High school contest math problem

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:

$\lim_{x \to \infty}\frac{f(x)}{x^n} = \frac{2^n}{n!} \lim_{x \to \infty}f(x) = \infty$

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.

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

$\lim_{x \rightarrow \infty} \dfrac{f(x)}{x^{n+1}}=\infty$.