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Let $f_n(x)=n^2 x(1-x^2)^n$ ($0 \le x \le 1, n=1,2,3...$)

For $0<x\le1$, we have $\lim_{n→\infty}f_n(x)=0$ by the theorem.

Theorem: If $p>0$ and $\alpha$ is real, then $\lim_{n→\infty}\frac{n^\alpha}{(1+p)^n}=0$

I have no idea how to apply this theorem to this limit value.
I think it's impossible to apply that theorem, becaus in here,
if we put $n=-n, \alpha=2, p=-x^2$ then $-x^2<0$.

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Use the theorem in the following form:

If $0\leq r < 1$ then $n^2\cdot r^n \xrightarrow[n\to \infty]{}0$ (i.e. $a=2$ and $p=\frac{1}{r}-1$, for $r\neq 0$).

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Wait. Use the following: $$1-x^2 \leq \frac{1}{1+x^2}$$ Then your problem is:

$$0\leq n^2x(1-x^2)^n \leq n^2x\left(\frac{1}{1+x^2}\right)^n$$

Now take limits, use your theorem and you are done.

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The hint follows because $1-x^4 \leq 1$ – Gautam Shenoy Nov 29 '12 at 11:24

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