Question: **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 $C=1$?

Answer: 
If you can use derivative-based methods:
$$
\lim\_{x\to\infty}\frac{\ln x}{x^r}=
\lim\_{x\to\infty}\frac{1/x}{r x^{r-1}}=
\lim\_{x\to\infty}\frac{1}{r x^{r}}=0
$$
In my answer here
([How to show that $\sqrt{x}$ grows faster than $\ln{x}$.](https://math.stackexchange.com/questions/2473780/how-to-show-that-sqrtx-grows-faster-than-lnx/2473999#2473999)
),

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'.