If $\lim_{x \to +\infty} f'(x) = L$, then $\lim_{x \to \infty} \frac {f(x)}{x} = L$ I'm trying to solve this question:

Let $f:[0,+\infty) \to \mathbb{R}$ be derivable and $\lim_{x \to +\infty} f'(x) = L$, then $\lim_{x\to \infty}\frac {f(x)}{x}=L$.

I'm trying to solve this question using l'Hôpital rule, but I couldn't use it, because I don't know if $\lim_{x\to +\infty} f(x)=+\infty$.
I need help.
Thanks a lot.
 A: As mentioned in the comments,  L'Hôpital's rule is applicable, and readily yields the result.

Here's a proof that does not appeal to L'Hôpital's rule (though it essentially uses the same proof as that result).
First assume $L$ is finite.
Let $\epsilon>0$.  
By the Mean Value Theorem and the hypothesis that 
$\lim\limits_{x\rightarrow\infty} f'(x)=L$, there is a number $a$
so that for all $x>a$
$$\tag{1}
  L-\epsilon < {f(x)-f(a)\over x-a} <L+\epsilon.
$$
Since ${x-a\over x}=1-{a\over x}>0 $ for $x>a$ it follows from $(1)$ that 
$$\tag{2}
  (L-\epsilon)\Bigl(1-{a\over x}\Bigr) < {f(x)\over x}- {f(a)\over x} < (L+\epsilon)\Bigl(1-{a\over x}\Bigr),
$$
for $x>a$.
Since $\lim\limits_{x\rightarrow\infty} {f(a)\over x}=\lim\limits_{x\rightarrow\infty }{a\over x}=0$,
it follows from $(2)$ that 
$$
 \textstyle \limsup\limits_{x\rightarrow\infty} {f(x)\over x}\le L+\epsilon
  \ \ \text{and}\ \ 
  \liminf\limits_{x\rightarrow\infty} {f(x)\over x}\ge L-\epsilon.
$$
Since $\epsilon$ was arbitrary, we have 
$$\textstyle
  \limsup\limits_{x\rightarrow\infty} {f(x)\over x}\le L 
  \ \ \text{and}\ \ 
  \liminf\limits_{x\rightarrow\infty} {f(x)\over x}\ge L;
$$ 
whence the result follows.

If $L=\infty$, the argument is similar; but one starts with an arbitrary $M>0$, and then finds an $a$ so that for all $x>a$,
$$\tag{1}
  {f(x)-f(a)\over x-a} >M.
$$
Then
$$\ 
   {f(x)\over x}- {f(a)\over x} >M\Bigl(1-{a\over x}\Bigr).
$$
From this it follows that for $x$ sufficiently large ${f(x)\over x}\ge M\cdot(1/2)-1/2. $
Since $M$ was arbitrary, we have  $\lim\limits_{x\rightarrow\infty}{f(x)\over x}=\infty$.
The argument for $L=-\infty$ proceeds as in the argument for $L=\infty$.
