Question:

<p><strong>Step 1:</strong> Rewrite the denominator as a nested exponent:</p><p>$$\left(1 + \frac{1}{n}\right)^{n^2} = \left[\left(1 + \frac{1}{n}\right)^n\right]^n$$</p><p><strong>Step 2:</strong> Use the fundamental limit $\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e$. For large $n$, let $\left(1 + \frac{1}{n}\right)^n = e - \delta_n$ where $\delta_n \to 0$.</p><p><strong>Step 3:</strong> Then:</p><p>$$\left(1 + \frac{1}{n}\right)^{n^2} = \left[\left(1 + \frac{1}{n}\right)^n\right]^n = (e - \delta_n)^n$$</p><p><strong>Step 4:</strong> As $n \to \infty$, $(e - \delta_n)^n \approx e^n \cdot \left(1 - \frac{\delta_n}{e}\right)^n$. More precisely, for large $n$:</p><p>$$\ln\left[\left(1 + \frac{1}{n}\right)^{n^2}\right] = n \ln\left(1 + \frac{1}{n}\right)^n = n\ln(e) + o(1) = n + o(n)$$</p><p><strong>Step 5:</strong> More carefully, using $\left(1 + \frac{1}{n}\right)^n = e^{1 - \frac{1}{2n} + O(n^{-2})}$:</p><p>$$\left(1 + \frac{1}{n}\right)^{n^2} = e^{n\left(1 - \frac{1}{2n} + O(n^{-2})\right)} = e^{n - \frac{1}{2} + O(n^{-1})}$$</p><p><strong>Step 6:</strong> Therefore:</p><p>$$\lim_{n \to \infty} \frac{e^n}{\left(1 + \frac{1}{n}\right)^{n^2}} = \lim_{n \to \infty} \frac{e^n}{e^{n-1/2}} = \lim_{n \to \infty} e^{1/2} = \sqrt{e}$$</p><p><strong>∴ Answer:</strong> The limit equals $\sqrt{e}$, which corresponds to option D.</p>