High school contest math problem 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.
 A: 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.
A: Credit to user622002.
Show $\lim_{x \rightarrow \infty} \dfrac{f(x)}{x^n}=\infty$, $n \in \mathbb{N}$.
Induction:
Base case: $n=0$ √.
Hypothesis:
$\lim_{x \rightarrow \infty} \dfrac{f(x)}{x^n}=\infty$;
L'Hospital:
$\lim_{x \rightarrow \infty}\dfrac{f(x)}{x^{n+1}}=$
$\lim_{x \rightarrow \infty}\dfrac{f(x)(f'(x)/f(x))}{(n+1)x^n}$;
For large enough $x$: $f'(x)/f(x)>1$;
$(1/(n+1))\dfrac{f(x)}{x^n}\lt$
$ \dfrac{f(x)(f'(x)/f(x)}{(n+1)x^n}$.
Taking limits, invoking the hypothesis for the left hand side, we get
$\lim_{x \rightarrow \infty} \dfrac{f(x)}{x^{n+1}}=\infty$.
A: I think you can just use de l’Hopital inductively:
Given $n \in \Bbb{N}$, we have (since the numerator and denominator diverge):
$\lim_{x \to \infty}\frac{f(x)}{x^n} = \lim_{x \to \infty} \frac{f’(x)}{nx^{n-1}} = \lim_{x \to \infty} \frac{f(x) \frac{f’(x)}{f(x)}}{nx^{n-1}} = \frac{2}{n} \lim_{x \to \infty} \frac{f(x)}{x^{n-1}} $
So inductively we get:
$\lim_{x \to \infty}\frac{f(x)}{x^n} = \frac{2^n}{n!} \lim_{x \to \infty}f(x) = \infty$
