If a non-decreasing function $f: \mathbb{R}\rightarrow (0,+\infty)$ satisfies $\lim\inf (f(n+1)-f(n))>0$, then $\lim \sup \frac{f(x)}{x}>0$ Prove if a non-decreasing function $f: \mathbb{R}\rightarrow (0,+\infty)$ satisfies $\lim \inf_{n\rightarrow \infty} (f(n+1)-f(n))>0$, then $\lim \sup_{x\rightarrow \infty} \frac{f(x)}{x}>0$ .
Here is my trying: If $\lim \sup_{x\rightarrow \infty} \frac{f(x)}{x}\leq0$. Because $f\geq 0$, so $\lim \inf_{x\rightarrow \infty} \frac{f(x)}{x}\geq0$, so we get $\lim \sup_{x\rightarrow \infty} \frac{f(x)}{x}=\lim \inf_{x\rightarrow \infty} \frac{f(x)}{x}=0$, i.e $\lim_{x\rightarrow \infty}  \frac{f(x)}{x}=0$.
Then by Hospital's Rule, $\lim_{x\rightarrow \infty}  \frac{f(x)}{x}=0=\lim_{x\rightarrow \infty} f'(x)=0$. In the following, I don't know how to prove when $n$ is large enough, $f(n+1)-f(n)$ can be very small and tends to 0, then contradictory to the known condition.
So anyone can give me some idea?

You don't know $f$ to be differentiable. But if indeed it were: if for all $x > N$, you have $f'(x)<c$, then $f(n+1)-f(n)<c$ for all $n> N$ by mean value theorem, and you can derive a contradiction with the lim inf being positive.
Try something else for the general case: if the lim inf is $c$ then for $n\geqslant N$, $f(n+1)-f(n)>c/2$, which implies (why ?) $f(n-N)>c(n-N)/2$ for $n>N$ and...
