Find $\sup_{x\in[0,1]} \frac{x}{x^2+n^2+1}$ We have $f_n:[0,1]\to \mathbb{R},\:f_n(x)=\frac{x}{x^2+n^2+1}$ and we need to prove that is uniform convergence using formula: 

$\lim _{n\to \infty } \sup_{x\in[0,1]} |f_n(x)-f(x)| =0$


First step I prove that the sequence $f_n$ converges pointwise to $f(x)=0$. After it: $|f_n(x)-f(x)|=\frac{x}{x^2+n^2+1}$ and $(\frac{x}{x^2+n^2+1})'=\frac{-x^2+n^2+1}{(x^2+n^2+1)^2}$ but I don't understand what will help us to find supremum, in this case.
I need some help to find $\sup_{x\in[0,1]} \frac{x}{x^2+n^2+1}$.
P.S: The author says that  $\sup_{x\in[0,1]} \frac{x}{x^2+n^2+1}$ is equal with $\frac1{2n}$, but I think he's wrong, I'm not sure.

$f_n(x)$ is a positive function over $(0,1)$ and by the AM-GM inequality
$$ f_n(x)=\frac{x}{x^2+(n^2+1)}=\frac{1}{x+\frac{n^2+1}{x}}\leq\frac{1}{2\sqrt{n^2+1}}\tag{1} $$
with equality attained in $x=\sqrt{n^2+1}$. It follows that $f_n$ is an increasing function over $[0,1]$ and its supremum is simply given by $f_n(1)=\frac{1}{n^2+2}$.
