How to evaluate this limit?? 
Evaluate: $$\ f(x)= \lim_{n\rightarrow \infty}\left( \dfrac{n^n(x+n)\left( x+\dfrac{n}{2}\right)\left( x+\dfrac{n}{3}\right)... \left( x+\dfrac{n}{n}\right)}{n!(x^2+n^2)\left( x^2+\dfrac{n^2}{4}\right)\left( x^2+\dfrac{n^2}{9}\right)...\left( x^2+\dfrac{n^2}{n^2}\right)}\right)^{\dfrac{x}{n}}$$

$x\in R^+$
Find the coordinates of the maxima of $f(x)$.
 A: We solve it using Riemann sum. You can write it as $$\left(\frac {(\frac {x}{n}+1)\cdots (\frac {nx}{n}+1)}{((\frac{x}{n})^2+1)\cdots ((\frac {nx}{n})^2+1)}\right)^{x/n}$$
Now take logs and use the fact of Riemann sum. Put $\dfrac {1}{n}=\mathrm dt,\dfrac {r}{n}=t$ in numerator and denominator where $r$ varies from $1-n $ thus the limit changes to integration. The equation becomes $\displaystyle \ln f (x))=x\int _0 ^1\ln \left(\frac {xt+1}{x^2t^2+1}\right)\mathrm dt $
Note we are integrating wrt $t $ so treat $x$ as constant. Put $xt=p $ thus $x\, \mathrm dt=\mathrm dp $ . Now limits change from $0$ to $x$ so we have $\displaystyle \ln (f (x))=\int _0 ^x \ln \left(\frac {1+p}{1+p^2}\right)\, \mathrm dp. $ Now use log property to separate the integral and then by parts . So we get $\displaystyle \ln (f (x))=(x+1)(\ln (x+1)-1)-\int _0 ^1 \frac {2p^2+2-2 }{1+p^2}\, \mathrm dp $ thus the result is $\ln (f (x))=(x+1)(\ln (x+1)-1)+1-x\ln (1+x^2)+2x-2\arctan (x)$ .
Now you can see that $f'(x)=\ln (1+x)-\ln (1+x^2) $ thus maxima is achived either at $0$ or $1$ . Using second derivative test we see the maxima is at $1$ thus putting $x=1$ we have $f (1)=e^{\ln (2)+1-\frac {\pi}{2}}=2.e.e^{-pi/2}=2ei^i\approx 1.12$ thus coordinates are $(1,1.13)$.
