Suppose I want to increase the ratio:
$$ S(x) = \frac{f(x)}{g(x)} $$
I know that
$$ f(x) = \hat{f}(x) + O(x^{-p}) $$ $$ g(x) = \hat{g}(x) + O(x^{q}) $$
Where $p,q \in \mathbb{Z}_{++}$ and $f,g$ are positive real functions. Ie we have:
$$ S(x) = \frac{f(x)}{g(x)} = \frac{\hat{f}(x) + O(x^{-p})}{\hat{g}(x) + O(x^{q})} $$
Now if I have the ratio:
$$ \hat{S(x)} = \frac{\hat{f}(x)}{\hat{g}(x)} $$
I think that:
$$ S(x) \leq \hat{S(x)} \quad \forall x $$
Ie throwing out these higher order terms increases the ratio. How can I show this?
I looked at:
Big-O notation in division General Big-O operations. Check my proof: Big O notation
But I still am confused. I tried calculating:
$$ \frac{f(x) + O(x^{-p})}{g(x) + O(x^{q})} - \frac{f(x)}{g(x)} = \frac{O(x^{q})}{g^{2}(x) + O(x^{q})} $$
But that seems to just evaluate to 1.
Any insight?