Let f(x)=limlimits_n → ∞≤ft(n^n(x+n)≤ft(x+n/2) ...≤ft(x+n/n))/(n !≤ft…

We have,

$f(x)=\lim\limits_{n \rightarrow \infty}\left\{\frac{n^n(x+n)\left(x+\frac{n}{2}\right) ...\left(x+\frac{n}{n}\right)}{n !\left(x^2+n^2\right)\left(x^2+\frac{n^2}{4}\right) ...\left(x^2+\frac{n^2}{n^2}\right)}\right\}^{x / n}$

$\Rightarrow \log f(x)=\lim\limits_{n \rightarrow \infty} \frac{x}{n} \log \left\{\prod_{r=1}^n \frac{\left(x+\frac{n}{r}\right)}{\left(x^2+\frac{n^2}{r^2}\right)} . \frac{n}{r}\right\}$

$\Rightarrow \log f(x)=\lim\limits_{n \rightarrow \infty} \frac{x}{n} \sum\limits_{r=1}^n \log \left\{\frac{1+\frac{r}{n} x}{1+\left(\frac{r}{n} x\right)^2}\right\}$

$\Rightarrow \log f(x)=x \lim\limits_{n \rightarrow \infty} \sum\limits_{r=1}^n \log \left\{\frac{1+\frac{r}{n} x}{1+\left(\frac{r}{n} x\right)^2}\right\} \frac{1}{n}$

$\Rightarrow \log f(x)=x \int\limits_0^1 \log \left\{\frac{1+t x}{1+(t x)^2}\right\} d t$

$\Rightarrow \log f(x)=\int\limits_0^x \log \left(\frac{1+u}{1+u^2}\right) d u$, where $u=t x$

$\Rightarrow \frac{f'(x)}{f(x)}=\log \left(\frac{1+x}{1+x^2}\right)$          .....(i)

We observe that $f(x)>0$ for all $x>0$.

Also, $\log \left(\frac{1+x}{1+x^2}\right)>0 \Leftrightarrow \frac{1+x}{1+x^2}>1 \Leftrightarrow x>x^2 \Leftrightarrow 0<x<1$

∴  $\frac{f'(x)}{f(x)}>0$ for $0<x<1$ and $\frac{f'(x)}{f(x)}<0$ for $x>1$

$\Rightarrow f'(x)>0$ for $0<x<1$ and $f'(x)<0$ for $x>1$          ...(ii)

$\Rightarrow f(x)$ is increasing on $(0,1)$ and decreasing on $(1, \infty)$.

$\Rightarrow f\left(\frac{1}{2}\right) \leq f(1)$ and $f\left(\frac{1}{3}\right) \leq f\left(\frac{2}{3}\right)$

So, option (b) is correct and option (a) is incorrect.

From (ii), we obtain

$f'(2)<0$. So, option (c) is correct.

From (i), we obtain

$\frac{f'(3)}{f(3)}-\frac{f'(2)}{f(2)}=\log \frac{4}{10}-\log \frac{3}{5}=\log \left(\frac{2}{3}\right)<0$

$\Rightarrow \frac{f'(3)}{f(3)}<\frac{f'(2)}{f(2)}$.

So, option (d) is not correct.