The growth rate of $(\ln(x))^n$ is a lot slower than I expected

Question

Obviously, the growth rate of $(\ln(x))^a$ is less than the growth rate of $(\ln(x))^b$ as long as $a>b$.

Also, the growth rate of $(\ln(x))^n$ is apparently always less than the growth rate of $x$.

Isn't it surprising that $(\ln(x))^n$ has a slower growth rate than $x$? And a proof? All I've managed to do is see that for increasingly large $x$, $x>(\ln(x))^n$, regardless of how large $n$ is.

Answer 1

Letting $y=\log x$, this is equivalent to stating that $e^{y}>y^n$ for $y$ large enough.

Since $e^{y} > \frac{y^{n+1}}{(n+1)!}$ when $y>0$, we have that when $y>(n+1)!$, that $e^{y}>y^n$.

Answer 2

Though it doesn't exactly give an intuitive answer, L'Hopital's rule does tell you the correct answer:

$$\lim_{x \to \infty} \frac{\ln(x)^n}{x} = \lim_{x \to \infty} \frac{n \ln(x)^{n-1}}{x} = \dots = \lim_{n \to \infty} \frac{n!}{x} = 0.$$

Another approach is to prove that for any $\epsilon>0$, $\lim_{x \to \infty} \frac{\ln(x)}{x^\epsilon} = 0$ (by a single L'Hopital application). Then $\frac{\ln(x)^n}{x}=\left ( \frac{\ln(x)}{x^{1/n}} \right )^n$ so the result follows by continuity.

Still another approach is to introduce $u=\ln(x)$, which reduces the problem to showing that exponentials grow faster than powers of $x$.

Answer 3

I think it is better to go in the details to see why $(\log x)^{n}$ tends to $\infty$ much slower than $x$. By taking $n^{\text{th}}$ roots we see that the problem is equivalent to showing that $\log x$ tends to $\infty$ much slower than $x^{1/n}$.

Clearly $$(\log x)' = \frac{1}{x},\,(x^{1/n})' = \frac{1}{nx^{(n - 1)/n}}$$ and we can see that $$\frac{1}{x} < \frac{1}{nx^{(n - 1)/n}}$$ for large $x$ (should be obvious after some manipulation of the inequality). Hence the rate of growth of $\log x$ is less than that of $x^{1/n}$ for large $x$.

Without using derivatives it is possible to establish directly that $$\lim_{x \to \infty}\frac{\log x}{x^{a}} = 0$$ for any $a > 0$ so that $\log x$ tends to $\infty$ much slower than any positive power of $x$. To establish this we need the fundamental inequality $$\log x < x - 1$$ for $x > 1$. The above inequality seems to be too simple to give any idea about the growth of $\log x$ as $x \to \infty$. But its power increases many fold if we combine it with another fundamental property of $\log x$ namely that $\log x^{b} = b\log x$ (so that the effect of exponentiation is reduced to just plain multiplication when we take logs). Thus let us take a number $b$ such that $0 < b < a$ so that $a - b > 0$. If $x > 1$ then $x^{b} > 1$ and hence we have $$0 < \log x^{b} < x^{b} - 1 < x^{b}$$ or $$0 < b\log x < x^{b}$$ or $$0 < \frac{\log x}{x^{a}} < \frac{1}{bx^{a - b}}$$ and now using Squeeze theorem as $x \to \infty$ and noting that $(a - b) > 0$ we get $$\lim_{x \to \infty}\frac{\log x}{x^{a}} = 0$$