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How to prove that 1/(1+g(x)) = 1 - g(x) + o(g(x)). Moreover, what exactly does this equality mean in this particular case?

user516517
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1 Answers1

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You can do a geometric series: $1/(1+g(x))=1/(1-(-g(x))=1-g(x)+g(x)^2-g(x)^3+\dots=1-g(x)+\mathcal o(g(x)^2)$.

The equality means that the tail end of the series goes to zero faster than $g(x)^2$.