Doubts on Big O definition

Question

I have doubts on big-o $O$.

Considering two function $f$ and $g$ , if

$\lim_{x\to c} \frac{f(x)}{g(x)}=l \in \mathbb{R}$

Then we say that $f=O(g)$ as $x \to c$

Is it true in general that $f=O(g) \iff g=O(f)$?

I don't see why it shouldn't be, since, following the definition,

$\lim_{x\to c} \frac{f(x)}{g(x)}=l \in \mathbb{R} \iff \lim_{x\to c} \frac{g(x)}{f(x)}=m \in \mathbb{R} $

If this is not true, in this particular case is the following implication correct?

$f\sim g \implies f=O(g) \wedge g=O(f)$

Thanks a lot for your help

Answer 1

No.

The usual definition of $f=O(g)$ is that there exist $a,b>0$ such that $|f(x)|<a\cdot |g(x)|$ whenever $|x-c|<b$ (or for the exceptional case $c=+\infty$: whenever $x>b$). Then limit $\lim_{x\to c}\frac{f(x)}{g(x)}$ need not even exist; for example $\sin x = O(1)$ as $x\to +\infty$ (or in fact for any other $c$ as well).

Answer 2

No, it is not true since it is allowed that the limit $l$ may be zero. -- Note that the limit should properly be a limes superior, if you want to define the big-O using limits. This then makes the behavior of the reciprocal even less automatic.

There is notation for asymptotic equivalence, i.e. when the limit exists and is different from zero (or if the lim sup is finite in both directions), with Omega or Theta letters, consult the wikipedia page on Landau notation.