Show that a series is not uniformly convergent

Question

How can we show that $\sum_{n=1}^\infty x(1-x)^n$ is not uniformly convergent on $[0,1]$?

From the definition, if $f$ is uniformly convergent on a set $E$ of values of $x$, then for each $\epsilon>0$, an integer $N$ can be found such that $|f_n(x)-f(x)| < \epsilon$ for all $x \in E$.

Answer 1

Note that for $x\in (0,1)$ \begin{align*} \sum_{n=1}^{\infty}x(1-x)^{n} &=x\sum_{n=1}^{\infty}(1-x)^{n}=x\Big(\frac{1}{1-(1-x)}-1\Big)=1-x. \end{align*} And at $x=0$ and $x=1$ the series is equal to zero. Now recall that the uniform limit of continuous functions is a continuous function. Can you argue from here that the convergence can not be uniform?