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I would appreciate help understanding how the Big-O term is handled in taking the inverse of this expression and in general how to go about it:

The inverse of this:

$log (2) (s - 1) + \mathcal{O}((s - 1)^2)$ as $s \rightarrow 1$

was given as:

$(log (2) (s - 1))^{- 1} + \mathcal{O}(1)$ as $s \rightarrow 1$

Thanks very much.

EDIT: In looking at this further, it looks as if the Big-O term is just $(s - 1)$ to a degree higher than the explicit term. So in the expression for the inverse, it becomes $\mathcal{O}((s - 1)^{0})$.

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