# 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.

• Can't we simply say that f(x) is equivalent to exp(2x) when x goes to infinity? – Jeanba Dec 31 '19 at 13:43
• Marios.I think it is fine.+1. One remark .While correct I personally try to avoid something like $2(+\infty)$. – Peter Szilas Dec 31 '19 at 16:09
• @Jeanba Maybe, if you can work out how to be sufficiently exact about what "equivalent when $x$ goes to infinity" means. It's certainly a good intuitive argument for why it ought to be true. – aschepler Dec 31 '19 at 22:13

## 3 Answers

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:

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

• user622002.Nice answer. – Peter Szilas Dec 31 '19 at 14:35
• This is really slick. – Randall Jan 1 at 1:35

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.

• +1.... before the constest,no theory of integration was covered but nice solution though,i will keep it in mind..is my approach correct? – Marios Gretsas Dec 31 '19 at 15:00
• Matematleta.Nice. – Peter Szilas Dec 31 '19 at 17:03
• @Matematleta Not sure if the assumption $f \to \infty$ really is superfluous. Couldn't $f$ and $f'$ be both negative so that you can not even write down the expression $\operatorname{ln} f (x)$? – Bruno Krams Jan 2 at 18:31
• By the way and in response to @MariosGretsas' objection: There is no need to use an integral in this proof. One could argue that - under appropriate conditions - the mean value theorem yields $$\operatorname{ln} f(x) - \operatorname{ln} f(x_0) = \frac{f'(\xi)}{f(\xi)} \cdot ( x - x_0) \qquad \text{for some } \xi \in (x_0, x)$$ and from this equation one can deduce your estimate. – Bruno Krams Jan 2 at 18:36
• @Matematleta +1 for your l’Hospital-free proof. I appreciate that – Bruno Krams Jan 3 at 9:52

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

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

• Is my solution ok though? – Marios Gretsas Dec 31 '19 at 14:36
• @PeterSzilas, you wrote hypothesis: $\displaystyle\lim_{x\to\infty}\frac{f(x)}{x^n}=0$ and it should be $n\in\mathbb N_0$ because $0\in\mathbb N_0$ – Invisible Dec 31 '19 at 15:50
• VerkovtsevaKatya. Thanks for your comment. Actually as far as I know $0 \in \mathbb{N}$, if you want the positive integers one could write for example $\mathbb{Z^+}$. – Peter Szilas Dec 31 '19 at 16:04
• According to the original Peano axiomatisation that we use at our uni, $1$ isn't a successor, therefore we define $\mathbb Z^+:=\mathbb N=\{1,2,3,\ldots\}$. However, I searched the internet to check it and saw the modern version includes $0$. My mistake for correcting you, your statement was legitimate as well. – Invisible Dec 31 '19 at 16:46
• VerkovtsevaKatya. No mistake correcting me:) Happy New Year. – Peter Szilas Dec 31 '19 at 16:59