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?

First let us observe that $x(1-x^2)^n$ is smaller than $(1-x^2)^n$ on the interval, since $|x|<1$.
The largest point of $(1-x^2)$ is at $x=0$ which is $1$ and the smallest is $0$ at $x=1$ (easy to check) and monotonically decreasing. So the function $(1-x^2)^n$ will be bounded by 1. Any $f(x_0) = (1-{x_0}^2)^n$ will decrease exponentially with $n$ for any $x_0\neq 0$ so the crossing $y=\epsilon>0$ will be pushed towards 0 with increasing $n$. So we see that the function $(1-x^2)^n$ will shrink towards $0$ for all values on the interval, except for $x=0$, but as our function of interest is actually $x$ times the function we investigated, so it will of course be even more suppressed.
