TITLE: $\lim_{x\rightarrow\infty}\frac{\log f(x^2)}{f(x)}$
QUESTION [5 upvotes]: 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.

REPLY [3 votes]: 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.