Prove this limit as $x\to\infty$ Let $f:(a,\infty)\to\mathbb{R}$ be a function. Suppose for each $b>a$, $f$ is bounded on $(a,b)$ and $\lim_{x\to\infty}f(x+1)-f(x)=A$. Prove
$$\lim_{x\to\infty}\frac{f(x)}{x}=A.$$
Here we assume $f$ to be arbitrary and no further conditions. $f$ could be continuous,discontinuous,as long as it fits all assumptions. I have some trouble with the intermediate steps.

We have that $$\lim_{x\to\infty}f(x+1)-f(x)=A$$ which means that for any $\varepsilon\in\mathbb R^+$ there is an $n\in\mathbb N$ such that for each $x>n$ then $A-\varepsilon<f(x+1)-f(x)<A+\varepsilon$.  (if we find some $n<a$ for this condition then we can chose $n=\lceil a\rceil$ for the remind of the proof.)
For simplicity I will define $g(x):=f(x)-Ax$, then we have that $g(x+1)-g(x)=f(x+1)-f(x)-A$ and for $x>n$: $|g(x+1)-g(x)|<\varepsilon$.
As $f$ is bounded in $(a,b)$ for any $b>a$, then $f$ is bounded in $(a,n+2)$ and as we have find $n\ge a$, then $f$ is bounded in $(n,n+1]$, let's call $\beta$ that boundary: $|f(x)|<\beta$ for any $x$ in $(n,n+1]$.
Also, let $\gamma=\beta+(n+1)|A|$, then  $g$ is bounded and $|g(x)|<\gamma$ for $x\in(n,n+1]$.
Now, we can prove that 
\begin{align}
|g(x)|&\le\gamma+(\lfloor x\rfloor-n)\varepsilon,&&\text{and}\\
\frac{|g(x)|}x &\le\frac{\gamma+(\lfloor x\rfloor-n)\varepsilon}x =\frac\gamma x+\frac{\lfloor x\rfloor-n}x\varepsilon
\end{align}
For any fixed positive $\varepsilon$, we have:
$$\lim_{x\to\infty}\frac\gamma x+\frac{\lfloor x\rfloor-n}x\varepsilon=\varepsilon$$
therefor
$$\lim_{x\to\infty}\left|\frac{g(x)}x\right|=\lim_{x\to\infty}\frac{|g(x)|}x\le\varepsilon$$
This should prove that $\lim_{x\to\infty}\frac{g(x)}x=0$, now the limit for $\frac{f(x)}x$ should follow by having $f(x)=Ax+g(x)$.
