Related limits of a function. Let $f:[0,\infty]\rightarrow \mathbb{R}$ a bounded function in each bounded interval. If $\lim_{x\rightarrow \infty} [f(x+1)-f(x)]=L$ then prove that $\lim_{x\rightarrow \infty}\displaystyle\frac{f(x)}{x}=L$


I appreciate any hint to solve this problem. 

Thanks a lot!
For any $a &gt; 0$:

&emsp; Let $y$ be such that $f(x+1) - f(x) \in L + [-a,a]$ for any $x \ge y$

&emsp; Then $f(y+t+k) - f(y+t) \in k ( L + [-a,a] )$ for any $k \in \mathbb{N}$ and $t \in [0,1]$

&emsp; Also $f(y+t)$ is bounded over all $t \in [0,1]$

&emsp; Thus ${\large \frac{f(y+t+k)}{y+t+k} } \in {\large \frac{f(y+t)+k(L+[-a,a])}{y+t+k} } \to L+[-a,a]$ uniformly for all $t \in [0,1]$ as $k \to \infty$

&emsp; Thus ${ \large \frac{f(x)}{x} } \in L + [-2a,2a]$ as $x \to \infty$

Therefore ${ \large \frac{f(x)}{x} } \to L$ as $x \to \infty$



To explain what happens in the last two lines inside "For any $a&gt;0$", we basically have $\large\frac{p(t)+k(L+q(t))}{y+t+k}$ for some $p$ that is bounded on $[0,1]$ and $q$ that is in $[-a,a]$. As $k$ increases, $p(t)$ becomes insignificant because it is bounded, and also $\frac{k}{y+t+k}$ approaches 1 uniformly over all $t \in [0,1]$, hence the expression approaches something in $L+[-a,a]$ uniformly over all $t \in [0,1]$. "Uniformly" means that for any desired margin of error there is a common cutoff point for $k$ beyond which the expression is within the error margin for all $t$. The next line follows from this because eventually the expression will be within $a$ of $L+[-a,a]$.
This is another proof:

Given $\epsilon&gt;0$ there exist $y&gt;0$ such that if $x&gt;y$ then:

$$L-\epsilon&lt;f(x+1)-f(x)&lt;L+\epsilon$$

we can conclude the following inequalytis valid for every $n$

$$L-\epsilon&lt;f(x+2)-f(x+1)&lt;L+\epsilon$$

$$L-\epsilon&lt;f(x+2)-f(x+1)&lt;L+\epsilon$$

$$\vdots$$

$$L-\epsilon&lt;f(x+n)-f(x+n-1)&lt;L+\epsilon$$

adding each of this inequalitys 

$$nL-n\epsilon&lt;f(x+n)-f(x)&lt;nL+n\epsilon $$

taken $n=[x]$ and $h(x)=x-[x]$ then

$$[x]L-[x]\epsilon&lt;f(x)-f(h(x))&lt;[x]L+[x]\epsilon $$

since $0\leq h(x) &lt;1$ and $[0,1)$ is bounded, then there exist $M&gt;0$ such that $|f(h(x))|&lt;M$. This implies that $\lim_{x\rightarrow\infty}\displaystyle\frac{f(h(x))}{x}=0$. Dividing by $x$:

$$\displaystyle\frac{[x]}{x}L-\displaystyle\frac{[x]}{x}\epsilon&lt;\displaystyle\frac{f(x)-f(h(x))}{x}&lt;\displaystyle\frac{[x]}{x}L+\displaystyle\frac{[x]}{x}\epsilon $$

and recalling that $\lim_{x\rightarrow\infty}\displaystyle\frac{[x]}{x}=1$. We take  $x\rightarrow\infty$ then

$$L-\epsilon\leq\lim_{x\rightarrow\infty}\displaystyle\frac{f(x)}{x}\leq L+\epsilon$$

since $\epsilon&gt;0$ is aritrary we get the result.