# $f(x+1)-f(x)$ converges $\Rightarrow\frac{f(x)}x$ converges

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)

• 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? – Zach L. Dec 24 '14 at 16:30
• 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. – abnry Dec 24 '14 at 16:39
• 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 $\lim_{x \to \infty}f(x+1)/(x+1)-f(x)/x=\lim_{x \to \infty}[f(x+1)-f(x)]/x=l/x=0$ – Curious Dec 24 '14 at 17:09
• If $f'$ is also continuous , then one has a simpler proof by using $\lim f'(x) = l.$ – mick Feb 4 '15 at 23:30

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 $$\frac{m_{n_x}}{n_x}\le\frac{f(x)}{x}\le\frac{M_{n_x}}{n_x+1} .$$

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

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