How do we know which terms are of higher order?

Question

From Asymptotic analysis and perturbation theory by Paulsen:

Find the behavior of the function defined implicitly by $$x^2+xy-y^3=0$$ as $x\to\infty$.

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At this point, we have shown that $y ∼ x^{2/3}$ is consistent, but this alone is not proof that the leading behavior is indeed $x^{2/3}$. In order to show that this is the correct leading behavior, we must find the next term in the series, and show that it is smaller than $x^{2/3}$.

Letting $y = x^{2/3} + g(x)$ gives us $$x^2 + x(x^{2/3} + g(x)) = (x^{2/3} + g(x))^3,$$ or $$x^2+ x^{5/3}+ xg(x) ∼ x^2+ 3g(x)x^{4/3}+ O(x^{2/3}g(x)^2).$$

The $x^2$’s cancel, and keeping the higher order terms, we get $x^{5/3} ∼ 3g(x)x^{4/3}$, which simplifies to $g(x) ∼ x^{1/3}/3$. This is indeed smaller than $x^{2/3}$ as $x → ∞$, so we have confirmed the leading behavior, along with the next term $y ∼ x^{2/3} + x^{1/3}/3$.

I don't understand the part where it says

and keeping the higher order terms, we get $x^{5/3} ∼ 3g(x)x^{4/3}$

Okay, after the $x^2$’s cancel we get $$x^{5/3}+ xg(x) ∼ 3g(x)x^{4/3}+ O(x^{2/3}g(x)^2).$$ but how do we know that $x^{5/3}$ is of higher order than $xg(x)$ and thus the latter is the one to be dropped? Isn't is what we wanted to show in the first place?

The reasoning seems circular.

How do we know which terms are of higher order?

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Note: In case it turns out to be relevant, here is my other question on this example: When is a balance assumption consistent?

Answer 1

On the right hand side we have after cancellation of $x^2$ \begin{align*} O\left(g(x)x^{4/3}\right) \end{align*} The order of the left hand side has to be the same as the order of the right hand side. Since the order of $x$ is less than the order of $x^{4/3}$ when $x$ tends to infinity we obtain \begin{align*} O\left(g(x)x^{4/3}\right)=O\left(x^{5/3}+xg(x)\right)=O\left(x^{5/3}\right) \end{align*}

We conclude \begin{align*} x^{5/3} \sim 3g(x)x^{4/3} \end{align*}