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 use that $\ln x=\frac1r\ln x^r$ so that all cases reduce to the $r=1$ case $\ln x\in O(x)$.
[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 $\quad(C=1)$