Limit at infinity for sequence $ n^2x(1-x^2)^n$ I'm supposed to prove that this sequence goes to zero as n goes to infinity.
$$\lim_{n\to \infty} {n^2x (1-x^2)^n}, \mathrm{where~} 0 \le x \le 1$$
I've been trying a few things (geometric formula, rewriting $(1-x^2)^n$ as $\sum_{k=0}^n \binom{n}{k} (-1)^k x^{2k} $) and messing around with that. But I can't seem to get anywhere. I could be missing something key that I've forgotten. Can somebody point me in the right direction?

If $x=0$ or $x=1$ we are done so you can assume that $1>x>0.$ Then $0<1-x^2<1.$ 
Now put $p:=\dfrac{x^2}{1-x^2}.$ For $n\geqslant4$ we have 
$$
\begin{aligned}
(1+p)^n&>\binom{n}{3}p^3\\\\&=\dfrac{n(n-1)(n-2)}{3!}p^3\\\\&>\dfrac{(n/2)(n/2)(n/2)}{3!}p^3\\\\&=\dfrac{n^3\cdot p^3}{48} 
\end{aligned}
$$
and hence $n^2x(1-x^2)^n=\dfrac{n^2x}{(1+p)^n}\;<\;\dfrac{48x}{p^3}\cdot\dfrac{1}{n}$ and since $\dfrac{48x}{p^3}\cdot\dfrac{1}{n}\to0$ as $n\to\infty$ then $$\lim_{n\to\infty}n^2x(1-x^2)^n=0.$$
In general, you can prove that given any real number $\alpha$ and any real $a\in(0,1)$ then $\lim\limits_{n\to\infty}n^\alpha a^n=0$ using a similar argument.  
