Given two real valued funtions $f, g$, where $g$ is strictly positive, defined in some unbounded subset of $[0,\infty)$, we say that $f\in O(g)$ if there is some constant $C\in \mathbb{R}$ such that $$ |f(x)|\leq Cg(x) $$ for all large enough values of $x$.
We say that $f\in o(g)$ if $$ \lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}=0. $$ So clearly if $f\in o(g)$ then $f\in O(g)$, where $C$ in the definition can be taken to be any positive constant.
Now consider functions $x^r$ where $r\in [0,\infty)$.
My question is whether some function $f$ can satisfy $f\in o(x^r)$ for some $r$ but also $f\notin O(x^s)$ for every $0\leq s<r$.
Consider function $f(x) = \frac{x^r}{\ln(x)}$. Clearly, then $\lim_{x \to \infty} \frac{f(x)}{x^r} = \lim_{x \to \infty} \frac{1}{\ln(x)} = 0$.
Now, take any $s \in [0,r)$. Then $ t= r-s > 0$.
So we get: $f(x) = g(x) \cdot \frac{x^t}{\ln(x)}$
One can show, that $\frac{x^t}{\ln(x)} \to \infty$ as $x \to \infty$, so there doesn't exist any constant $C \in \mathbb R$, such that $f(x) \le C\cdot g(x)$