Please help me to understand why $\lim\limits_{n\to\infty } {nx(1-x^2)^n}=0$ This is the exercise: 

$$f_{n}(x) = nx(1-x^2)^n,\qquad n \in {N}, f_{n}:[0,1] \to {R}.$$
  Find ${f(x)=\lim\limits_{n\to\infty } {nx(1-x^2)^n}}$.

I know that $\forall x\in (0,1]$ $\Rightarrow (1-x^2) \in [0, 1) $ but I still don't know how to calculate the limit.  $\lim\limits_{n\to\infty } {(1-x^2)^n}=0$ because  $(1-x^2) \in [0, 1) $ and that means I have $\infty\cdot0$. 
I tried transformation to $\frac{0}{0} $ and here is where I got stuck. 
I hope someone could help me. 

Please see $\textbf{9.10}$ at this given link: The Solution is actually worked out for a more general case and I think that should help.


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*http://www.csie.ntu.edu.tw/~b89089/book/Apostol/ch9.pdf
