Big O - Equivalence: "Standard" definition vs Limit definition
If $f = O(g)$ then there exists $C&gt;0$ such that for large enough $n$, we have $f(n)/g(n) \leq C$. Hence under mild assumptions,
$$
f = O(g) \Longleftrightarrow \limsup_{n\to\infty} \frac{f(n)}{g(n)} &lt; \infty.
$$
As Raphael mentions in a comment to your post, your definition is symmetric in $f,g$ and so is more appropriate for $\Theta$.