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

Question

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 1

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.

Answer 2

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

$$x_0(\epsilon) = \left(\frac{1}{b \epsilon}\right)^{1/(a-b)}$$

Answer 3

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!

Answer 4

Hint: Divide and multiply the entire limit by a, using $a\ln x=\ln x^a$, then let $t=x^a$.