When looking for the approximate roots of $\epsilon^2x^6-\epsilon x^4-x^3+8=0$, since this is a single perturbation problem, we need to track down the three missing roots, so we consider all possible dominant balances between pairs of terms as $\epsilon\to 0$.
Now suppose $\epsilon^2x^6\sim\epsilon x^4$ ($\epsilon\to 0$) is the dominant balance. Then $x=O(\epsilon^{-1/2}) (\epsilon\to 0)$. It follows that the terms $\epsilon^2x^6$ and $\epsilon x^4$ are both $O(\epsilon^{-1})$. But $\epsilon x^4\ll x^3=O(\epsilon^{-3/2})$ as $\epsilon\to 0$.
Instead if writing the entire comparison, I will write the comparison using shorthand $\ll$. Now I am trying to understand why $O(\epsilon^{-1})\ll O(\epsilon^{-3/2})$ as $\epsilon\to 0$.
I am finding it hard to get the logic behind it or at least understand it in a more general setting.
I tried to reason as follows:
We know $O (\epsilon^{-1/3})\ll O(\epsilon^{-1})$ is obvious, then I guess it is also the case that $O (\epsilon^{-1/2})\ll O(\epsilon^{-1})$ because taking the root (square and cube) of $\epsilon$ makes it bigger for small $\epsilon$, since $\epsilon$ is in the denominator, the whole fraction gets smaller.
But $O (\epsilon^{-3})\gg O(\epsilon^{-1})$, because for small $\epsilon$ taking power of it makes it smaller, since $\epsilon$ is in the denominator, hence the whole fraction gets larger.
Then if we "combine" the operation of taking square root and raising to the power of 3. how can we conclude $O(\epsilon^{-1})\ll O(\epsilon^{-3/2})$? Why not $O(\epsilon^{-1})\gg O(\epsilon^{-3/2})$?
Is there any general rule to compare $O(\epsilon^{-1})$ with any $O(\epsilon^{-n/m})$? and to make it even more general to compare $O(\epsilon^{-n/m})$ with $O(\epsilon^{-r/s})$?
Thanks!
Let $p$ and $q$ be positive real numbers with $p < q$. Then
$$ \frac{\epsilon^{-p}}{\epsilon^{-q}} = \epsilon^{q-p} \to 0 $$
as $\epsilon \to 0$ with $\epsilon > 0$.