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? 

Uniquesolution's comments answer the question, but in the hope a picture helps:

You're correct that if a sequence $(f_{n})$ of continuous functions converges to a discontinuous limit $f$, the convergence is not uniform.
Note carefully, however, that discontinuity of the limit is not necessary; a sequence of continuous functions can converge non-uniformly to a continuous limit. For example:


*

*If $f_{n}(x) = x^{n}$ for $0 \leq x < 1$, then $(f_{n}) \to 0$ pointwise, but the convergence is not uniform. (As the animation loop indicates, if $0 < a < 1$ is fixed, the convergence is uniform on $[0, a]$.)

*If $f_{n}(x) = x/n$ for $0 \leq x$, then $(f_{n}) \to 0$, but the convergence is not uniform.
(I leave to you the fun of finding a sequence of continuous functions $(f_{n})$ that converge pointwise to the continuous function $0$ on the compact set $[0, 1]$, but the convergence is not uniform.)
