Question:

Step 1: Set up the limit and take logarithm. We need to find: $$L = \lim_{n\to\infty} \frac{n^2}{\left(\prod_{r=1}^{n}(n^2+r^2)\right)^{1/n}}$$ Taking the natural logarithm of both sides: $$\ln L = \lim_{n\to\infty}\left[2\ln n - \frac{1}{n}\sum_{r=1}^{n}\ln(n^2+r^2)\right]$$ Step 2: Factor out $n^2$ from inside the logarithm. Rewrite each term in the sum by factoring: $$\ln(n^2+r^2) = \ln\left(n^2\left(1+\frac{r^2}{n^2}\right)\right) = 2\ln n + \ln\left(1+\frac{r^2}{n^2}\right)$$ Substituting this back: $$\ln L = \lim_{n\to\infty}\left[2\ln n - \frac{1}{n}\sum_{r=1}^{n}\left(2\ln n + \ln\left(1+\frac{r^2}{n^2}\right)\right)\right]$$ Step 3: Simplify by canceling terms. Distribute the sum: $$\ln L = \lim_{n\to\infty}\left[2\ln n - \frac{2n\ln n}{n} - \frac{1}{n}\sum_{r=1}^{n}\ln\left(1+\frac{r^2}{n^2}\right)\right]$$ $$= \lim_{n\to\infty}\left[2\ln n - 2\ln n - \frac{1}{n}\sum_{r=1}^{n}\ln\left(1+\frac{r^2}{n^2}\right)\right]$$ Step 4: Convert the Riemann sum to an integral. The remaining sum is a Riemann sum. As $n \to \infty$, with $x = \frac{r}{n}$: $$\frac{1}{n}\sum_{r=1}^{n}\ln\left(1+\frac{r^2}{n^2}\right) \to \int_0^1 \ln(1+x^2)\,dx$$ Therefore: $$\ln L = -\int_0^1 \ln(1+x^2)\,dx$$ Step 5: Evaluate the integral using integration by parts. Let $u = \ln(1+x^2)$ and $dv = dx$. Then $du = \frac{2x}{1+x^2}dx$ and $v = x$. $$\int_0^1 \ln(1+x^2)\,dx = \left[x\ln(1+x^2)\right]_0^1 - \int_0^1 \frac{2x^2}{1+x^2}\,dx$$ $$= \ln 2 - \int_0^1 \frac{2x^2}{1+x^2}\,dx$$ Step 6: Simplify the remaining integral. Rewrite the integrand by dividing: $$\frac{2x^2}{1+x^2} = 2\left(\frac{x^2+1-1}{1+x^2}\right) = 2\left(1 - \frac{1}{1+x^2}\right)$$ Therefore: $$\int_0^1 \ln(1+x^2)\,dx = \ln 2 - 2\int_0^1\left(1 - \frac{1}{1+x^2}\right)dx$$ $$= \ln 2 - 2\left[x - \tan^{-1}x\right]_0^1$$ $$= \ln 2 - 2\left(1 - \frac{\pi}{4}\right) = \ln 2 - 2 + \frac{\pi}{2}$$ Step 7: Find $\ln L$ and solve for $L$. Since $\ln L = -\int_0^1 \ln(1+x^2)\,dx$: $$\ln L = -\left(\ln 2 - 2 + \frac{\pi}{2}\right) = -\ln 2 + 2 - \frac{\pi}{2}$$ Exponentiating both sides: $$L = e^{-\ln 2 + 2 - \pi/2} = e^{-\ln 2} \cdot e^{2 - \pi/2} = \frac{1}{2}e^{2-\pi/2}$$ **Final Answer:** The limit equals $\boxed{\dfrac{1}{2}e^{2-\frac{\pi}{2}}}$, which corresponds to **Option 3**. <div class="key-concept"><strong>Key Concept:</strong> Converting a limit of a product to a Riemann integral using logarithms</div> <div class="trap-box"><strong>Trap:</strong> Incorrectly factoring out $n^2$ from the product or making sign errors when computing the integral by parts.</div>