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Let $f:[o,+\infty)\rightarrow \Bbb R$ a function bounded on each finite interval.

i want to try that if $\lim\limits_{x\rightarrow+\infty}[f(x+1)-f(x)]= L$ then also $\lim\limits_{x\rightarrow+\infty}\dfrac{f(x)}x = L$

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    $\begingroup$ I've tried to make the title more informative and more searchable, but I feel there's still room for improvement. $\endgroup$ – Gerry Myerson Sep 9 '12 at 0:54
  • $\begingroup$ thanks, the title is very important. $\endgroup$ – Camilo Acevedo. Sep 9 '12 at 1:07
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    $\begingroup$ You want to "try that"? Isn't a "to show" missing in there? $\endgroup$ – Rod Carvalho Sep 9 '12 at 1:49
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Since $\lim\limits_{x\rightarrow+\infty}[f(x+1)-f(x)]= L$, for any $\epsilon$ we have some $y>0$ such that $$x> y\implies |f(x+1)-f(x)-L|<\epsilon$$ so for any $x>y$ we have $$\begin{align} \frac{f(x)}{x} &=\frac{f(x)-f\left(x-\lfloor x-y\rfloor\right)}{x}+\frac{f\left(x-\lfloor x-y\rfloor\right)}{x}\\ &=\frac{1}{x}\left(\sum\limits_{i=1}^{\lfloor x-y\rfloor}(f(x-i+1)-f\left(x-i\right))+f\left(x-\lfloor x-y\rfloor\right)\right)\\ &\leq\frac{1}{x}\left(\sum\limits_{i=1}^{\lfloor x-y\rfloor}(L+\epsilon)+\sup\limits_{z\in [0,y+1)}|f(z)|\right)\\ &\leq\frac{\lfloor x-y\rfloor}{x}(L+\epsilon)+\frac{1}{x}\sup\limits_{z\in [0,y+1)}|f(z)|\\ &\leq L+\epsilon+\frac{1}{x}\sup\limits_{z\in [0,y+1)}|f(z)|\\ \end{align}$$ which as $x\to+\infty$ approaches $L+\epsilon$. Letting $\epsilon\to 0$ gives the us $\limsup\limits_{x\to+\infty} \frac{f(x)}{x}\leq L$. A similar technique shows that $\liminf\limits_{x\to+\infty} \frac{f(x)}{x}\geq L$, finishing the proof.

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