Order of growth of logarithms, compared to linear

Question

I think it is true that any power of a logarithm, no matter how big, will eventually grow slower than a linear function with positive slope.

Is it true that for any exponent $m>0$ (no matter how big we make $m$), the function $f(x)$ $$f(x)=(\ln x)^m$$

will eventually always be less than $g(x) = x$ ?

I am pretty sure this is true because I have tried large values of $m$ and it always eventually slows down. But I don't know how to prove it.

Answer 1

Yes, this is true. This is equivalent to proving that, for any $a > 0$, we have $$ \frac{\ln x}{x^a} \xrightarrow[x\to\infty]{}0 $$ (you can see it by setting $a=\frac{1}{m}$ from your question).\; which itself is equivalent to showing $$ \frac{a\ln x}{x^a} = \frac{\ln x^a}{x^a} \xrightarrow[x\to\infty]{}0 $$ so, at the end of the day, it is sufficient to show ("setting $t = x^a$") that $$ \frac{\ln t}{t} \xrightarrow[t\to\infty]{}0. $$ Now, that you can show in many ways (some would say "L'Hopital's rule"). One of them is to do (yet another) change of variable, $t=e^u$, and show the equivalent statement $$ \frac{u}{e^u} \xrightarrow[t\to\infty]{}0. $$ Now, this is obvious given the series definition of the exponential, for instance, as $e^u = 1+u+\frac{u^2}{2}+\sum_{n=3}^\infty \frac{u^n}{n!} > \frac{u^2}{2}$ (the last inequality for $u > 0$), so that $\frac{u}{e^u} < \frac{2}{u}$ for $u>0$.

Answer 2

No calculus required.Taking logs to any base $b=1+r$ with $r>0,$ then for $x>(1+r)^2$ we have $\log_b x>2.$ So for $x>(1+r)^2$ let $n_x$ be the positive integer such that $$n_x\leq \log_bx<n_x+1.$$ We have then $b^{n_x}\leq x<b^{n_x+1}$. And since $n_x\geq 2$ and $r>0$, we have, by the Binomial Theorem, $$x\geq b^{n_x}=(1+r)^{n_x}=1+r n_x+r^2 \binom {n_x}{2}+...>r^2 \binom {n_x}{2}>r^2(n_x-1)^2/2.$$ So we have $x> r^2(n_x-1)^2/2$, equivalently $1+\sqrt {2 x}/r>n_x.$ But since $\log_bx<n_x+1$ we have $\log_bx<\sqrt {2x}/r+2$ and we conclude that $\lim _{x\to \infty}(\log_bx)/x=0.$ As shown in the first part of the answer by Clement C., this is sufficient to imply $\lim_{x\to \infty}(\log_bx)^m/x=0$ for any $m>0$.