Q:

Reciprocal of an asymptotic quantity

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?

A:

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.

