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

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

• 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? – abiessu Nov 18 '13 at 22:00
• What is the definition of $\lim_{x \to \infty} \left[ f(x + 1) - f(x) \right] = 0$ ? – CompuChip Nov 18 '13 at 22:07
• @Lordoftheinf : this is very easy if you use the MVT. But the wording of your question suggests you're not allowed to. – 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.

• 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. – Stefano Nov 18 '13 at 22:28
• 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 $\pm\infty$. – Beni Bogosel Nov 18 '13 at 22:36
• Sorry, I assumed you intended $L$ as finite. My fault. – Stefano Nov 18 '13 at 22:40
• I think you want to use the mean value theorem, not the intermediate vale theorem. – Potato Nov 18 '13 at 22:41
• 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. – 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

$$f'(c_x)=\frac{f(x)-f(x+1)}{x+1-x}$$

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

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

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