How do I show $f_n(x)=n^2 x^n(1-x)$ pointwise converges to $0$ on $[0,1]$? 
How do I show $f_n(x)=n^2 x^n(1-x)$ pointwise converges to $0$ on $[0,1]$?


I first started with the case $x=1/2$ to try out. We see that $n^2(1/2)^n(1/2)=n^2/2^{n+1}$ which goes to $0$ as $n$ goes to infinity since $n^2 < 2^{n+1}$.
Now I need to generalize. If I take the derivative of $f_n(x)$ then I get, after simplifying,
$$f_n'(x)= n^2x^{n-1}(n-(n+1)x).$$
So, we know $x=0$ and $x=n/(n+1)$ are critical points.
I plotted the graph and I see that as $n$ gets larger, the maximum of the function on $[0,1]$ grows and gets closer to $x=1$.

For $0 < x < 1$ we have $y = 1/x - 1 > 0$ and $ x = 1/ (1 + (1/x - 1))= 1/(1+y).$
Thus,
$$n^2 x^n  = \frac{n^2}{(1 + y)^n}.$$
Using the binomial expansion $(1+y)^n = 1 + ny + \frac{1}{2}n(n-1)y^2 + \frac{1}{6}n(n-1)(n-2)y^3 + \ldots,$we have
$$0 \leqslant n^2 x^n  = \frac{n^2}{(1 + y)^n} < \frac{n^2}{n(n-1)(n-2)}\frac{6}{y^3},$$
and $\lim_{n \to \infty} n^2x^n = 0$ by the squeeze theorem.
