$f_n(x):=nx(1-x)^n$ Determine whether the sequence $(f_n)$ converges uniformly on $[0,1]$ I am having a bit of trouble on this revision question.
To determine pointwise convergence: $\lim_{n\rightarrow\infty} = nx(1-x)^n $. For $x=0, x=1$, it's clear that the limit is $0$. How can I determine the limit for $0 \le x \le 1$? (My limit finding skills are rusty). I'm quite sure it is $0$ as well, but how do I go about showing this properly?
Supposing this is true, then to determine uniform convergence, let $d_n(x):= |f_n(x)-f(x)|= nx(1-x)^n - 0$
Then $$ d_n'(x)= n(1-x)^n - n^2(1-x)x = 0$$ if $x=1$. So the maximum of $f_n(x)$ occurs at $x=1$. It follows that:
$$0 \le d_n(x)= |f_n(x)-f(x)|< d_n(1) = 0$$ So $(f_n)$ converges uniformly 
on $[0,1]$. Would this be correct?
Thanks for the help in advance!

Observe that $$\sup_{x \in [0,1]} |f_n(x)-f(x)| \geq \left|f_n \left(\frac{1}{n} \right) -f\left( \frac{1}{n} \right) \right|=\left(1-\frac{1}{n} \right)^n \to \frac{1}{e}$$
as $n \to \infty$ .So that we can't have $\lim_{n \to \infty} \sup_{x \in [0,1]} |f_n(x)-f(x)|=0$. i.e. there is no uniform convergence.
