grasping uniform convergence for function series I would like some help with my grasping the idea of uniformly convergence for a series of function.
I know that if $\lim_{n\rightarrow\infty}\sup|f_n(x)-f(x)| = 0$ than the series $ {f_n(x)}$ converges uniformly, where $f(x)$ is the limit function.
If we look at the function $x^n$ when $0\le x\le1$ than:
$f(x)$ is $0$ when $0\le x\lt1$ and $1$ when $x=1$
on one hand, the function $f(x)$ is not continuous so I think it's safe to assume the function series is not uniform convergence. is that enough to answer the question?
on the second hand, when i tried to use the notation above I got different result when I look at the two situations separately.   
when $0\le x\lt1$ than $\lim_{n\rightarrow\infty}\sup|f_n(x)-f(x)| =0$
when $x=1$ than $\lim_{n\rightarrow\infty}\sup|f_n(1)-f(1)| =0$
how to use the notation above for that question? 

The third line from bottom is false.
$$\sup_{[0,1)}|x^n|=1$$
