Showing that $f_n:[0,1]\to\mathbb{R}$ when $f_n=nx(1-x)^n$ converges pointwise but not uniformly Suppose $f_n:[0,1]\to\mathbb{R}$ such that $f_n=nx(1-x)^n$, then I need to show that $f_n$ converges pointwise but not uniformly. I can see that since $f$ is defined on $[0,1]$, the part $(1-x)^n$ is smaller than $1$ and therefore will converge, but what about $nx$? This will go to infinity...
Therefore, I can't even see that it converges pointwise, but let's suppose it converges to some function, how do I show that it doesn't converge uniformly? I should find an $\epsilon$ such that it's not valid for all $n>n_0(\epsilon)$ that $|f_n-something|<\epsilon$, right? 

Clearly $f_n(x)\to f(x)=0$ as $n\to\infty$ in [0,1]. Let $f_n'(x)=0$ and then $x=\frac{1}{n+1}$. It is easy to see that $f''_n(\frac{1}{n+1})<0$ and hence $f_n(x)$ reaches the maximum $f(\frac{1}{n+1})=(\frac{n}{n+1})^{n+1}$ at  $x=\frac{1}{n+1}$.
Thus
$$ \max |f_n(x)-f(x)|=(\frac{n}{n+1})^{n+1}\to\frac1e\neq0$$
as $n\to\infty$. Thus $f_n(x)$ does not converge uniformly to $0$.
