How to prove "Logarithms grow more slowly than polynomials"

Question

"Logarithms grow more slowly than polynomials. That is, $Θ(\lg n)$ grows more slowly than $Θ(n^a)$ for any positive constant a."

if $$y_1= x^{\frac{1}{2}}$$ and $$y_2 = \log (x)$$

by comparing the graphs can say $y_1 \gt y_2$

But, is there a way to prove this?

Answer 1

May use $$\lim _{x\to0}\frac{\log (\frac{1}{x})}{\sqrt{\frac{1}{x}}}=0$$

Answer 2

Actually, even more is true. For $p>0$ and $q>0$, we have that $$ \lim_{x\to\infty}\frac{\log^px}{x^q} =\lim_{x\to\infty}\Bigl(\frac{\log x}{x^{q/p}}\Bigr)^p =\Bigl(\lim_{x\to\infty}\frac{\log x}{x^{q/p}}\Bigr)^p =\Bigl(\lim_{x\to\infty}\frac1{(q/p)x^{q/p}}\Bigr)^p=0 $$ using continuity and l'Hôpital's rule.