Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I got a good answer to this question over on MathOverflow a while ago. Harald Hanche-Olsen claimed that, if $f, g: D\to \mathbb{R}^+$, then $$ f(x) \sim g(x) \implies f(x) \asymp g(x) \qquad \qquad (*) $$ holds whenever $D\subseteq\mathbb{C}$ is closed in $\mathbb{C}$, and is false whenever it is not closed.

However, something bugs me about this. Most instances I've seen this, it does hold when $D$ is the strictly positive reals, which I believe is not closed in $\mathbb{C}$.

So this is my question. Does $(*)$ hold when $f, g: \mathbb{R}^+ \to \mathbb{R}^+$? That is, when $f$ and $g$ are positive real-valued and defined on the positive reals?

(Edit: For a detailed explanation of which definitions I am using, see the link. I believe they are standard.)

share|cite|improve this question
Claimed in the sense of proved. – Did Jun 10 '11 at 17:48

No, for basically the same reason it fails for other non-closed sets. Just because $\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}=1$, it doesn't mean that $\lim_{x\rightarrow 0} \frac{f(x)}{g(x)}$ even exists.

So you could, for example, take $f(x) = \frac{1}{x}$ and $g(x) = \frac{1}{x+1}$. Then certainly $\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}=1$, but there is no $A$ such that $f(x) < Ag(x)$ for all $x$, since $f$ has a vertical asymptote at 0.

share|cite|improve this answer
... which is why the definitions in the link are not useful ... – GEdgar Jun 10 '11 at 19:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.