$\lim_{x\rightarrow\infty}\frac{\log f(x^2)}{f(x)}$

Question

I have a feeling that $\lim_{x\rightarrow\infty}\frac{\log f(x^2)}{f(x)}=0$ for any positive monotone increasing function such that $\lim_{x\rightarrow\infty}f(x)=\infty$, (in reality I am thinking mostly about super-exponential growth here but I do not think it might make a difference). The idea, intuitively, is that the logarithm takes away more growth than the square in the argument boosts. Though, I am not sure if this is true (maybe adding some more hypotheses) because I find it hard to prove neatly, having $f$ in between mixing these two contributions up. Tried L'Hospital but did not work and standard limit stuff neither, unsuccessfully. Anybody sees which way of approaching this could be productive? Don't need detailed answers, if you have a hunch that'll be enough.

Answer 1

This is not always the case because it would imply that $$\exp(f(x))>f(x^2)$$ for all $x$ large enough, which can't be the case for all $f$ that satisfy your conditions. More precisely, it is not true if $f$ grows very fast.

For example, take any monotonic (for example piecewise linear) continuation of the tetration $$f:\mathbb N\to\mathbb N, n\mapsto 3\uparrow\uparrow n.$$

Then we have $f(n^2)\gg\exp(f(n))$ for all $n\in\mathbb N_{>2}$ and I think that your limit actually equals $\infty$.

Answer 2

Define a sequence $a_n$ with arbitrary $a_0>1$ and $a_{n + 1}=e^{a_n}$.

Define a function $f$ such that $f(x)= a_0$ for $x< 2$ and $f(x) = a_n$ for $2^{2^{n - 1}}\leq x < 2^{2^n}$.

Then $f$ is a counter example.

It can be modified (by "smoothing out" the jump points) to an example of continuous or even $\mathcal C^\infty$ function.