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 $L\not=-\infty$?

If $L = -\infty$, then for sufficiently large $x$, $f'(x) < -1$, and so $f(x) < k - x$ where $k$ is some constant. This implies $\frac{f(x)}{x^2} < \frac{k - x}{x^2}$ for large $x$. Therefore, $\limsup_{x\to\infty} \frac{f(x)}{x^2} \le 0$.
