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?
 A: You might want to forget L'Hospital and differentiability and proofs by contradiction, and go for a direct, hands-on, proof. 
Let $3\ell=\liminf\limits_{n\to\infty}f(n+1)-f(n)$ (yes, the factor $3$ is weird, but it simplifies things afterwards). Then $\ell\gt0$ hence $2\ell\lt3\ell$ and, by definition of liminf, $f(n+1)-f(n)\geqslant2\ell$ for every $n$ large enough, say every $n\geqslant N$. 
By concatenation, for every $n\geqslant N$, $f(n)\geqslant f(N)+2\ell\cdot(n-N)=2\ell\cdot n-C$, where $C=2\ell\cdot N-f(N)$ does not depend on $n$. 
Thus, $f(n)/n\geqslant2\ell-(C/n)$ and $2\ell-(C/n)\geqslant\ell$ for every $n$ large enough, say, every $n\geqslant\max\{N,C/\ell\}$. 
In particular, $\liminf\limits_{n\to\infty}f(n)/n\geqslant\ell$, which implies that $\limsup\limits_{n\to\infty}f(n)/n\geqslant\ell$, and finally that $\limsup\limits_{x\to\infty}f(x)/x\geqslant\limsup\limits_{n\to\infty}f(n)/n\geqslant\ell$ hence $\limsup\limits_{x\to\infty}f(x)/x\gt0$. 
The method above allows to prove that $\limsup\limits_{x\to\infty}f(x)/x\geqslant\liminf\limits_{n\to\infty}f(n+1)-f(n)$.
The hypothesis that $f$ is nondecreasing is not useful here. If it holds, one can strengthen the result to  $\liminf\limits_{x\to\infty}f(x)/x\geqslant\liminf\limits_{n\to\infty}f(n+1)-f(n)$.
A: 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...
