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.
 A: You are on the right track.
For $x \in [0,1]$, you have
$$
f'_n(x)=\frac{-x^2+n^2+1}{(x^2+n^2+1)^2}>0
$$ then
$$
\sup_{x\in[0,1]} \left(\frac{x}{x^2+n^2+1}\right)= \left.\left(\frac{x}{x^2+n^2+1}\right)\right|_{x=1}=\frac{1}{n^2+2} \to 0, \quad \text{as}\,\, n \to +\infty,
$$ and the convergence is uniform on $[0,1]$.
A: Actually you don't need to find the supremum.
Note that, since $0 \leq x \le1$ you and $x^2+n^2+1 \ge n^2+1$, you have
$$0 \le \frac{x}{x^2+n^2+1} \le \frac{1}{n^2+1}$$
Hence $\sup_{x \in [0,1]} |f_n(x)| \le \frac{1}{n^2+1} \to 0$ as $n \to + \infty$
A: $f_n(x)$ 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 $f_n(1)=\frac{1}{n^2+2}$.
A: Since the only roots of the derivative are $\pm\sqrt{n^2+1} \notin [0,1]$, the supremum is attained at the boundary. The left endpoint yields $0$ and the right one gives $\frac1{n^2+2} \to 0$.
$$\sup_{x\in[0,1]} |f_n(x)-f(x)| = \frac1{n^2+2}$$
