Find the pointwise limit and determine if convergence is uniform 
Find the pointwise limit of $f_n(x)=nx^n(1-x^n)$ and determine if the convergence is uniform.

Solution
For $x=1$
$$\lim_{n\to\infty}(nx^n(1-x^n))\to 0$$
For $0\le x<1$
$$\lim_{n\to\infty}(nx^n(1-x^n))\to0$$
as $x^n\to0$ as $n\to\infty$.
Thus the pointwise limit of $f_n(x)$ is $f(x)=0$.
Determine if the convergence is uniform:
We use,
$$\sup\vert f_n(x)-f(x)\vert<\epsilon$$
$$\sup\vert nx^n(1-x^n)-0\vert$$
$$\sup\vert nx^n(1-x^n)\vert$$
$$\sup \vert nx^n-nx^{2n}\vert$$
We want too find the largest possible value of this function, we take the derivative set to $0$.
$$\frac{d}{dx}(nx^n-nx^{2n})=0$$
$$n^2x^{n-1}-2n^2x^{2n-1}=0$$
$$x^{n-1}-2x^{2n-1}=0$$
$$x^{n-1}(1-2x^{n})=0$$
We can then find the trivial solution $x^{n-1}=0$ and the solution,
$$x=\sqrt[n]{\frac{1}{2}}$$
Is this a correct solution? Or have I made an error?
As $n\to\infty$, $\sqrt[n]{\frac{1}{2}\to1$, which is obviously not the largest value.

It is always interesting to plot graphical representations concretizing algebraic/analytic proofs. Here it is for the 8 first curves with coordinates of maximum (red stars) 
$$(\sqrt[n]{\frac12}, \ \frac{n}{4}).$$

