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

  • $\lim_{n \rightarrow \infty} x_n=0$
  • 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.

  • $\begingroup$ 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

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 $\ell = 0$.


$\quad$ $\bullet$ 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 .


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.