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. 

Three ways for showing that, for any $a \in (0,1)$ fixed,
$$
\mathop {\lim }\limits_{n \to \infty } na^n  = 0
$$
(which implies that $f(x)=0$).
1) Fix $b \in (a,1)$. Since $\lim _{n \to \infty } n^{1/n}  = 1$, $n^{1/n}a < b$ $\forall n \geq N_0$. Hence,
$na^n  = (n^{1/n} a)^n  < b^n $ $\forall n \geq N_0$; the result thus follows from $\lim _{n \to \infty } b^n  = 0$.
2) If $b \in (a,1)$, then
$$
\frac{{b - a}}{a}na^n = (b - a)na^{n - 1}  = \int_a^b {na^{n - 1} \,dx}  \le \int_a^b {nx^{n - 1} \,dx}  = x^n |_a^b  = b^n  - a^n ,
$$
and so the result follows from $\lim _{n \to \infty } (b^n - a^n)  = 0$.
3) Similarly to 2), if $b \in (a,1)$, then by the mean-value theorem 
$$
b^n  - a^n  = nc^{n - 1} (b - a) \ge na^{n - 1} (b - a) = na^n \frac{{b - a}}{a},
$$
for some $c \in (a,b)$; hence the result.
