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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... – David Mitra Jan 3 '12 at 21:07
Doesn't it immediately follow from the limit for integers, and the fact that every $x$ lies between two integers? – Scaramouche Jan 3 '12 at 21:25
@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. – Austin Mohr Jan 3 '12 at 21:32
Does the limit always exist? I think there's counter example for which the limit doesn't exist. – bonnnnn2010 Jan 3 '12 at 22:01
up vote 2 down vote accepted

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.

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