Is $f_n(x)=\frac {2nx}{1+n^2x^2}$ uniformly convergent?

Question

$f_n(x)=\frac {2nx}{1+n^2x^2}, 0\leq x \leq1 $

we know $f_n(0)\to 0$ and $f_n(1)\to 0$

and $f'_n(x)=0\to x=\pm\frac1n$

since for x<1/n $f'_n(x)>0$ x=1/n is max. point.

it seems $f_n\to^{uc}0 $ how can we show this? is this enough. if it is, for $x=\frac 1n$ doenst it u.con. to 1 since $f_n(\frac 1n)=1\ne0$?

and how can we choose N such that $n\ge N, \forall\epsilon>0$ and $|f_n(x)-0|<\epsilon$

Answer 1

For uniform convergence, it is necessary that $$\lim_{n \rightarrow \infty}\sup_{0\leq x \leq 1}|f_n(x)|=0$$

As you showed, $f_n$ has a maximum at $x = \frac1{n}$ and $f_n(\frac1{n}) = 1$ for all $n$. Therefore, the supremum is always $1$ and the convergence is not uniform.