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: 
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.
[Proof of $\dfrac{\ln x}{x} \underset{x\to +\infty}{\longrightarrow}0$](https://math.stackexchange.com/a/2377432/460772)

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$\dfrac{\ln x}{x^r}=\dfrac{1}{r}\dfrac{\ln x^r}{x^r}\underset{x\to \infty}\to0$

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 $\quad(C=1)$