Suppose I have a quantity $M(x,\epsilon) = f(x) + O(\epsilon)$ where $o < \epsilon << 1$. So in other words, $M$ is expanded in small parameter $\epsilon$. If I take its reciprocal, what would be correct?
Either a): $$\frac{1}{M(x, \epsilon)} = \frac{1}{f(x)} + O(\epsilon).$$
or b):
$$\frac{1}{M(x, \epsilon)} = \frac{1}{f(x)} + O(\epsilon^{-1})$$
Common sense implies the correct statement is (a), but how to prove that?
Assuming that $f(x)$ is bounded away from zero,
$\begin{array}\\ \frac1{M(x,\epsilon)} & = \frac1{f(x) + O(\epsilon)}\\ & = \frac1{f(x)}\frac1{1 + O(\epsilon)/f(x)}\\ & = \frac1{f(x)}(1 - \frac{O(\epsilon)}{f(x)}) \qquad(*)\\ & = \frac1{f(x)}- \frac{O(\epsilon)}{f^2(x)}\\ \end{array} $
$(*)$ follows from $\frac1{1+z} = 1-\frac{z}{1+z} $ so, if $|z| < r < 1$, $\frac1{1+z} = 1-az $ where $\frac1{1+r} \lt a \lt \frac1{1-r} $.
For example, if $|z| < \frac12$, then $\frac23 < a < 2$.
Also, I left the sign of the big-oh term negative purposely.