MOTIVATION
I am investigating the behavior of monotonic functions defined in some neighborhood of $+\infty$ in terms of the limit $$\lim_{x\to +\infty}\frac{f(x)}{x^p},\tag{1}\label{1}$$ for $p> 0$.
Since the function is monotonic, the limit \eqref{1} always exists (either finite or infinite) for $p=0$.
In general, depending on $p$ the limit may be $0$, $\ell \neq 0$ (when $f(x) \sim kx^p$ for some $k$), it may not exist, and finally be infinite.
If the limit is $+\infty$ for some $\overline p$, it is infinite for also for $p< \overline p$, and, conversely, if the limit is $0$ for $\overline p$, it is $0$ for any $p>\overline p$. In between, we might have an interval (or a singleton) of values of $p$ for which the limit is either finite and non null or nonexistent.
It is easy to construct examples where the limit is either $0$, $+\infty$, or nonexistent for all $p>0$.
It is also possible to construct functions for which the limit is $+\infty$ for $p\leq \overline p$, and $0$ for $p> \overline p$. Consider, for example, for $x\geq 1$,
$$ f(x)= \begin{cases} x^{\frac{k+1}k} & \left(k^{k} \leq x < (k+1)^{k}\right)\\ (k+1)x & \left((k+1)^{k}\leq x < (k+1)^{k+1}\right) \end{cases}, \ \ \ k=1,2,\dots $$
See Figure below.
We have
$$\lim_{x\to \infty} \frac{f(x)}{x^p} = +\infty$$ when $p\leq 1$, and $$\lim_{x\to \infty} \frac{f(x)}{x^p} = 0,$$ for $p>1$.
QUESTION
I can only, think, however of non convex functions that have the above feature. Is this a limitation?
Thus we have:
Does there exist a convex function $f(x)$ and a positive number $\overline p$ such that $$\lim_{x\to \infty}\frac{f(x)}{x^p} = +\infty$$ for $p\leq \overline p$ and and $$\lim_{x\to \infty}\frac{f(x)}{x^p} = 0$$ for $p>\overline p$?
Do you have any hints on how to constuct it (if the answer is yes) or to prove that the answer is no (if that is the case)?