TITLE: $f(x+1)-f(x)$ converges $\Rightarrow\frac{f(x)}x$ converges
QUESTION [11 upvotes]: 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)

REPLY [6 votes]: 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 
$$|f(y)-ly|\le$$ $$|f(y)-f(y-1)-l|+|f(y-1)-f(y-2)-l|+\dots+|f(y-k_y+1)-f(y-k_y)-l|+|f(y-k_y)|+|lk_y-ly|\le k_y\varepsilon+M+|l|(y-k_y)\le  3y\varepsilon. $$