In many calculus exercises we encounter limits like $\lim_{x\rightarrow \infty} 1/f(x)$ where $f(x)$ is some unbounded and strictly increasing function of $x$.
I'm wondering if there's a way to prove that the above limit is $0$ for any such $f(x)$. That is, we want to show that for all $\epsilon>0$ there is an $M= M(\epsilon)$ such that $\forall x>M$ we have $|1/f(x)|<\epsilon$.
The tricky part seems to be how to precisely define $M(\epsilon)$ as its value depends on the actual growth of $f(x)$, which we don't know.
So is it possible to prove the above statement without knowing $f(x)$?