How to show the limit of the derivative is positive infinity, as x aproaches positive infinity? Here is the exercise that I can not find the proof at present: Let $f:[0,\infty)\to\mathbb{R}$ be a differentiable function satisfying $\lim_{x\to+\infty}\frac{f(x)}{x^2}=1, \text{ and } \lim_{x\to+\infty}f'(x)=L$. Show that $L=+\infty$.
This exercise is extracted from Giaqinta & Modica's TextBook: Mathematical Analysis, Foundations of one variable, Page143, 3.86. 
I have tried like the following: If $L\in\mathbb{R}$, then there exist $M>0, X>0,$ such that $|f'(x)|\leq M, \forall x\geq X.$ Thus for each $x>X$, $\exists \eta\in (X,x)$, such that $f(x)-f(X)=f'(\eta)(x-X), $ and then 
$$\frac{f(x)}{x^2}=\frac{f(X)}{x^2}+(\frac{1}{x}-\frac{X}{x^2})f'(\eta)\to 0, \text{ as } x\to+\infty,$$
which contradicts the condition $\frac{f(x)}{x^2}\to 1, \text{as} x\to+\infty.$ Therefore $L\in\{+\infty, -\infty\}.$ But How to conclude that $L\not=-\infty$?

Added Much simpler argument:
It is clear that $\lim_{x \to \infty} f(x)= \infty$.
Since $\lim_{x \to \infty} f'(x)=L$, exists, and e are in teh $\frac{\infty}{\infty}$ case, by L'Hospital we get that 
$$\lim_{x \to \infty} \frac{f(x)}{x} = \lim_{x \to \infty} f'(x) =L$$
Then
$$L=\lim_{x \to \infty} \frac{f(x)}{x} = \lim_{x \to \infty} \left( \frac{f(x)}{x^2} \right)x = 1 \cdot \infty=\infty$$  

You can use the Mean Value Theorem:
For each $n$ there exists a $c_n$ so that 
$$\frac{f(2n) -f(n)}{n}= f'(c_n)$$
Now, with $\epsilon=\frac{1}{5}$, there exists an $N$ so that for all $n >N$, we have
$$f(2n)>4n^2(1-\frac{1}{5}) \,;\, f(n)< n^2 (1+\frac{1}{5})$$
Thus, for all $n >N$we have 
$$f'(c_n) >2n (*)$$
Letting $n \to \infty$, since $c_n >n$ we have $c_n \to \infty$, and hence $f'(c_n) \to L$. Then, (*) implies that $L=\infty$.
P.S. This argument can probably be simplified.
