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?
 A: If $x=0$ or $x=1$ it is trivial, so $0<x<1$. Define $a=1-x^2$, then $0<a<1$.
$$\lim\limits_{n\to\infty}n^2x(1-x^2)^n = x\lim\limits_{n\to\infty}n^2a^n \le \lim\limits_{n\to\infty}n^2a^n = \lim\limits_{n\to\infty}\frac{n^2}{(1/a)^n} = \lim\limits_{n\to\infty}\frac{2n}{\ln(1/a)(1/a)^n}=\lim\limits_{n\to\infty}\frac{2}{\ln^2(1/a)(1/a)^n}=\frac{2}{\ln^2(1/a)}\lim\limits_{n\to\infty}a^n=0$$
Using l'Hopital twice and the fact that $0 < a < 1$.
Also all of the terms in the sequence are positive, $0\le\lim\limits_{n\to\infty}n^2x(1-x^2)^n\le 0$.
A: This one's example 7.6 on Rudin's wonderful book Principles Of Mathematical Analysis. He solves it by "Theorem 3.20(b)", which states that "If $p>0$, and $\alpha$ is real, then $lim_{n\to\infty} \frac{n^\alpha}{(1+p)^n}=0$. ", so you need to rewrite $f_{n}(x)$ in that form.
Ignore $x$ and put $n^{2}(1-x^2)^{n}=\frac{n^\alpha}{(1+p)^n}$, clearly $\alpha = 2$ and $(1+p)^n (1-x^2)^n = 1$, take the $n$th root on both sides (you can take the positive root, since you need a value that works for both even and odd n), so with a little bit of algebra you can show that $p = \frac{x^2}{1-x^2}$ fits the bill, since $0<x<1$ implies that $p = \frac{x^2}{1-x^2} > 0$, so indeed, by theorem 3.20(b) we have:
$lim_{n\to\infty}n^2 x(1-x^2)^n = x lim_{n\to\infty} \frac{n^2}{(1+\frac{x^2}{1-x^2})^n}=0$.
A: $$ f_n(x) = x(1-x^2)^n $$
is a positive function on $(0,1)$. By computing $f_n'$, it is easy to check that $x=\frac{1}{\sqrt{2n+1}}$ is the only stationary point of $f_n(x)$ over $(0,1)$, hence:
$$ 0 \leq f_n(x) \leq f_n\left(\frac{1}{\sqrt{2n+1}}\right) = \frac{1}{\sqrt{2n+1}}\left(1-\frac{1}{2n+1}\right)^n\leq \frac{1}{\sqrt{2en}}.$$
Moreover, $f_n(x)$ is exponentially small on the interval $\left[\frac{1}{n^{1/3}},1\right]$. 
By putting all together, we have that $n^2 f_n(x)$ is a sequence of functions that pointwise converge to the zero function on $[0,1]$. Pointwise but not uniformly, also because:
$$ \int_{0}^{1} n^2 f_n(x)\,dx = \frac{n^2}{2n+2}\to +\infty.$$
A: 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.
A: 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 $x=1$.
A: 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.  
