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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.

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    $\begingroup$ Can't we simply say that f(x) is equivalent to exp(2x) when x goes to infinity? $\endgroup$ – Jeanba Dec 31 '19 at 13:43
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    $\begingroup$ Marios.I think it is fine.+1. One remark .While correct I personally try to avoid something like $2(+\infty)$. $\endgroup$ – Peter Szilas Dec 31 '19 at 16:09
  • $\begingroup$ @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. $\endgroup$ – aschepler Dec 31 '19 at 22:13
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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:

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

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

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  • $\begingroup$ +1.... before the constest,no theory of integration was covered but nice solution though,i will keep it in mind..is my approach correct? $\endgroup$ – Marios Gretsas Dec 31 '19 at 15:00
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    $\begingroup$ Matematleta.Nice. $\endgroup$ – Peter Szilas Dec 31 '19 at 17:03
  • $\begingroup$ @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 $\operatorname{ln} f (x)$? $\endgroup$ – Bruno Krams Jan 2 at 18:31
  • $\begingroup$ 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. $\endgroup$ – Bruno Krams Jan 2 at 18:36
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    $\begingroup$ @Matematleta +1 for your l’Hospital-free proof. I appreciate that $\endgroup$ – Bruno Krams Jan 3 at 9:52
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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

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

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  • $\begingroup$ Is my solution ok though? $\endgroup$ – Marios Gretsas Dec 31 '19 at 14:36
  • $\begingroup$ @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 $0\in\mathbb N_0$ $\endgroup$ – Invisible Dec 31 '19 at 15:50
  • $\begingroup$ 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 $\mathbb{Z^+}$. $\endgroup$ – Peter Szilas Dec 31 '19 at 16:04
  • $\begingroup$ 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. $\endgroup$ – Invisible Dec 31 '19 at 16:46
  • $\begingroup$ VerkovtsevaKatya. No mistake correcting me:) Happy New Year. $\endgroup$ – Peter Szilas Dec 31 '19 at 16:59

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