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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.

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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
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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 $|f(x+1)-f(x)|=|f'(c)|<\epsilon$.

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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
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