$$f(x)=\lim_{n\rightarrow\infty}\left(\frac{n^n(x+n)(x+\frac{n}{2})....(x+\frac{n}{n})}{n! (x^2+n^2)(x^2+\frac{n^2}{4})...(x^2+\frac{n^2}{n^2})}\right)^\frac{x}{n} $$
max I could proceed was to take $\ln$ at both sides but not further. I need a reduced form of this $f(x)$ (which will be found by solving the lmit?)
Please help
Using the Riemann integral, one can get \begin{eqnarray} \ln f(x)&=&\lim_{n\rightarrow\infty}\ln\bigg[\left(\frac{n^n(x+n)(x+\frac{n}{2})....(x+\frac{n}{n})}{n! (x^2+n^2)(x^2+\frac{n^2}{4})...(x^2+\frac{n^2}{n^2})}\right)^\frac{x}{n}\bigg]\\ &=&\lim_{n\rightarrow\infty}\frac{x}{n}\bigg[n\ln n-\ln(n!)+\sum_{k=1}^n\ln(x+\frac{n}{k})-\sum_{k=1}^n (x^2+\frac{n^2}{k^2})\bigg]\\ &=&x+\lim_{n\rightarrow\infty}x\bigg[\sum_{k=1}^n\frac1n\ln(x+\frac{n}{k})-\sum_{k=1}^n \frac1n(x^2+\frac{n^2}{k^2})\bigg]\\\ &=&x+x\bigg[\int_0^1\ln(x+\frac1t)dt-\int_0^1\ln(x+\frac1{t^2})dt\bigg]. \end{eqnarray} It is easy to get this two integral and I omit the detail. Here $$ \lim_{n\rightarrow\infty}\frac{1}{n}\bigg[n\ln n-\ln(n!)\bigg]=1 $$ is used.
