TITLE: Calculate the limit of $nx_n^2$
QUESTION [6 upvotes]: 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:
$$\mathop {\lim }\limits_{n \to \infty } n{x_n} \cdot \mathop {\lim }\limits_{n \to \infty } {x_n}$$

REPLY [3 votes]: 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.