Does $x^n$ converges uniformly on $ [0,1)?$ Doing the limit we can see that in the open interval it converges pointwise to the constant function $f(x) = 0$. In the closed interval it doesn't converge uniformly because in $x=1$ $f(x) =1$ and when $0<x<1$ then $f(x)=0$. It isn't a continuous function although $f_n(x)$ is continuous for all $n$. So it can't be uniform. 
However,let $0<ε<1$, we know the function increases as $x$ increases. So, 
$|f_n(x) - f(x)| \leq ε^{n}$
And ε^(n) goes to 0 as n tends to infinity. Doesn't it mean that the convergence is uniform? 

No, $x^n$ does not converge to zero uniformly on $[0,1)$, because
$$\lim_{n\to\infty}\sup_{0\leq x<1}|x^n|=1\neq 0$$
