Question:

Step 1: Express the limit in exponential form. Let the given limit be $L$. We have: $$ L = \lim_{n\to\infty}\left(\prod_{k=1}^{n}\dfrac{n^2}{n^2+k^2}\right)^{1/n} $$ This limit can be evaluated by first taking the natural logarithm: $$ \ln L = \lim_{n\to\infty}\dfrac{1}{n}\ln\left(\prod_{k=1}^{n}\dfrac{n^2}{n^2+k^2}\right) $$ Step 2: Simplify the logarithm of the product. The product term can be rewritten as: $$ \prod_{k=1}^{n}\dfrac{n^2}{n^2+k^2} = \prod_{k=1}^{n}\dfrac{1}{1+(k/n)^2} $$ Taking the logarithm, the product becomes a sum: $$ \ln\left(\prod_{k=1}^{n}\dfrac{1}{1+(k/n)^2}\right) = \sum_{k=1}^{n}\ln\left(\dfrac{1}{1+(k/n)^2}\right) = \sum_{k=1}^{n}-\ln\left(1+\left(\dfrac{k}{n}\right)^2\right) $$ Step 3: Convert the sum to a definite integral. Substituting this back into the expression for $\ln L$: $$ \ln L = \lim_{n\to\infty}\dfrac{1}{n}\sum_{k=1}^{n}-\ln\left(1+\left(\dfrac{k}{n}\right)^2\right) $$ This is a Riemann sum for the function $f(x) = -\ln(1+x^2)$ over the interval $[0,1]$. Therefore, the limit can be expressed as a definite integral: $$ \ln L = \int_0^1 -\ln(1+x^2) dx $$ Step 4: Evaluate the definite integral. We use integration by parts, $\int u\,dv = uv - \int v\,du$. Let $u = -\ln(1+x^2)$ and $dv = dx$. Then $du = -\dfrac{2x}{1+x^2}\,dx$ and $v = x$. $$ \int_0^1 -\ln(1+x^2) dx = \left[-x\ln(1+x^2)\right]_0^1 - \int_0^1 x\left(-\dfrac{2x}{1+x^2}\right) dx $$ $$ = \left(-1\cdot\ln(1+1^2) - 0\cdot\ln(1+0^2)\right) + \int_0^1 \dfrac{2x^2}{1+x^2} dx $$ $$ = -\ln(2) + \int_0^1 \dfrac{2(1+x^2)-2}{1+x^2} dx $$ $$ = -\ln(2) + \int_0^1 \left(2 - \dfrac{2}{1+x^2}\right) dx $$ $$ = -\ln(2) + \left[2x - 2\arctan(x)\right]_0^1 $$ $$ = -\ln(2) + (2(1) - 2\arctan(1)) - (2(0) - 2\arctan(0)) $$ $$ = -\ln(2) + 2 - 2\left(\dfrac{\pi}{4}\right) $$ $$ = 2 - \dfrac{\pi}{2} - \ln(2) $$ Step 5: Substitute the integral result back into the exponential form. Since $\ln L = 2 - \dfrac{\pi}{2} - \ln(2)$, the limit $L$ is: $$ L = e^{2 - \frac{\pi}{2} - \ln 2} $$ This can be further simplified using exponent properties: $$ L = e^{2 - \frac{\pi}{2}} \cdot e^{-\ln 2} = e^{2 - \frac{\pi}{2}} \cdot \dfrac{1}{e^{\ln 2}} = \dfrac{1}{2}e^{2 - \frac{\pi}{2}} $$