Min/Max of $f_{n}(x)=\lim_{n\to\infty}\frac{(n^{x+1}+1^{x})(n^{x+1}+2^{x})\cdots(n^{x+1}+n^{x})}{(n^{x+1}-1^{x})(n^{x+1}-2^{x})\cdots(n^{x+1}-n^{x})}$ Let's consider the function $f_{n}(x)$ with $x>0$ defined as:
$$f_{n}(x)=\lim_{n\to\infty}\frac{(n^{x+1}+1^{x})(n^{x+1}+2^{x})\cdots(n^{x+1}+n^{x})}{(n^{x+1}-1^{x})(n^{x+1}-2^{x})\cdots(n^{x+1}-n^{x})}$$
I'd like to know what is the way to follow such that I may find out for what values of $x$ the function reaches its minimum and maximum , and then to compute these  values. It's a
problem that came to my mind after studying another limit. However, I have more questions regarding this function, but for the moment it's enough if I find an answer for the part with minimum/maximum. Thanks.
 A: First write
\begin{align*}
\log f_n(x) &= \sum_{j=1}^n \log(n^{x+1}+j^x) - \sum_{j=1}^n \log(n^{x+1}-j^x) \\
&= \sum_{j=1}^n \log\bigg( 1+\frac{j^x}{n^{x+1}} \bigg) + \sum_{j=1}^n \log\bigg( 1- \frac{j^x}{n^{x+1}} \bigg)^{-1}.
\end{align*}
The following inequalities are valid for all $0<y<\frac12$:
\begin{align*}
y-y^2 &< \log(1+y) < y \\
y &< \log(1-y)^{-1} < y+y^2
\end{align*}
Therefore when $n\ge2$,
\begin{align*}
\log f_n(x) &> \sum_{j=1}^n \bigg( \frac{j^x}{n^{x+1}} - \bigg( \frac{j^x}{n^{x+1}} \bigg)^2  \bigg) + \sum_{j=1}^n \frac{j^x}{n^{x+1}} \\
&> 2\sum_{j=1}^n \frac1n \bigg( \frac jn \bigg)^x - \sum_{j=1}^n \frac1{n^2} \\
&= 2\sum_{j=1}^n \frac1n \bigg( \frac jn \bigg)^x - \frac1n
\end{align*}
and
\begin{align*}
\log f_n(x) &< \sum_{j=1}^n \frac{j^x}{n^{x+1}} + \sum_{j=1}^n \bigg( \frac{j^x}{n^{x+1}} + \bigg( \frac{j^x}{n^{x+1}} \bigg)^2 \bigg) \\
&< 2\sum_{j=1}^n \frac1n \bigg( \frac jn \bigg)^x + \sum_{j=1}^n \frac1{n^2} \\
&= 2\sum_{j=1}^n \frac1n \bigg( \frac jn \bigg)^x + \frac1n.
\end{align*}
By the squeeze theorem,
$$
\lim_{n\to\infty} \log f_n(x) = 2 \lim_{n\to\infty} \sum_{j=1}^n \frac1n \bigg( \frac jn \bigg)^x = 2 \int_0^1 u^x\, du = \frac2{x+1},
$$
because the sum is a Riemann sum for the integral. In other words,
$\lim_{n\to\infty} f_n(x) = e^{2/(x+1)}$
for all $x>0$ (since exponentiation is continuous).
