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Let $f:\mathbb{R}\to\mathbb{R}$ continuous such that $\lim_{x\to\infty}f(x+1)-f(x)=l$.

How can I prove that $\dfrac{f(x)}x$ converges for $x\to+\infty$ ?

I am just trying to prove the convergence, not that it converges to $l$ (that part be deduced immediately by Stolz-Cesaro and the sequential characterization of limits)

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  • $\begingroup$ Since you know Stolz-Cesaro, what happens when you try to turn the proof you have for that result into the proof of this fact? $\endgroup$ – Zach L. Dec 24 '14 at 16:30
  • $\begingroup$ Intuitively, $f(x) = \sum f(n+1+x)-f(n+x) = (L+\epsilon) \lfloor x \rfloor +A(x)$ for some constant $A(x)$. $A(x)$ will be bounded since it will inherit some continuity from $f(x)$. With this you can see why we have convergence. $\endgroup$ – abnry Dec 24 '14 at 16:39
  • $\begingroup$ if $\lim_{x \to \infty}f(x+1)/(x+1)-f(x)/x=0$ then $f(x)/x$ converges. But$f(x+1)/(x+1) \to f(x+1)/x$, as n goes to infinity. So $\lim_{x \to \infty}f(x+1)/(x+1)-f(x)/x=\lim_{x \to \infty}[f(x+1)-f(x)]/x=l/x=0$ $\endgroup$ – Curious Dec 24 '14 at 17:09
  • $\begingroup$ If $f'$ is also continuous , then one has a simpler proof by using $\lim f'(x) = l.$ $\endgroup$ – mick Feb 4 '15 at 23:30
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Here is a similar but more interesting statement.

Let $f$ defined on $(a,+\infty)$ and bounded on all bounded interval $(a,b)$. If $\lim_{x\to\infty}\bigr(f(x+1)-f(x)\bigl)=l$ then $\lim_{x\to\infty}\frac{f(x)}{x}=l$ as well.

Let $\lim_{x\to\infty}\bigr(f(x+1)-f(x)\bigl)=l$ and introduce $M_n=\sup_{[n,n+1)} f(x)$ and $m_n=\inf_{[n,n+1)} f(x)$. The sequences $\{M_n\}$ and $\{m_n\}$ are well defined for $n\ge \lfloor a\rfloor +1$. By definition of $\sup$ and $\inf$ we have have a sequence $\{x_n\}\in[n,n+1)$ sand $f(x_n)>M_n-\varepsilon$. Now conclude that $\lim_{n\rightarrow +\infty} (M_{n+1}-M_n)=l$ and $\lim_{n\rightarrow +\infty} (m_{n+1}-m_n)=l$ as well.

By Stolz-Cesaro theorem $$\lim_{n\rightarrow +\infty}\frac{M_n}{n}=\lim_{n\rightarrow +\infty}\frac{m_n}{n+1}.$$ Then there exist $n_0$ such that $n>n_0$ we have $$-\varepsilon<\frac{M_n}{n}-l<\varepsilon\quad\text{and}\quad -\varepsilon<\frac{m_n}{n+1}-l<\varepsilon.$$ It follows that $f(x)>0$ for $x$ large enough if $l>0$. Then if $n_x=\lfloor x\rfloor$ then $$ \frac{m_{n_x}}{n_x+1}\le\frac{f(x)}{x}\le\frac{M_{n_x}}{n_x} .$$ It's pretty easy to conclude for $l\ne 0$. I leave the rest of the job to you for $l=0$ and for $l<0$ the preceding inequality becomes $$ \frac{m_{n_x}}{n_x}\le\frac{f(x)}{x}\le\frac{M_{n_x}}{n_x+1} .$$

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It seems the following.

Let $1>\varepsilon>0$ be an arbitrary number. There exists a number $N>0$ such that $$|f(x+1)-f(x)-l|\le \varepsilon$$ for each $x\ge N$. Since the function $f$ is continuous, $$\sup \{f(y): y\in [N;N+1]\}=M<\infty.$$ Let $y\ge \max \{M, (|l|+1)(N+1)\}/\varepsilon $ be an arbitrary number. There exists a nonegative integer $k_y$ such that $N\le y-k_y<N+1$. Then $k_y\le y$, $M\le y\varepsilon$, and $|l|(y-k_y)<|l|(N+1)\le y\varepsilon$. So

$$|f(y)-ly|\le$$ $$|f(y)-f(y-1)-l|+|f(y-1)-f(y-2)-l|+\dots+|f(y-k_y+1)-f(y-k_y)-l|+|f(y-k_y)|+|lk_y-ly|\le k_y\varepsilon+M+|l|(y-k_y)\le 3y\varepsilon. $$

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