Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Prove: if $f$ is a real-valued differentiable function such that $\lim_{x\to \infty}f'(x) = 0$ then $\lim_{x\to \infty}(f(x+1)-f(x))=0$

Proof: We know $\lim_{x\to \infty}f'(x) = 0$. (1)

By the mean value theorem there exists $c\in [x, x+1]$ such that $f'(c) = \dfrac{f(x+1)-f(x)}{1}$

$\iff \lim_{x\to \infty} f'(c) = \lim_{x\to \infty} f(x+1)-f(x) = 0$ by (1).

End of Proof.

share|cite|improve this question
You should indicate that $c$ depends on $x$; call it $c_x$, for instance. Also, your final line is badly phrased. Say "Since $\lim_{x\rightarrow\infty }f'(x)=0$, it follows from (1) that $\lim_{x\rightarrow\infty }\bigl( f(x+1)-f(x)\bigr)=0$" (and perhaps explain why by pointing out that $c_x\rightarrow\infty$). – David Mitra Dec 10 '12 at 1:01
up vote 3 down vote accepted

Your proof as written is not too right as you have not stated where your $x$ lies when you say that there is some $c$ between $x$ and $x+1$. We need to fix $x$ first and then find the corresponding $c$ which depends on $x$.

If $\lim_{x\to +\infty} f'(x)=0$, then for any $\epsilon>0$, there is an $M>0$ such that for any $x>M$ we have $|f'(x)|=|f'(x)-0|<\epsilon$.

Let $x>M$. By the mean value theorem, $\frac{f(x+1)-f(x)}{(x+1)-x}=f'(c)$ for some $x<c<x+1$. But $c>x>M$ so that $|f(x+1)-f(x)|=|f'(c)|<\epsilon$.

share|cite|improve this answer
Technically aren't you using the definition of the limit of a sequence and not the limit definition of a function? – CodeKingPlusPlus Dec 10 '12 at 5:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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