# Landau notation division

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1. Nov 8, 2014

### 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 $\mathcal{O}(1)$.

2. Nov 13, 2014

### Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. Nov 14, 2014

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