I'm trying for too long time solve this but without success. I'm tired and need help.
Show that sequence of funcions converges pointwise but not uniformly convergent: $$ f_n(x)=\frac{n^2x}{1+n^3x^2} $$ Where $f_n(x): [0,1] \rightarrow R$
MY ATTEMPT
First I whot that conerges pointwise:
if $x\neq 0$ $$ \lim_{n\rightarrow+\infty}{\frac{n^2x}{1+n^3x^2}}=\lim_{n\rightarrow+\infty}{\frac{2nx}{3n^2x^2}}=\lim_{n\rightarrow+\infty}{\frac{2x}{6nx^2}}=0 $$
if $x = 0$ $$\lim_{n\rightarrow+\infty}{\frac{n^2x}{1+n^3x^2}}=0$$ So $f_n \rightarrow 0$.
Now i wanna show that convergence is not uniformly:
$$ |f_n(x)-f(x)|=|f_n(x)-0|=|f_n(x)|=f_n(x)=\frac{n^2x}{1+n^3x^2}>\frac{x}{1+n^3x^2} $$ ...
I don't know continue this!
Another approach from the comments above is to find the maximum value of the function $f_n(x)$ by setting its derivative to zero. I find that the maximum occurs where $x=n^{-3/2}$ and the maximum value is $\sqrt{n}/2$. So even as the function goes pointwise to zero the maximum value goes to infinity -- therefore there's no way to find $\epsilon$ that traps $f(x)$ within $\epsilon$ of zero for $x\in [0,1]$.