Show that for any $r>0$, $\ln x=O(x^r)$ as $x\to \infty$ 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 $C=1$?
 A: If you can use derivative-based methods:
$$
\lim_{x\to\infty}\frac{\ln x}{x^r}=
\lim_{x\to\infty}\frac{1/x}{r x^{r-1}}=
\lim_{x\to\infty}\frac{1}{r x^{r}}=0
$$
A: You can use that $\ln x=\frac1r\ln x^r$ so that all cases reduce to the $r=1$ case $\ln x\in O(x)$.
A: 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.
A: 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'.
A: 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 $\quad(C=1)$
