Inequality in the limit

Question

Given that we have the following conditions:

$f = O(\delta)$, $g = O(\delta^2)$, $f > 0, \delta > 0$,

can we conclude that as $\delta \to 0^+$, $f+g>0$?

Answer 1

No. You could have $f(\delta)=-g(\delta)$ for example. In general, $f=O(\delta)$ is only an upper bound, but you're trying to deduce a lower bound - that won't work.

You would need a stronger statement like $f(\delta) = \Omega(\delta)$ or $f(\delta) \ne O(\delta^2)$ - some lower bound on $f$ - to conclude that $f+g>0$ near $0$.