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?
 A: 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.
A: If $x > 1$ then choosing any $b$ such that $0 < b < a$ we have as $x \to \infty$
$$ 0 < \frac{\ln x}{x^a} = \frac{\ln x^b}{bx^a}< \frac{x^b}{bx^a} \to 0.$$
Apply the $\epsilon$ argument to the RHS to find $x_0(\epsilon)$ where
$$  \frac{\ln x}{x^a}  < \frac{x^b}{bx^a} < \epsilon,$$
when $x > x_0(\epsilon).$  
In this case we can use
$$x_0(\epsilon) = \left(\frac{1}{b \epsilon}\right)^{1/(a-b)}$$
A: Here is a way without l'Hospital, just by changing variables several times until the problem becomes much simpler.
Because $e^x$ is an increasing function,
$$
\lim_{x\to\infty}\frac{\log x}{x^a}=0
$$
is equivalent to
$$
\lim_{x\to\infty}\frac{e^{x^a}}{x}=\infty,
$$
which in turn is equivalent to
$$
\lim_{x\to\infty}\frac{e^x}{x^b}=\infty,\quad b=1/a,
$$
after replacing $x$ with $x^b$.
Taking $b^{th}$ roots of both sides, the last statement is equivalent to
$$
\lim_{x\to\infty}\frac{e^{x/b}}{x}=\infty,
$$
which is equivalent to
$$
\lim_{x\to\infty}\frac{e^{x}}{bx}=\infty,
$$
at which point we can multiply through by $b$ to reduce the problem to showing
$$
\lim_{x\to\infty}\frac{e^x}{x}=\infty.
$$
From the inequality $e^x\geq x$, we get that $e^{x/2}\geq \frac{x}{2}$ and therefore $e^x\geq \frac{x^2}{4}$. Hence
$$
\lim_{x\to\infty}\frac{e^x}{x}\geq \lim_{x\to\infty}\frac{x}{4}=\infty.
$$
The best part is that the choice of $N,\epsilon$ is clear in the last limit. Now reverse all of our variable changes to get an $N,\epsilon$ choice that works for the original problem!
A: Hint: Divide and multiply the entire limit by a, using $a\ln x=\ln x^a$, then let $t=x^a$.
