If $f$ is $O(g)$ over some base, this means that $f(x) = \beta(x)g(x)$, where $\beta$ is eventually bounded. So this means that eventually, $f$ is at most $c$ times $g$, where $c$ is some constant.
But I thought $O$ was a way of expressing that two functions are eventually approximately the same. In that case, this seems like a poor definition. If $g$ is unbounded, then the difference between $f$ and $g$ is also unbounded, so the two functions might be $O$ of one another and yet still diverge wildly.
I would have used an additive definition: $f(x)=g(x)+\beta(x)$, where $\beta$ is eventually bounded. So then we know that $f(x)=g(x)\pm \epsilon$, where $\epsilon$ is some positive constant. We now have a statement on the error incurred by assuming $f=g$.
Similar objections can be raised against the definition of asymptotic equivalence. $x^2\sim x^2+x$ as $x\rightarrow+\infty$ even though their difference grows without bound. So in what sense are they equivalent?