 2: Apply Taylor expansion to $\cos(\alpha^n) - 1$. As $\alpha \to 0$, we use the Taylor series expansion of cosine: $$\cos(\alpha^n) = 1 - \frac{(\alpha^n)^2}{2} + O(\alpha^{4n})$$ Therefore: $$\cos(\alpha^n) - 1 \approx -\frac{\alpha^{2n}}{2}$$ Step 3: Apply Taylor expansion to the exponential term. For small values of $x$, we know that $e^x - 1 \approx x$. Since $\cos(\alpha^n) - 1 \to 0$ as $\alpha \to 0$: $$e^{\cos(\alpha^n)-1} - 1 \approx \cos(\alpha^n) - 1 \approx -\frac{\alpha^{2n}}{2}$$ Step 4: Substitute the approximation into the limit. The limit now becomes: $$\lim_{\alpha \to 0} \frac{e \cdot \left(-\frac{\alpha^{2n}}{2}\right)}{\alpha^m} = \frac{-e}{2} \lim_{\alpha \to 0} \alpha^{2n-m}$$ Step 5: Determine the condition for the limit to be finite and nonzero. For the limit to equal $\frac{-e}{2}$ (a finite, nonzero value), the exponent of $\alpha$ must be zero. This means: $$2n - m = 0$$ Therefore: $$m = 2n$$ Step 6: Calculate the ratio $\frac{m}{n}$. $$\frac{m}{n} = \frac{2n}{n} = 2$$ The value of $\dfrac{m}{n}$ is $\boxed{2}$, which corresponds to **Option 1**.