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Let $\{ x_n \}_{n=1}^{\infty}$ 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.

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  • $\begingroup$ The first limit if exists have to be $0$. I didn't do the computation, but I'd try prove it by induction the sequence is decreasing. I'd try the same with the second. For the third, I'd try to calculate $f(n+1)-f(n)$ in the limit, and the mean value? $\endgroup$ – Rafa Budría Mar 8 '17 at 8:51
  • $\begingroup$ @RafaBudría For the third, by MVT I get $f'(c)=-x_n^3$ and then when taking limit, I get $\lim_{x \rightarrow \infty} f'(c)=0$, Now with some more explanations, I can say it's true for all $x$? $\endgroup$ – Itay4 Mar 8 '17 at 9:08
  • $\begingroup$ You need not worry about that, since it is assumed that $\lim_{x\to\infty} f'(x)$ exists. Let me show you in my answer. $\endgroup$ – Sangchul Lee Mar 8 '17 at 9:15
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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 $\ell = 0$.

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$\quad$ $\bullet$ 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 .

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