Let $\{ x_n \}_{n=1}^{\infty}$ such that $x_{n+1}=x_n-x_n^3$ and $0Let $\{ x_n \}_{n=1}^{\infty}$ such that $x_{n+1}=x_n-x_n^3$ and $0<x_1<1, \ \forall n\in \mathbb{N}.$
Prove:


*

*$\lim_{n \rightarrow \infty} x_n=0$

*Calculate$\ \lim_{n \rightarrow \infty} nx_n^2$

*Let $f$ be a differentiable in $\mathbb{R}$ such that $f(n)=x_n,\forall n\in \mathbb{N}.$ Prove that if $\lim_{x \rightarrow \infty} f'(x)$ exists, then it equals to $0$.


I proved the first, but struggling with the next two.
My intuition tells me that $\lim_{n \rightarrow \infty} nx_n^2=0$, and I tried to squeeze it but got stuck.
I tried to prove by contradiction the third, and I managed to contradict the limit is greater than $0$, but couldn't get further.
Any help appreciated.
 A: For the second one, the usual trick is to consider $y_n = 1/x_n^2$. Then by the Stolz–Cesàro theorem,
$$ \lim_{n\to\infty} \frac{y_n}{n} =  \lim_{n\to\infty} (y_{n+1} - y_n) = \lim_{n\to\infty} \frac{2 - x_n^2}{(1-x_n^2)^2} = 2 $$
and hence $n x_n^2 \to \frac{1}{2}$ as $n\to \infty$.

For the third one, for each $n$ we pick $\xi_n \in (n, n+1)$ so that $ f(n+1) - f(n) = f'(\xi_n)$. This is possible from the mean value theorem. Since $\xi_n \to \infty$, the assumption on the existence of the limit $\ell := \lim_{x\to\infty} f'(x)$ tells that
$$ \ell = \lim_{n\to\infty} f'(\xi_n) = \lim_{n\to\infty} (x_{n+1} - x_n). $$
Of course, the last limit is zero and therefore $\ell = 0$.
A: $\quad$ $\bullet$ For the first one, we recall $t \in ]0,1[$ then $t > t^3 > 0$. So from $x_1 \in ]0,1[$ and $x_{n+1} = x_n - x_n^3, \forall n \geq 2$, we can prove that $ \forall n \in \mathbb{N}, x_n > 0$ and $$x_1 > x_2 > x_3 > \ldots > 0\ .$$ 
Hence $\{x_n\}_{n \in \mathbb{N}}$ converges. Suppose that $\lim\limits_{n \to + \infty} x_n = x$, then $x = x - x^3$ and then $x = 0$.
$\quad$ $\bullet$ For the second and third one, you can do like Sangchul Lee .
