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Let's suppose $f$ a continuous function where $$\lim_{x\to+\infty} { f(x+1)- f(x)}=l.$$ How to prove that $$\lim_{x\to +\infty}\frac{f(x)}{x}=l\,?$$

I've already proove by Cesaro theorem that $$\lim_{n\to +\infty}\frac{f(n) }{n}=l,$$ where $n$ is an integer. How to continue it?

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    I think this was asked before, recently...2012-01-03
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    Doesn't it immediately follow from the limit for integers, and the fact that every $x$ lies between two integers?2012-01-03
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    @Scaramouche Define $f(x)$ to equal $0$ when $x$ is an integer and equal $1$, otherwise. The limit over the integers is $0$, but the limit over the reals does not exist.2012-01-03
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    @Scaramouche I think it's a bit dangerous, as for example sin(2n*pi) -> 0, but sin(x) doesn't have a limit2012-01-03
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    Hmm, you're right. What if we define $$l(x) = \lim_{n \rightarrow + \infty} f(n+x) / (n+x)$$ and then deduce that $l(x)$ is continuous...?2012-01-03
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    Does the limit always exist? I think there's counter example for which the limit doesn't exist.2012-01-03
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    Anyway, after doing some research, i think that we have to use the fact that f is continuous that i didn't use since the beginning. Maybe we can use the sequential characterization of limits & continuity. That for all sequence $x(n)$ where $$\lim_{n\rightarrow oo}{x(n) =+oo},$$ then $$\lim_{n\rightarrow oo}\frac{f(x(n)) }{x(n)}=l$$ But, using Cesaro again, i only find that $$\lim_{n\rightarrow oo}\frac{f(x(n)) }{n}=l$$2012-01-03
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    http://math.stackexchange.com/questions/95079/asymptote-problem2012-01-03

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Put $g(x)=f(x)-lx$. Then $g$ is continuous, $\lim_{x\to +\infty}g(x+1)-g(x)=0$ and we have to show that $\lim_{x\to+\infty}\frac{g(x)}x=0$. Fix $\varepsilon>0$, and $n_0$ an integer such that $|g(x+1)-g(x)|\leq\varepsilon$ if $x\geq n_0$. Let $M:=\sup_{x\in [x_0,x_0+1]}|g(x)|$, which is finite since $g$ is continuous. Let $x\geq n_0$. We can find $N=N(x)$ such that $x-N\in [x_0,x_0+1[$. Since $$g(x)-g(x-N)=\sum_{k=0}^{n-1}g(x-k)-g(x-k-1),$$ we have $\frac{|g(x)-g(x-N)|}N\leq\varepsilon$, hence $|g(x)-g(x-N)|\leq N\varepsilon$. We get $$\left|\frac{g(x)}x\right|\leq \frac{N\varepsilon}x+\frac{|g(x-N)|}x\leq \frac{N\varepsilon}x+\frac Mx,$$ and since $x_0\leq x-N$, $\frac{x_0}x\leq 1-\frac Nx$, so $\frac Nx\leq 1-\frac{x_0}x\leq 1$ and we got $$\forall \varepsilon>0,\,\forall x\geq n_0,\quad \left|\frac{g(x)}x\right|\leq \varepsilon+\frac Mx,$$ so for all $\varepsilon>0,\: \limsup_{x\to+\infty}\left|\frac{g(x)}x\right|\leq\varepsilon$, which is the wanted result.