This is the sequence $f_n(x) = nxe^{−nx}$ on $(0,\infty)$.
Well, i can see that $f_n$ tend to $0$ pointwise for every $x \in (0,\infty)$.
But if i want to check if $f_n$ tend to $0$ uniformly in $(0,\infty)$, i used the technique and found that for every $n$ the sup of $f_n$ is in $x = \dfrac{1}{n}$.
For every n, $f_n(1/n)$ = $1/e$, So:
$ \sup| f_n(x)-f(x) | = \dfrac{1}{e} \ne 0$
And therefore, $f_n$ does not converge uniformly to $0$ in $(0,\infty)$.
This is fine proof or am i entirely wrong?
Second question:
Thanks guys. Here is another question: $f_n(x)=\frac{nx}{1 + nx}$ on $(0,1)$
If i want to use the same technique i used above, i get $f'_n(x)$ = n for every $n$ and for every x. What can i do to show that is NOT uniformly converges?
For your second question:
Hint: if a sequence of continuous functions converges uniformly to some $f$, then $f$ is continuous.