$$\lim_{n \to \infty} \frac{n^2}{((n^2 + 1^2)(n^2 + 2^2) \dots (n^2 + n^2))^\frac{1}{n}}$$ equals:
Step-by-Step Solution
Key Concept: First factor \(n^2\) from each term:
\[
n^2+k^2=n^2\left(1+\frac{k^2}{n^2}\right).
\]
The \(n^2\) in the numerator cancels with the \(n^2\) coming from the \(n\)-th root of the product. The remaining expression becomes
\[
\left[
\prod_{k=1}^{n}\left(1+\frac{k^2}{n^2}\right)
\right]^{-1/n}.
\]
Now take logarithm and use the Riemann sum
\[
\frac{1}{n}\sum_{k=1}^{n}
\ln\left(1+\frac{k^2}{n^2}\right)
\longrightarrow
\int_0^1 \ln(1+x^2)\,dx.
\]
\subsection*{Question 2: Solution}
Let
\[
L=\lim_{n\to\infty}
\frac{n^2}{
\left(
(n^2+1^2)(n^2+2^2)\cdots(n^2+n^2)
\right)^{1/n}
}.
\]
For each \(k=1,2,\ldots,n\),
\[
n^2+k^2=n^2\left(1+\frac{k^2}{n^2}\right).
\]
Therefore
\[
\prod_{k=1}^{n}(n^2+k^2)
=n^{2n}\prod_{k=1}^{n}
\left(1+\frac{k^2}{n^2}\right).
\]
Taking the \(n\)-th root,
\[
\left[
\prod_{k=1}^{n}(n^2+k^2)
\right]^{1/n}
=n^2
\left[
\prod_{k=1}^{n}
\left(1+\frac{k^2}{n^2}\right)
\right]^{1/n}.
\]
Hence
\[
L=\lim_{n\to\infty}
\left[
\prod_{k=1}^{n}
\left(1+\frac{k^2}{n^2}\right)
\right]^{-1/n}.
\]
Taking logarithm,
\[
\ln L
=-\lim_{n\to\infty}
\frac{1}{n}\sum_{k=1}^{n}
\ln\left(1+\frac{k^2}{n^2}\right).
\]
This is a Riemann sum, so
\[
\ln L
=-\int_0^1 \ln(1+x^2)\,dx.
\]
Now
\[
\int \ln(1+x^2)\,dx
=x\ln(1+x^2)-2x+2\tan^{-1}x.
\]
Thus
\[
\int_0^1 \ln(1+x^2)\,dx
=\ln 2-2+\frac{\pi}{2}.
\]
So
\[
\ln L
=-\left(\ln 2-2+\frac{\pi}{2}\right)
=2-\frac{\pi}{2}-\ln 2.
\]
Therefore
\[
L=e^{2-\pi/2-\ln 2}
=\frac{1}{2}e^{2-\pi/2}.
\]
\[
\boxed{\frac{1}{2}e^{2-\pi/2}}
\]
Correct Answer: 4