If $f:R\to R$ is bounded on bounded intervals, and $\lim f(x+1) - f(x) = L$ for $x\to +\infty$ then $\lim f(x)/x = L$ when $x\to +\infty$ 
If $f:R\to R$ is bounded on bounded intervals, and $\lim f(x+1) - f(x) = L$ when $x\to +\infty$ then $\lim f(x)/x = L$ when $x\to +\infty$

I found answers here at the site when $f$ is differentiable, but I just don't know what to use in this case, where we don't have differentiability (at least not explicitly stated). Any help would be appreciated.
 A: My attempt :)
Notation: $\Delta g(x)=g(x+1)-g(x)$.
Theorem:
Let $f, g: R^+ \rightarrow R$ bounded in  each bounded interval , $g$ crescent, with
$$\lim_{x \rightarrow \infty} \frac{\Delta f(x)}{\Delta g(x)}=L\;\;\; \lim_{x \rightarrow \infty} g(x)= \infty $$
so
$$\lim_{x \rightarrow \infty} \frac{ f(x)}{ g(x)}=L .$$
Proof:
Given $\varepsilon>0$ exists  $x>M$ such that
$$ \varepsilon- L< \frac{\Delta f(x)}{\Delta g(x)}< \varepsilon +L $$
$g$ is crescent then $\Delta g(x)>0$ so  we can multiply the inequality by that term. Change $x$ by $x+k$ where $k$ is natural and apply the sum $\sum^{n-1}_{k=0}$, which results in
$$ (\varepsilon- L) (g(x+n)-g(x))+f(x)< f(x+n)< (\varepsilon +L) (g(x+n) - g(x)) +f(x)$$
by the telescopic sum.
Now divide by $g(x+n)$, which is positive because $g \rightarrow \infty$
$$ (\varepsilon- L) (1-\frac{g(x)}{g(x+n)})+\frac{f(x)}{g(x+n)}< \frac{f(x+n)}{g(x+n)}< (\varepsilon +L) (1-\frac{g(x)}{g(x+n)}) +\frac{f(x)}{g(x+n)}$$
now we use sequences.
We take $x=y_n$ in $[M, M+1]$ and $x_n=n+y_n$ ( $x_n$ is an arbitrary sequence with infinite limit) $g$ and $f$ are bounded in $[M, M+1]$ so
$$(\varepsilon- L) (1-\frac{g(y_n)}{g(x_n)})+\frac{f(y_n)}{g(x_n)}< \frac{f(x_n)}{g(x_n)}< (\varepsilon +L) (1-\frac{g(y_n)}{g(x_n)}) +\frac{f(y_n)}{g(x_n)} $$
the limit ensures  $\lim \frac{f(x_n)}{g(x_n)} = L $ because  $g(y_n)$ and $f(y_n)$ are bounded  and $\lim g(x_n)=\infty$  $\qed$.
A: Just write for any $x>0$,
$$\frac{f(x)}{x}=\frac{(f(x)-f(x-1))+(f(x-1)+f(x-2))+\cdots+(f(x-\lfloor x\rfloor+1)-f(x-\lfloor x\rfloor))}{x}+\frac{f(x-\lfloor x\rfloor)}{x}$$
where $\lfloor x\rfloor$ is the integer part.
As $f$ is bounded on $[0,1]$, the summand $\frac{f(x-\lfloor x\rfloor)}{x}$ goes to $0$ as $x$ goes to infinity. Hence we only have to take into account the limit of $$\frac{(f(x)-f(x-1))+(f(x-1)+f(x-2))+\cdots+(f(x-\lfloor x\rfloor+1)-f(x-\lfloor x\rfloor))}{x}\text{.}$$
Now, the proof is similar to that of the Stolz–Cesàro theorem. The idea is to divide the average in those difference that are near $L$, so their average remains near $L$, and those who are far but we can control by the hypothesis on $f$ being bounded, which will vanish in the limit.
By hypotesis on $f$, for each $\varepsilon>0$, there is an integer $C_\varepsilon>0$ such that for all $x\geq C_\varepsilon$,
$$|f(x+1)-f(x)-L|<\varepsilon\text{,}$$
which is the limit definition written out for $\lim_{x\to\infty}f(x+1)-f(x)=L$. And again by hypothesis, for each $\varepsilon>0$, there is $M_\varepsilon>0$ such that
$$|f(x)|\leq M_\varepsilon$$
for all $x\in[0,C_\varepsilon]$.
Write $\frac{(f(x)-f(x-1))+(f(x-1)+f(x-2))+\cdots+(f(x-\lfloor x\rfloor+1)-f(x-\lfloor x\rfloor))}{x}$ as the sum of 
$$\frac{(f(x)-f(x-1))+(f(x-1)+f(x-2))+\cdots+(f(x-\lfloor x\rfloor+C_{\varepsilon/3}+1)-f(x-\lfloor x\rfloor+C_{\varepsilon/3}))}{\lfloor x\rfloor-C_{\varepsilon/3}}\frac{\lfloor x\rfloor-C_{\varepsilon/3}}{x}$$
and
$$\frac{(f(x-\lfloor x\rfloor+C_{\varepsilon/3})-f(x-\lfloor x\rfloor+C_{\varepsilon/3}-1)+\cdots+(f(x-\lfloor x\rfloor+1)-f(x-\lfloor x\rfloor))}{C_{\varepsilon/3}}\frac{C_{\varepsilon/3}}{x}\text{.}$$
In the first one, all the summands $f(x-k)-f(x-k-1)$ satisfy $x-k-1\geq C_{\varepsilon/3}$ and so $|f(x-k)-f(x-k-1)|<\varepsilon/2$. By the triangular inequality, this gives
$$\left|\frac{(f(x)-f(x-1))+\cdots+(f(x-\lfloor x\rfloor+C_{\varepsilon/3}+1)-f(x-\lfloor x\rfloor+C_{\varepsilon/3}))}{\lfloor x\rfloor-C_{\varepsilon/3}}-L\right|<\varepsilon/3\text{.}$$
This implies that
$$\begin{align*}
&\left|\frac{(f(x)-f(x-1))+\cdots+(f(x-\lfloor x\rfloor+C_{\varepsilon/3}+1)-f(x-\lfloor x\rfloor+C_{\varepsilon/3}))}{\lfloor x\rfloor-C_{\varepsilon/3}}\frac{\lfloor x\rfloor-C_{\varepsilon/3}}{x}-L\right|\\
=&\left|\frac{(f(x)-f(x-1))+\cdots+(f(x-\lfloor x\rfloor+C_{\varepsilon/3}+1)-f(x-\lfloor x\rfloor+C_{\varepsilon/3}))}{\lfloor x\rfloor-C_{\varepsilon/3}}-L\frac{\lfloor x\rfloor-C_{\varepsilon/3}}{x}\right|\left|\frac{x}{\lfloor x\rfloor-C_{\varepsilon/3}}\right|\\
=&\left|\frac{(f(x)-f(x-1))+\cdots+(f(x-\lfloor x\rfloor+C_{\varepsilon/3}+1)-f(x-\lfloor x\rfloor+C_{\varepsilon/3}))}{\lfloor x\rfloor-C_{\varepsilon/3}}-L-L\frac{\lfloor x\rfloor-x-C_{\varepsilon/3}}{x}\right|\left|\frac{x}{\lfloor x\rfloor-C_{\varepsilon/3}}\right|\\
\leq &\left|\frac{(f(x)-f(x-1))+\cdots+(f(x-\lfloor x\rfloor+C_{\varepsilon/3}+1)-f(x-\lfloor x\rfloor+C_{\varepsilon/3}))}{\lfloor x\rfloor-C_{\varepsilon/3}}-L\right|\left|\frac{x}{\lfloor x\rfloor-C_{\varepsilon/3}}\right|+\left|L\frac{\lfloor x\rfloor-x-C_{\varepsilon/3}}{x}\right|\left|\frac{x}{\lfloor x\rfloor-C_{\varepsilon/3}}\right|\\
<&\frac{\varepsilon}{3}\left|\frac{x}{\lfloor x\rfloor-C_{\varepsilon/3}}\right|+|L|\left|\frac{\lfloor x\rfloor-x-C_{\varepsilon/3}}{\lfloor x\rfloor-C_{\varepsilon/3}}\right|
\end{align*}$$
But taking $x>\max\{9C_{\varepsilon/3},C_{\varepsilon/3}+8|L|(1+C_{\varepsilon/3})/\varepsilon\}$, this gives
$$<3\varepsilon/8+\varepsilon/8\leq \varepsilon/2$$
as $\left|\frac{x}{\lfloor x\rfloor-C_{\varepsilon/3}}\right|$ is monotone decreasing converging to $1$ as $x\to\infty$ and $|L|\left|\frac{\lfloor x\rfloor-x-C_{\varepsilon/3}}{\lfloor x\rfloor-C_{\varepsilon/3}}\right|\leq |L|(1+C_{\varepsilon/3})/(\lfloor x\rfloor-C_{\varepsilon/3})$.
In the second one, each $f(x-k)-f(x-k-1)$ satisfies that $x-k,x-k-1\in[0,C_{\varepsilon/3}]$ and so
$$|f(x-k)-f(x-k-1)|\leq 2M_{\varepsilon/3}\text{.}$$
Because of this,
$$\begin{align*}
&\frac{(f(x-\lfloor x\rfloor+C_{\varepsilon/3})-f(x-\lfloor x\rfloor+C_{\varepsilon/3}-1)+\cdots+(f(x-\lfloor x\rfloor+1)-f(x-\lfloor x\rfloor))}{C_{\varepsilon/3}}\frac{C_{\varepsilon/3}}{x}\\
\leq& \frac{2M_{\varepsilon/3}C_{\varepsilon/3}}{x}
\end{align*}$$
which for $x>4M_{\varepsilon/3}C_{\varepsilon/3}/\varepsilon$ satisfies
$$<\varepsilon/2\text{.}$$
Combining these bounds, we obtain that for every $\varepsilon>0$, if
$$x>\max\{9C_{\varepsilon/3},C_{\varepsilon/3}+8|L|(1+C_{\varepsilon/3})/\varepsilon,4M_{\varepsilon/3}C_{\varepsilon/3}/\varepsilon\}\text{,}$$
then
$$\left|\frac{(f(x)-f(x-1))+(f(x-1)+f(x-2))+\cdots+(f(x-\lfloor x\rfloor+1)-f(x-\lfloor x\rfloor))}{x}-L\right|<\varepsilon$$
by the triangular inequality. This means that
$$\lim_{x\to\infty}\frac{(f(x)-f(x-1))+(f(x-1)+f(x-2))+\cdots+(f(x-\lfloor x\rfloor+1)-f(x-\lfloor x\rfloor))}{x}=L$$
finishing the proof that
$$\lim_{x\to\infty}f(x)/x=L\text{.}$$
