big O inequality

Question

In $e^{-\omega/\epsilon}\leq10^{-9}$ why is $\omega=O(\epsilon)$ and in $e^{-\omega/\epsilon}\leq \epsilon$ why is $\omega=O(\epsilon\ln1/\epsilon)$ where $\epsilon$ is a very small parameter.

Answer 1

(I'll post my comment above as an answer since it seems to address what the OP was looking for.)

In the reference given for the source of this question it states

The width of the boundary layer [...] could be taken to be the smallest value $\omega$ such that $e^{-\omega/\epsilon} \leq 10^{-9}$, say. In this case it is clear that $\omega = O(\epsilon)$.

As shown in this related answer, the smallest such $\omega$ is $\omega = (9\log 10)\epsilon$, which is indeed $O(\epsilon)$.

It's not true that all $\omega$ which satisfy $e^{-\omega/\epsilon} \leq 10^{-9}$ are $O(\epsilon)$ (for example take $\omega = (9\log 10)\sqrt{\epsilon}$), but the smallest one, the one you care about, certainly is. Dealing with the other inequality is similar.