0
$\begingroup$

I would very much like to say that $f + h = o(\text{whichever is bigger between g and p})$ but am a little bit worried if we have $g$ and $p$ oscillating as (say) $x \to \infty$. If we add the restriction that $g = O(p)$ OR $p = O(g)$ is this sufficient to show my desired statement?

$\endgroup$
  • $\begingroup$ You are right. In general, you can only say $o(\max\{g,p\})$. But for the usual items used in asymptotic analysis (Hardy's field), any two of them are comparable, so that maximum is equal to one of them. Why not try to find a counterexample, where $\limsup g/p = +\infty$ and $\liminf g/p = 0$ where both $f+g=o(p)$ and $f+g = o(g)$ fail. $\endgroup$ – GEdgar May 30 at 12:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.