calculus - Show that for any $r>0$, $ln x=O(x^r)$ as $xto infty$

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 $C=1$?

Answer

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