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.

$\quad$ $\bullet$ 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 .
