Limit of ratio ${f(x)\over x}$ equal to limit of difference $f(x+1)-f(x)$. Using my right to ask for help once again. I can't solve the following problem for quite a long time. Just have no idea how to do that. Here it is.
Let $f(x)$ be bounded on any interval $(1,b), b>1$. Then $\lim_{x\to+\infty}{f(x) \over x}=\lim_{x\to+\infty}(f(x+1)-f(x))$.
It is from the Russian book "Lekcii po matematicheskomu analizu" by G.I.Arkhipov, V.A.Sadovnichy and V.N.Chubarikov, 2nd ed., Moscow, 2000, page 676. Please, keep in mind, that times, when I did homework had passed long time ago. Now I'm solving problems for pleasure.
 A: The right statement is as follows:


*

*If $\lim\limits_{x\to\infty}\left(f(x+1)-f(x)\right)=\ell\in\mathbb{R}$ then $\lim\limits_{x\to\infty}\dfrac{f(x)}{x}=\ell$.


Indeed, suppose that $\lim\limits_{x\to\infty}\left(f(x+1)-f(x)\right)=\ell\in\mathbb{R}$. Let $\varepsilon>0$, then there exists $b_\varepsilon>1$ such that 
$$\forall\,t\ge b_\varepsilon,\qquad -\varepsilon<f(t+1)-f(t)-\ell<\varepsilon.\tag{1}$$
Define $$M_\varepsilon=\sup\{ |f(x)|:1\le x\le b_\varepsilon+1\}.\tag{2}$$
Now, consider $x\ge b_\varepsilon$, and define $k(x)=\lfloor x-b_\varepsilon\rfloor$, it follows that 
$$k(x)+b_\varepsilon\le x<k(x)+1+b_\varepsilon\tag{3}$$
Applying (1) with $t=x-j-1$ for $j=0,1,\ldots,k(x)-1$ and adding the resulting inequalities we get
$$(\ell-\varepsilon) k(x)\le f(x)-f(x-k(x))\le(\ell+\varepsilon) k(x)$$
Using (2) we get
$$(\ell-\varepsilon) k(x)-M_\varepsilon\le f(x)\le(\ell+\varepsilon) k(x)+M_\varepsilon$$
Or
$$(\ell-\varepsilon) \frac{k(x)}{x}-\frac{M_\varepsilon}{x}\le \frac{f(x)}{x}\le(\ell+\varepsilon) \frac{k(x)}{x}+\frac{M_\varepsilon}{x}\tag{4}$$
But according to (3) we have $\lim\limits_{x\to\infty}\dfrac{k(x)}{x}=1$, so, from (4) we conclude that
$$\ell-\varepsilon \le \liminf_{x\to\infty}\frac{f(x)}{x}\le
\limsup_{x\to\infty}\frac{f(x)}{x}
\le\ell+\varepsilon$$
But $\varepsilon>0$ is arbitrary. Therefore
$$\ell \le \liminf_{x\to\infty}\frac{f(x)}{x}\le
\limsup_{x\to\infty}\frac{f(x)}{x}
\le\ell$$
That is the limit $\lim\limits_{x\to\infty}\dfrac{f(x)}{x} $ exists and is equal to $\ell$.


*

*Note that the example $f(x)=\sin(\pi x)$ shows that 
$\lim\limits_{x\to\infty}\dfrac{f(x)}{x}$ might exist while $\lim\limits_{x\to\infty}\left(f(x+1)-f(x)\right)$   does not.

A: If both exists :
$f$ bounded imply that $\lim_{x\to + \infty} \frac{f(x)}{x} = 0$
Now, suppose that $f(x)-f(x+1)$ doesn't converge to $0$
We have that 
$$f(n) - f(0) = \sum_{k=0}^{n-1} f(k+1)-f(k)$$
As $ \lim_{k\to \infty}  f(k+1)-f(k) $ exists (without loss of generality suppose it greater than $0$) and is not equal to $0$, there exists $\eta >0$ and $M>0$ such that $\forall n > M, f(n+1)-f(n) \geq \eta$
Hence 
$$f(n) = \underbrace{f(0) + \sum_{k=0}^M f(k+1)-f(k)}_{Constant} + \sum_{k=M+1}^{n-1} f(k+1)-f(k)$$
$$f(n) \geq  C_{\eta} + \sum_{k=M+1}^{n-1} \eta = C_\eta + (n-M+1) \eta$$
The left side is bounded and the right side diverge to $+\infty$ : contradiction
Now, it's easy to construct functions $f$ where $\lim_{x\to + \infty} \frac{f(x)}{x} $ exists and not $\lim_{x\to + \infty} f(x+1)-f(x) $
