Show that $\dfrac{f(x)}{x}=\infty$. 
Let $f$  defined on $(a,\infty)$ be bounded below on each finite interval $(a,b)$ .
Show that if $\lim _{x\to \infty} (f(x+1)-f(x))=\infty $ then $\lim_{x\to \infty}\dfrac{f(x)}{x}=\infty$.

TRY:
If $\lim _{x\to \infty} (f(x+1)-f(x))\implies x>G,|f(x+1)-f(x)|>M$ where $G,M\in \Bbb R$ are large.
But how to use it to show $\dfrac{f(x)}{x}=\infty$.
Please help
 A: Given $M\gt a$, pick $\varepsilon\gt 0$, and choose $G$ so that $f(x+1)-f(x)\gt M+\varepsilon$ for $x\gt G$. Now, let $N$ be a lower bound of $f(x)$ on $(G,G+1]\subseteq(G,G+\frac 32)$, which is assumed to exist. Thus, for every $n\in\mathbb N$ and $x\in(G,G+1]$ you have $f(x+n)\gt f(x)+n(M+\varepsilon)\ge n(M+\varepsilon)+N$ (induction on $n$). Now we use that to estimate $f(\xi)$ for $\xi\gt G+1$. Given $\xi$, find $x\in(G,G+1]$ such that $n=\xi-x$ is a positive integer. We have $n\ge\xi-G-1$, so $\frac{f(\xi)}{\xi}\gt\frac{n(M+\varepsilon)+N}{\xi}\ge\frac{(\xi-G-1)(M+\varepsilon)+N}{\xi}\to M+\varepsilon$ when $\xi\to\infty$. That means, for big enough $\xi$ we can make $\frac{f(\xi)}{\xi}$ bigger than $M$. As $M$ was arbitrarily chosen to start with, this implies that $\frac{f(x)}{x}\to\infty$ when $x\to\infty$.
A: For any $M > 0$
there is an $x(M)$ such that,
for $x > x(M)$
we have
$f(x+1)-f(x) > M$.
Then,
$f(x(M)+n)
\gt f(x(M))+nM
$
so
$\begin{array}\\
\dfrac{f(x(M)+n)}{x(M)+n}
&\gt \dfrac{f(x(M))+nM}{x(M)+n}\\
&= \dfrac{f(x(M))}{x(M)+n}+\dfrac{nM}{x(M)+n}\\
&= \dfrac{f(x(M))}{x(M)+n}+\dfrac{x(M)M+nM-x(M)M}{x(M)+n}\\
&= \dfrac{f(x(M))}{x(M)+n}+\dfrac{x(M)M+nM}{x(M)+n}-\dfrac{x(M)M}{x(M)+n}\\
&= M+\dfrac{f(x(M))}{x(M)+n}-\dfrac{x(M)M}{x(M)+n}\\
\end{array}
$
By making $n$ large enough,
we can make the two terms
on the right
as small as we want,
so that,
for any $M > 0$
and $c > 0$
we will have
$\dfrac{f(x)}{x}
\gt M-c$
for all large enough $x$.
Therefore
$\lim_{x \to \infty} \dfrac{f(x)}{x}
=\infty$.
