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 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.