Calculate the limit of $nx_n^2$ 
Let a sequence, $\{x_n\}$ such that: $x_{n+1}=x_n-x_n^3$ and $0<x_1<1$.
  1) Prove $\mathop {\lim }\limits_{n \to \infty } {x_n} = 0$
  2) Calculate $\mathop {\lim }\limits_{n \to \infty } n{x_n}^2$  

So, section (1) is very easy. I didn't really bother to write it down - just show the sequence is monotonically decreasing and bounded bellow by zero.  
Section (2) is the real fun, I do familiar with the Lemma says: "If $a_n$ limit is $0$ and $b_n$ is bounded then the limit of $a_nb_n$ is also zero" - But I don't think it can work here. 
I tried separating the limit using limits-arithmetic into two limits, but then I got:
$$\mathop {\lim }\limits_{n \to \infty } n{x_n} \cdot \mathop {\lim }\limits_{n \to \infty } {x_n}$$
 A: *

*$\lim_{x \to \infty } {x_n}=0$. Because the sequence is monotone and bounded by  $0$ and $1$. Last statement is proved by mathematical induction. Passing to the limit in the recurrence relation is obtained limit $0$. Dividing the recurrence relation by $x_n$ and then passing to the limit we find $\lim_{x \to \infty}\frac{x_{n+1}}{{x_n}} = 1$

*Applied Cesaro-Stolz's Lemma:
$$\lim_{x \to \infty } n{x_n}^2 = \lim_{x \to \infty }\frac{n}{(\frac{1}{x_n})^2}=\lim_{x \to \infty }\frac{n+1-n}{(\frac{1}{x_{n+1}})^2-(\frac{1}{x_n})^2}= \lim_{x \to \infty }\frac{(x_n)^2(x_{n+1})^2}{(x_n)^2-(x_{n+1})^2} =$$$$=\lim_{x \to \infty }\frac{(x_n)^4(1-(x_{n})^2)^2}{(x_n)^4(1+\frac{x_{n+1}}{x_n})} =\frac{1}{2}$$
A: Suppose that $0\lt x_k\lt1$, then $0\lt x_k^2\lt1$ and since $x_{k+1}=x_k(1-x_k^2)$, we know that $0\lt x_{k+1}\lt1$. Furthermore, since $x_k^3\gt0$, we have that $x_{k+1}=x_k-x_k^3\lt x_k$.
Thus, $x_k$ is decreasing and bounded below by $0$. Therefore, $\lim\limits_{k\to\infty}x_k$ exists and we have
$$
\lim_{k\to\infty}x_k=\lim_{k\to\infty}x_{k+1}=\lim_{k\to\infty}x_k-\left(\lim_{k\to\infty}x_k\right)^3\tag{1}
$$
Equation $(1)$ implies that $\lim\limits_{k\to\infty}x_k=0$.

Since $x_{k+1}=x_k-x_k^3$, dividing by $x_k$ and taking limits, we get that
$$
\begin{align}
\lim_{k\to\infty}\frac{x_{k+1}}{x_k}
&=\lim_{k\to\infty}\left(1-x_k^2\right)\\
&=1\tag{2}
\end{align}
$$
Next notice that
$$
\begin{align}
\frac1{x_{k+1}^2}-\frac1{x_k^2}
&=\frac{x_k^2-x_{k+1}^2}{x_k^2x_{k+1}^2}\\
&=\frac{(x_k-x_{k+1})(x_k+x_{k+1})}{x_k^2x_{k+1}^2}\\
&=\frac{x_k^3(x_k+x_{k+1})}{x_k^2x_{k+1}^2}\\
&=\frac{x_k^2}{x_{k+1}^2}+\frac{x_k}{x_{k+1}}\tag{3}
\end{align}
$$
Combining $(2)$ and $(3)$ yields
$$
\lim_{k\to\infty}\frac1{x_{k+1}^2}-\frac1{x_k^2}=2\tag{4}
$$
Apply $\displaystyle\lim_{k\to\infty}a_k=b\implies\lim_{n\to\infty}\frac1n\sum_{k=1}^na_k=b$ to $(4)$ to get
$$
\lim_{n\to\infty}\frac1{nx_{n+1}^2}=2\tag{5}
$$
which leads to
$$
\lim_{n\to\infty}nx_n^2=\frac12\tag{6}
$$

Note that the justification for $(5)$ is saying that if a sequence converges, then its Cesaro means converge to the same limit.
A: For what it's worth:
Since $x_n > 0$ for all $n$, one can safely divide both sides of the recurrence relation by $x_n$ to get that
$$
\frac{x_{n+1}}{x_n} = 1 - x_n^2
$$
and thus
$$
n x_n^2 = n\left( 1 - \frac{x_{n+1}}{x_n} \right)
$$
Now, since $x_n\searrow 0$, $x_n^3=o(x_n)$ and $x_{n+1}\operatorname*{\sim}_{n\to\infty}x_n$; and in particular $\frac{x_{n+1}}{x_n} \xrightarrow[n\to\infty]{} 1$. Suppose you can get a second order expansion of the form $\frac{x_{n+1}}{x_n}\operatorname*{=}_{n\to\infty} 1 + \varepsilon_n + o(\varepsilon_n)$ for some "simple" $\varepsilon_n$ (I would bet on $Cn^{-\alpha}$ for some $\alpha > 0$). Then, 
$$
n x_n^2 \operatorname*{\sim}_{n\to\infty}n\varepsilon _n
$$
and depending on $\varepsilon_n$, that'd give you the limit...
