How about Why does the Landau notation for division work when h approaches 0?

[IniquiTrance]

Why is it that,

$\frac{a+\mathcal{O}(h^2)}{b+\mathcal{O}(h^2)} = \frac{a}{b}+\mathcal{O}(h^2)$

as $h\rightarrow 0$? It seems like the $\mathcal{O}(h^2)$ term should become $\mathcal{O}(1)$.

 

[statdad]

Why is it that,

a+(h2)b+(h2)=ab+(h2) \frac{a+\mathcal{O}(h^2)}{b+\mathcal{O}(h^2)} = \frac{a}{b}+\mathcal{O}(h^2)

as h→0h\rightarrow 0? It seems like the (h2)\mathcal{O}(h^2) term should become (1)\mathcal{O}(1).

Think about this very simple illustration. Consider the specific case

<br /> c = \frac{a + k1h^2}{b + k2h^2}<br />

with k1, k2 positive (specific choices for the second terms in numerator and denominator, but since h-squared is the key term it's find. Bear with me.)

Suppose that fraction is expanded into a power series in h. The first few terms are

<br /> \frac a b −\frac{((ak2−bk1)h^2)}{b^2}+\frac{((ak2^2−bk1k2)h^4)}{b^3}−\frac{((ak2^3−bk1k2^2)h^6)}{b^4}+\frac{((ak2^4−bk1k2^3)h^8)}{b^5}+... = \frac a b + O(h^2)<br />