For positive a,b, I have that $x‘(t)=e^{-ax(t)}-e^{-ay(t)}+O(e^{-(a+b)x(t)})$ and can show that this has the lower bound $x‘(t)\geq O(e^{-(a+b)x(t)})\geq O(e^{-aM})$ for some positive M. I would like to deduce that $x‘(t)>0$ as $t\to\infty$.
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$\begingroup$ It does not make sense, but people will understand what you mean (though they'll probably wince). A better statement would be $f(x) = \Omega(1)$ (if you're familiar with the $\Omega()$ counterpart to $O()$; careful, number theorists have a slightly different definition than computer scientists). $\endgroup$– Clement C.Mar 11, 2020 at 16:17
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$\begingroup$ (question edited, the above comment does no longer apply verbatim, but the underlying point about $O()$ v. $\Omega()$ still does) $\endgroup$– Clement C.Mar 11, 2020 at 16:23
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$\begingroup$ Bit why is $x‘$ eventually positive? $\endgroup$– ScuderiMar 11, 2020 at 16:35
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$\begingroup$ I commented on your use of $O()$, which was your original question; not on what you're trying to prove in the end (which is your edited question). $\endgroup$– Clement C.Mar 11, 2020 at 16:37
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Why are you comparing the lower limits with the Big O notation? It is highly incorrect. It is used only for upper limits (bounds). As for ypur question, the Big O notation is defined for positive values only, and it will always be positive.
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$\begingroup$ For positive a,b, I have that $x‘(t)=e^{-ax(t)}-e^{-ay(t)}+O(e^{-(a+b)x(t)}$ and can show that this has the lower bound $x‘(t)\geq O(e^{-(a+b)x(t)})\geq O(e^{-aM})$ for some positive M. I would like to deduce that $x‘(t)>0$ as $t\to\infty$. $\endgroup$– ScuderiMar 11, 2020 at 16:20