A function $f(x)$ is $O(g(x))$ if and only if there exists a real number $M$ such that there exists $x_0$ such that for every $x>x_0$ the inequality $|f(x)|\le M|g(x)|$.
It turns out the following definition is equivalent to the one above: $f(x)=O(g(x))$ iff there exists a bounded function $h(x)$ such that $f(x)=h(x)g(x)$.
I guess it's not entirely true, because in the example below $f(x)=O(g(x))$, but you certainly can't express $f(x)$ as $h(x)g(x)$ for small $x$ with a bounded $h(x)$. So I believe there's one more requirements needed for equivalence - the function has to be bounded past some $x_0$.
If so, how can one prove these definitions equivalent?