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Problem: Suppose $f$ is defined and differentiable for $x>0$, and $f'(x)\rightarrow 0$ as $x\rightarrow +\infty$. Put $g(x)=f(x+1)-f(x)$. Prove that $g(x)\rightarrow 0$ as $x\rightarrow +\infty$.

Source: W. Rudin, Principles of Mathematical Analysis, Chapter 5, exercise 5.

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    I presume you mean "*as* $x\to+\infty$".2012-06-30

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From the mean value theorem, we know that

$f(x+1)-f(x)=g(x)=f'(t)$

for some $t\in (x, x+1)$. Then, because $f'\rightarrow 0$, so does $g$.