Why does the sequence of functions $f_n(x)=nx^n(1-x)$ not converge uniformly? Let $(f_n)_n$ be a sequence of functions defined by $$f_n : [0,1] \rightarrow \mathbb{R}: x \mapsto nx^n(1-x). $$
We have $\lim_{n \to \infty} f_n(x) = 0$ for all $x$. My textbook says the convergence isn't uniformly though, and I don't understand why.
I computed $$\sup_{x \in [0,1]} |f_n(x) - 0 | = \sup_{x \in [0,1]} | nx^n (1-x) | = \ ? $$ 
How do I figure out this supremum? 
When I use the definition, I see that $$| nx^n(1-x) - 0 | = | nx^n - nx^{n+1}| \leq | nx^n | + | nx^{n+1}| \leq n + n = 2n. $$ And I can never get this smaller than $\epsilon$. But I couldn't find an explicit $\epsilon > 0$ and a $x \in [0,1]$ such that $ |f_n(x)| \geq \epsilon. $
Any help is appreciated.
 A: Hint:
Define
$$f_n(x)=nx^n(1-x)\implies f'_n(x)= n^2x^{n-1}(1-x)-nx^n=nx^{n-1}\left(n(1-x)-x\right)=$$
$$=nx^{n-1}\left(n-(1+n)x\right)$$
so if $\;x\in (0,1)\;$ , we get that
$$f_n'(x)=0\iff x=\frac n{1+n}$$
and it's easy to check the above is a maximum , so
$$\forall\;x\in (0,1)\;,\;\;f_n(x)\le f_n\left(\frac n{1+n}\right)=n\left(\frac n{1+n}\right)^n\left(1-\frac n{1+n}\right)=$$
$$=\frac n{n+1}\frac1{\left(1+\frac1n\right)^n}\xrightarrow[n\to\infty]{}e^{-1}$$
A: Hint: Show that the extremum of $f_n$ is attained at $x_n=\frac{n}{n+1}$ and
that $f_n(x_n)= n\ \left( \frac{n}{n+1} \right)^n (1-\frac{n}{n+1})$ converges as $n\rightarrow \infty$ to a non-zero value.
A: Hint. Observe that
$$
f_n\left(\frac{n}{n+1} \right)=\left(1+\frac1n \right)^{-n}\cdot \frac{n}{n+1}
$$ then let $n \to \infty$.
A: For this type of problems use that the maximum is achieved either on the boundary or where the derivative is zero (this is due to compactness). Computing the derivative gives
$$f_n'(x)=n^2x^{n-1}(1-x)-nx^n=(n(1-x)-x)nx^{n-1}=(n-(n+1)x)nx^{n-1}.$$
so the maximum is attained at $x_n=\frac{n}{n+1}\in(0,1)$.(The boundary points are not relevant here, as $f$ there is identically zero).
Now we have
$$\sup_{x\in[0,1]}|f_n(x)|=|f_n(x_n)|,$$
but this last expression can be easily computed explicitly.
Of course the same expression holds with "$\lim_{n\to \infty}$" in front of both sides, so everything reduces to computing the limit of the right hand side: if it is zero, then the convergence will be uniform, else it won't.
A: This is not a solution but it might be useful to others. I decided to investigate the functions geometrically, so I thought I would share in case it brings to light the methods which have been put forward by others.

This demonstrates (non-rigorously) that the function ought not to converge uniformly and it shows the maximum at $\frac{n}{n+1}$ approaching $\frac{1}{e}$ as $n$ grows.
