calculus - is it for granted that $log x$ grows slower than $x^a$?

When facing this problem $$\lim_{x\to \infty}\frac{\log x}{x^a}$$ with $a>0$, then I automatically think of l'Hoital's. But then can we approach this problem using the definition of limits? I mean can we find $x_0$ such that for $x>x_0(\epsilon)$ it holds that $\dfrac{\log x}{x^a}<\epsilon$ for any $\epsilon>0$?

Essentially is it for granted that $\log x$ grows slower than $x^a$? What is the simplest thing we could assume to have this proved?

Answer

Let $a > 0$; note that for all $c > 0$ we have
$$
\log x = \int_{t=1}^{x} \frac{1}{t} \leq \int_{t=1}^{x}t^{c-1} = \frac{x^{c}-1}{c} < \frac{x^{c}}{c}
$$
for all $x > 1$; if $c := a/2$, then
$$

\frac{\log x}{x^{a}} < \frac{x^{c-a}}{c} = \frac{2}{a}x^{-a/2} \to 0
$$
as $x$ grows indefinitely.

Now you may extend the argument above to an epsilon-argument easily.