Let $f: [0,\infty)\to\mathbb{R}$ be a function such that $f(x)$ is bounded if $x$ is bounded.
If $\lim_{x\to\infty} f(x+1)-f(x)=0$ , prove that: $$\lim_{x\to\infty} \frac{f(x)}{x}=0.$$
If $f$ is bounded the proof is trivial. Then I assume that $f$ is not bounded; by the first condition, $f(x_n)\to \infty$ only if $x_n \to \infty$; I need to show $\lim_{n\to\infty} \frac{f(x_n)}{x_n}=0$, but I can't. Any suggestion?
Let $\epsilon>0$ and $N>0$ be a natural number such that $|f(x+1)-f(x)|<\epsilon$ for all $x\ge N$. Since $[N,N+1]$ is bounded, let $M>0$ be such that $|f(z)|\le M$ for all $N\le z\le N+1$. Then we can write for $x=\lfloor x\rfloor +\alpha$, $\ \alpha\in [0,1)$, $$ \frac{f(x)}x=\frac{f(N+\alpha)}x+\frac1{x}\sum_{j=N+1}^{\lfloor x\rfloor}f(j+\alpha)-f(j-1+\alpha). $$ We find that $$ \left|\frac{f(N+\alpha)}x\right|\le \frac{M}x,\quad \frac1{x}\sum_{j=N+1}^{\lfloor x\rfloor}\left|f(j+\alpha)-f(j-1+\alpha)\right|\le \frac{\epsilon(\lfloor x\rfloor-N)}{x}\le\epsilon. $$ Thus $$ \limsup_{x\to\infty}\left|\frac{f(x)}x\right|\le \lim_{x\to\infty} \frac{M}x+\epsilon =\epsilon. $$ Since $\epsilon>0$ was arbitrary, it follows $\lim_{x\to\infty}\left|\frac{f(x)}x\right|=0$.
