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Let f(x) be a differentiable function such that that $\lim_{x\to \infty} f '(x) = 0$.

I have to prove that:

$\lim_{x\to \infty} [f(x+1)-f(x)] = 0 $ just by using definitions of limit and definition of derivative.

I have no idea how to begin this...any hints?

i found some posts similar to this but i need a more specific explanation.. any help ? I would be grateful.

EDIT may i use MVT in some specific space?

EDIT2 : im yet confused. is it better to use MVT insted of intermediate value theorem?

EDIT3 : still cant get to a conclusion .....


marked as duplicate by Guy Fsone, muaddib, Xam, Sahiba Arora, user223391 Feb 4 '18 at 2:33

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  • $\begingroup$ You are given that $\lim_{h\to 0}\frac {f(x+h)-f(x)}h=0$ by the definition of derivative. What else have you considered? $\endgroup$ – abiessu Nov 18 '13 at 22:00
  • $\begingroup$ What is the definition of $\lim_{x \to \infty} \left[ f(x + 1) - f(x) \right] = 0$ ? $\endgroup$ – CompuChip Nov 18 '13 at 22:07
  • $\begingroup$ @Lordoftheinf : this is very easy if you use the MVT. But the wording of your question suggests you're not allowed to. $\endgroup$ – Stefan Smith Nov 19 '13 at 0:51

Suppose the contrary, i.e. that $\lim_{x \to \infty }f(x+1)-f(x)\neq 0$. Then there exists a sequence $(x_n)$ which converges to $+\infty$ such that $\lim_{n \to \infty}f(x_n+1)-f(x_n) =L \neq 0$.

Apply now the mean value theorem on each interval $[x_n,x_n+1]$ and you obtain a sequence $c_n \to \infty $ for which $\lim_{n\to \infty} f'(c_n)=L \neq 0$. This gives a contradiction.

  • 4
    $\begingroup$ The contrary of having 0 as limit is not having another value as limit; limit could not exist or be $\infty$. I think it's more direct if you use the intermediate value theorem in a constructive way. $\endgroup$ – Stefano Nov 18 '13 at 22:28
  • $\begingroup$ I never said that $L$ is finite and I didn't assume that the limit exists. Even if the limit does not exist, there is a subsequence which converges or goes to $\pm\infty$. $\endgroup$ – Beni Bogosel Nov 18 '13 at 22:36
  • $\begingroup$ Sorry, I assumed you intended $L$ as finite. My fault. $\endgroup$ – Stefano Nov 18 '13 at 22:40
  • $\begingroup$ I think you want to use the mean value theorem, not the intermediate vale theorem. $\endgroup$ – Potato Nov 18 '13 at 22:41
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    $\begingroup$ In my opinion, it should be a last resource, in general. In this specific case, it actually makes the argument appear longer and somewhat more involved than it is: First you need to find a sequence with $y_n=|f(x_n+1)-f(x_n)|$ bounded away from $0$. Second, you need to pass to a subsequence $(y_{n_k})_k$ that converges in the extended reals. Only then, you apply the mean value theorem. Two thirds of this strategy are not needed. $\endgroup$ – Andrés E. Caicedo Nov 19 '13 at 16:14

By mean value theorem, for each $x$ there exists $c_x\in (x,x+1)$ such that


Hence since $x<c_x<x+1$ we have $c_x\to\infty$ as $x\to\infty$ therefore,

$$0=\lim_{x\to \infty}f'(x)=\lim_{x\to \infty}f'(c_x) =\lim_{x\to \infty}[f(x)-f(x+1)] $$

  • $\begingroup$ Please avoid posting duplicate answers (this is not the first warning you got about that). $\endgroup$ – Jack D'Aurizio Jan 29 '18 at 17:17

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