# Landau notation division

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)##.

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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).

$$c = \frac{a + k1h^2}{b + k2h^2}$$
$$\frac a b −\frac{((ak2−bk1)h^2)}{b^2}+\frac{((ak2^2−bk1k2)h^4)}{b^3}−\frac{((ak2^3−bk1k2^2)h^6)}{b^4}+\frac{((ak2^4−bk1k2^3)h^8)}{b^5}+... = \frac a b + O(h^2)$$