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?

Yet another approach is as follows.  Let $z=\left|\log(1-x^2)\right|$.  Then, we have for $x(1-x) \ne 0$
$$\begin{align}
n^2x(1-x^2)^{n}&=n^2xe^{n\log(1-x^2)}\\\\
&=\frac{n^2x}{e{nz}}\\\\
&=\frac{n^2x}{1+nz+\frac12 n^2z^2+\frac16 n^3z^3+O(n^4)}
\end{align}$$
which clearly approaches $0$ as $n \to \infty$.  And we are done as the limit is $0$ for $x=0$ and $x=1$.
