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Why do we have this: For $x = o(1)$, $1 - x = \exp(-x-O(x^2))$? I know the idea is from the Taylor's formula, but it sounds weird to me. I would like to know a clear explanation for it?

To be more clear, I know that these two things are asymptotically equal, as a calculus problem, but I like to look at it from the reverse; I mean I want to know how can I create an equality like this? It is not clear here how can we take inside the exponential function the term $O(x^2)$?

user24175
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  • $x=o(1)$ is just a fancy way of saying $x\to 0$. The rest is $\log(1-x) = -x + O(x^2)$. –  Aug 05 '15 at 18:44
  • @ Normal Human : Thank you! I know that these two things are asymptotically equal, when $x$ tends to zero, but how should I know that I can take $O(x^2)$ into the exponential function?! It is not clear, since if we use the Taylor's series for the exponential function, then we will have simply $O(x^2) $, not $\exp(O(x^2))$!!!? Instead of looking at this like a calculus problem, I want to see how can I create an equality like this while working with asymptotic stuff? – user24175 Aug 06 '15 at 00:33

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