How to solve this challenging limit?

Question

I believe more than one answer may be correct. I thought about this for a while and I feel like there is a simplification that makes it look a whole lot easier but I can't figure out what it is.

I noticed that every term within the limit is multiplied so I figured that some terms might divide out but dividing $(x+n)$ by $(x^2+n^2)$ doesn't really seem to lead to any useful simplifications.

I thought about multiplying out everything on top and bottom of the inner expression but that again just seems to go nowhere and gets very messy and complicated without any reductions or simplifications.

Since they ask to take the derivative in one of the answer choices, surely there's some simplifications to be made to make this problem a lot easier.

$\mathbf 46.$ Let $\displaystyle f(x) = \lim_{n\to\infty} \left( \frac {\displaystyle n^n (x+n) \left(x+\frac n2\right) \ldots \left(x+\frac nn\right)} {\displaystyle n! (x^2+n^2) \left(x^2+\frac{n^2}4\right) \ldots \left(x^2+\frac{n^2}{n^2}\right)}\right)^{\dfrac xn}$, for all $x>0$. Then

(A) $f \left(\dfrac12\right) \geq f(1)$

(B) $f \left(\dfrac13\right) \leq f \left(\dfrac23\right)$

(C) $f'(2) \leq 0$

(D) $\dfrac {f'(3)}{f(3)} \geq \dfrac {f'(2)}{f(2)}$

Answer 1

\begin{align} \log f(x) &=\log\left(\frac{n^n}{n!}\prod_{k=1}^n\frac{x+\frac nk}{x^2+\frac{n^2}{k^2}}\right)^{\frac xn}\\ &=\frac xn\log\left(\frac{n!n^{2n}}{n!n^{2n}}\prod_{k=1}^n\frac{\frac{kx}n+1}{\left(\frac{kx}n\right)^2+1}\right)\\ &=\frac xn\sum_{k=1}^n\log\frac{\frac{kx}n+1}{\left(\frac{kx}n\right)^2+1}\\ &\xrightarrow{n\to\infty}\int_0^x\log\frac{t+1}{t^2+1}\mathrm dt \end{align} (Edit) Consequently, $f(x)>0$ for every $x>0$, $$\frac{f'(x)}{f(x)}=\log\frac{x+1}{x^2+1} \begin{cases} >0&0<x<1\\ =0&x=1\\ <0&x>1 \end{cases}$$ Thus $f$ is increasing for $0<x<1$, hence (A) is false, (B) is true, (C) is true and (D) is equivalent to $$\log\frac{4}{10}\ge\log\frac{3}{5}$$ which is false.