TITLE: High school contest math problem
QUESTION [6 upvotes]: This is a calculus problem from a high school math contest in Greece,from 2012.
I wish to know some solutions for this. I attempted to solve it.

Let $f:\Bbb{R} \to \Bbb{R}$ differentiable such that $\lim_{x \to +\infty}f(x)=+\infty$ and $\lim_{x \to +\infty}\frac{f'(x)}{f(x)}=2$.Show that $$\lim_{x \to +\infty}\frac{f(x)}{x^{2012}}=+\infty$$

Here is my attempt:
$\frac{f(x)}{x^{2012}}=e^{\ln{\frac{f(x)}{x^{2012}}}}$
Now $\ln{\frac{f(x)}{x^{2012}}}=\ln{f(x)}-2012\ln x=\ln{x}\left( \frac{\ln{f(x)}}{\ln{x}}-2012\right)$
Now from hypothesis we see that $\lim_{x \to +\infty}\ln{f(x)}=+\infty$
By L'Hospital's rule we have that $$\lim_{x \to +\infty}\frac{\ln{f(x)}}{\ln{x}}=\lim_{x \to +\infty}x \frac{f'(x)}{f(x)}=2(+\infty)=+\infty$$
Thus $$\lim_{x \to +\infty}\ln{x}\left( \frac{\ln{f(x)}}{\ln{x}}-2012\right)=+\infty$$
Finally $\lim_{x \to +\infty}\frac{f(x)}{x^{2012}}=+\infty$
Is this solution correct?
If it is,then are there also  better and quicker ways  to solve this?
Thank you in advance.

REPLY [6 votes]: Here is an approach that only uses the easily proved fact that 
$\underset{x\to\infty }\lim\frac{e^x}{p(x)}=\infty\  \text{whenever}\  p \ \text{is a polynomial}. \tag1$ 
Indeed, there is an $x_0\in \mathbb R^+$ such that $\frac{f'(x)}{f(x)}>1$ for all $x>x_0.$ Then, 
$\displaystyle\int^x_{x_0}\frac{f'(t)}{f(t)}dt>x-x_0\Rightarrow \ln f(x)-\ln f(x_0)>x-x_0\Rightarrow f(x)>(f(x_0)e^{-x_0})\cdot e^x \tag2.$
Divide $(2)$ by $x^{2012}$ and invoke $(1)$ to conclude.