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Show that for any $r>0$, $\ln x=O(x^r)$ as $x\to \infty$

I know that if $x_n=O(\alpha_n)$ then there is a constant $C$ and a natural number $n_0$ such that $|x_n|=C|\alpha_n|$ for all $n\geq n_0$. But in this case I do not have sequences, how can I work with these functions? In this case there would be no natural number? Would only the constant be demanded? One would not have to $\ln x\leq x$ for all $x>0$ and with this could not solve much of the problem with $C=1$?

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If you can use derivative-based methods: $$ \lim_{x\to\infty}\frac{\ln x}{x^r}= \lim_{x\to\infty}\frac{1/x}{r x^{r-1}}= \lim_{x\to\infty}\frac{1}{r x^{r}}=0 $$

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  • $\begingroup$ What is your definition of $\ln x=O(x^r)$ as $x\to \infty$? $\endgroup$ – user425181 Dec 22 '17 at 19:34
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    $\begingroup$ @user425181 I have proved that it is even $o(x^r)$, which means $\lim_{x\to\infty}\dfrac{f(x)}{x^r}=0$. $\endgroup$ – Przemysław Scherwentke Dec 22 '17 at 19:37
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You can use that $\ln x=\frac1r\ln x^r$ so that all cases reduce to the $r=1$ case $\ln x\in O(x)$.

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  • $\begingroup$ So you need to prove that. $\endgroup$ – marty cohen Dec 23 '17 at 2:26
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You can deduce this straight from the definition.We have

$\ln(t) = \int_{1}^{t} x^{-1} dx \leq \int_{0}^{t} x^{c-1} dx$, for any $0<c<r$.

Now, let's rewrite this as:

$\frac{\int_{1}^{t}x^{-1}dx}{\int_{0}^{t}x^{c-1}dx} \leq 1$, and notice that $\int_{0}^{t} x^{c-1}dx = \frac{1}{c} x^c$, so we have:

$c\times\frac{\int_{1}^{t}x^{-1}dx}{x^c} \leq 1$.

If we divide both sides by $t^{r-c}$, with $r-c > 0$, and take the limit, the result follows.

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  • $\begingroup$ In the answer I link to, I chose $c = r/2$ to make a definite choice. $\endgroup$ – marty cohen Dec 27 '17 at 5:39
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In my answer here (How to show that $\sqrt{x}$ grows faster than $\ln{x}$. ),

I showed that for any $a > 0$ we have $\dfrac{\ln(x)}{x^{a}} \lt\dfrac{2}{ax^{a/2}} \to 0 $.

My method uses the integral definition of $\ln(x)$, so it is similar to Elie Louis'.

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Proof of $\dfrac{\ln x}{x} \underset{x\to +\infty}{\longrightarrow}0$


$\dfrac{\ln x}{x^r}=\dfrac{1}{r}\dfrac{\ln x^r}{x^r}\underset{x\to \infty}\to0$

As$\dfrac{\ln x}{x^r}\underset{x\to \infty}\to0$, there exists $A\in\mathbb{R}_+,$ such that $\forall x>A\qquad\dfrac{\ln x}{x^r}\le1\iff \ln x\le x^r$

So we have shown that : $\exists C>0, \exists A>0, s.t.|\ln x|\le C|x^r|\iff\ln x=\mathcal{O}(x^r)$ with $\quad(C=1)$

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  • $\begingroup$ $\ln \ln x/\ln x \to 0$ is equivalent to $\ln x/x \to 0$, which is equivalent to what is being asked. So you have assumed the question. $\endgroup$ – marty cohen Dec 23 '17 at 2:26
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    $\begingroup$ @martycohen : I've added the proof of $\ln x/x \to 0$ $\endgroup$ – Stu Dec 23 '17 at 6:42

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