If two functions can be written as the sum of some expression and an error term of higher orders of error $\epsilon$: $$f(x+\epsilon)=f_0(x,\epsilon)+O(\epsilon^m)\quad \text{ and} \quad g(x+\epsilon)=g_0(x,\epsilon)+O(\epsilon^n)$$
What is the order of error of the fraction $\frac{f(x+\epsilon)}{g(x+\epsilon)}$? Is it simply $O(\epsilon^{m-n})$?
On
I think I figured it out:
$$\frac{f(x+e)}{g(x+e)}=\frac{f0(x,e)+O(e^m)}{g0(x,e)+O(e^n)} =\frac{f0(x,e)}{g0(x,e)+O(e^n)}+\frac{O(e^m)}{g0(x,e)+O(e^n)} $$
The order of error of the fraction is the sum of that of the first one, which is n, and that of the second term which is m, therefore the overall order of error is the smaller of the two.
But this does not seem to explain the last equal sign of the long derivation in this page:
If $g(x)=0$ you are in trouble. Let's assume $g(x)\ne 0$, and $g$ is continuous. Then $g$ is bounded away from $0$ in a neighborhood of $x$.
Since $t\mapsto 1/t$ is locally Lipschitz away from $0$, $$\frac{1}{g(x+\epsilon)}=\frac{1}{g_0(x,\epsilon)}+O(\epsilon^n)$$ Multiply this by $f(x+\epsilon) =f_0(x,\epsilon)+O(\epsilon^m)$ and clean it up: $$ \frac{f(x+\epsilon)}{g(x+\epsilon)} = \frac{f_0(x,\epsilon)}{g_0(x,\epsilon)} + O(\epsilon^{\min(m,n)}) $$
And that's the best you can get.