# Inequality including Big-Oh-notation

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 $$t\to\infty$$.

• 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). Mar 11, 2020 at 16:17
• (question edited, the above comment does no longer apply verbatim, but the underlying point about $O()$ v. $\Omega()$ still does) Mar 11, 2020 at 16:23
• Bit why is $x‘$ eventually positive? Mar 11, 2020 at 16:35
• 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). Mar 11, 2020 at 16:37

• 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 $t\to\infty$. Mar 11, 2020 at 16:20