Big O - Equivalence: "Standard" definition vs Limit definition This question is related to: Landau Notation, Definitions: Limits vs. Cormen's. 

Consider functions $f, \ g : N \rightarrow R^{\geq0}$.

For small-$o$, the definition:
$$f(n)\in o(g(n)) \iff \forall c&gt;0,\exists n_0&gt;0, \forall n\geq n_0: f(n) &lt; c \ g(n)$$

Can also be expressed as a limit (i guess this assumes the limit exists):
$$f(n)\in o(g(n)) \iff \lim_{n\to\infty}\frac{f(n)}{g(n)}=0 $$

This can be verified introducing the definition for $\lim_{n\to\infty}$ on $N$, and then both definitions are identical.

Now, the "standard" definition for big-O would be something like (near exactly to Cormen's, except the codomain here is $R^{\geq0}$):
$$f(n)\in O(g(n)) \iff \exists c&gt;0,\exists n_0&gt;0, \forall n\geq n_0: f(n) \leq c \ g(n)  \qquad  (1)$$

Im trying to see if the following is true (again, this assumes the limit exists):
$$f(n)\in O(g(n)) \stackrel{?}{\iff} \exists c&gt;0: \lim_{n\to\infty}\frac{f(n)}{g(n)}=c$$

Im not sure both directions of the equivalence are true. More specifically, i am having trouble with the $\Rightarrow$. According to this general answer, both directions should be true for $\Theta(g(n))$, wich is stronger than $O(\_)$. On the contrary, this answer (for roughly the same question, but in math SE) says big-$O$ cant be defined in that way (it was downvoted).

So, the questions are:


How exactly def (1) implies the $\lim$ definition?:


$$\forall \epsilon&gt;0,\exists n_0&gt;0, \forall n\geq n_0: \left| \frac{f(n)}{g(n)}-c\right| &lt; \epsilon$$


Is the above still true considering generic functions $f, \ g : X \rightarrow R^{\geq0}$, where $X$ is some reasonable domain, maybe $Z$, $N^k$ or even $R^k$ (although reals are not turing-computable, im thinking in algorithms in the more general sense possible).
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 $\Theta$.