Limit of $\frac{f(x)}{x}$ when $f'(x)$ tends to infinity While solving a problem, I found the statement below

If $f:[0,+\infty)\rightarrow \mathbb{R}$ and $f$ is differentiable on $(0,+\infty)$, then $\displaystyle \lim_{x\rightarrow +\infty} 
 f'(x)=+\infty$ if and only if $\displaystyle \lim_{x\rightarrow 
+\infty}\frac{f(x)}{x}=+\infty$

I don't know whether this statement is true or not, but I found a solution for it. Here is my solution:
$\bullet$ If $\displaystyle \lim_{x\rightarrow +\infty} 
 f'(x)=+\infty$ then there exists an $X$ such that for all $x>X:f'(x)>1$. By Lagrange theorem, for every $x>X$ there is $C_x\in(X,x)$ such that $$f(x)-f(X)=(x-X)f'(C_x)>x-X$$
Thus $f(x)$ tends to infinity when $x$ tends to infinity. By L'Hopital Rule: $$\displaystyle \lim_{x\rightarrow 
+\infty}\frac{f(x)}{x}=\displaystyle \lim_{x\rightarrow +\infty} 
 f'(x)=+\infty$$
$\bullet$ If $\displaystyle \lim_{x\rightarrow 
+\infty}\frac{f(x)}{x}=+\infty$ then it's easy to prove $\displaystyle \lim_{x\rightarrow +\infty}f(x)=+\infty$. By L'Hopital Rule, we get $\displaystyle \lim_{x\rightarrow +\infty} 
 f'(x)=+\infty$
From the above proof, we can also conclude that if $\displaystyle \lim_{x\rightarrow +\infty} 
 f'(x)=+\infty$ and by Lagrange theorem, for every $x>0$ there exists $c_x\in (0,x)$ such that $$f(x)-f(0)=xf'(c_x)$$
then the set of all $c_x$ (with all $x>0$) does not have a upper bound.
Is there any wrong with my statement and proof?
Thank you so much. 
 A: You are applying l'Hôpital's rule incorrectly. Rather, the conclusion should be that if $f'(x)\to\infty$, then indeed $\displaystyle\lim_{x\to\infty}\frac{f(x)}x=+\infty$. We are using here the form of l'Hôpital's rule (as presented, for instance, in Rudin's book, see Theorem 5.13) that states that if $a$ is in the extended reals, $\lim_{x\to a}g'(x)/h'(x)=L$ exists (in the extended reals), and  $\lim_{x\to a}h(x)=+\infty$, then also $\lim_{x\to a}g(x)/h(x)=L$. Note that there is no need to assume that $\lim_{x\to a}g(x)=\infty$ as well.
On the other hand, if $\displaystyle \lim_{x\to\infty}\frac{f(x)}{x}=\infty$, we cannot conclude that $\lim_{x\to\infty}f'(x)=\infty$. By l'Hôpital's rule, if $\lim_{x\to\infty}f'(x)$ exists, then it must be $\infty$. But the limit may fail to exist. For instance, we could have $f$ growing exponentially except that it is constant in little intervals around each integer. For any such $f$ we have $\lim_{x\to\infty}f(x)/x=+\infty$; however $\limsup_{x\to\infty}f'(x)=+\infty$ but $\liminf_{x\to\infty}f'(x)\le0$, so the limit does not exist. 
