Let $f_n$ be a sequence of real valued functions defined on $[0,1]$. $$f_n(x)=\frac{2x^2}{x^2+(1-2nx)^2}$$
I want to talk about convergence of this sequence of functions:
Dividing by $n^2$, we get
$$f_n(x)=\frac{2x^2/n^2}{x^2/n^2+(1/n-2x)^2}$$
Fix $x$, now taking limit as $n\to \infty$
we get $\lim f_n(x)=\frac{0}{0+(0-2x)}=0$
So the point wise limit of $f(x)=0$
For Uniform convergence,
Consider $M_n=\sup \left|\frac{2x^2}{x^2+(1-2x)^2}\right|$
This looks complicated. One way is to find critical points(keeping $n$ as constant) but this makes expression more complicated.
Is there an easy way to do it?
Hint: $$f_n\biggr(\frac1{2n}\biggr)=2.$$ Now use a proof by contradiction to show that the sequence is not uniformly convergent.