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:
$$
\lim\_{x\to\infty}\frac{\ln x}{x^r}=
\lim\_{x\to\infty}\frac{1/x}{r x^{r-1}}=
\lim\_{x\to\infty}\frac{1}{r x^{r}}=0
$$
You can use that $\ln x=\frac1r\ln x^r$ so that all cases reduce to the $r=1$ case $\ln x\in O(x)$.