# Limit of $\lim\limits_{s\to\infty}\big((2-e^{t/\sqrt{2s}})e^{t/\sqrt{2s}}\big)^{-s}$ [duplicate]

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Help with convergence in distribution

Could any one tell me how to calculate this limit, $$\lim_{s\rightarrow\infty}\left(\frac{1}{(2-e^{t/\sqrt{2s}})e^{t/\sqrt{2s}}}\right)^{s}$$

Thanks for helping me.

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## marked as duplicate by Did, Asaf Karagila, t.b., J. M., Willie WongDec 13 '11 at 9:31

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Is this question any different than one you asked about just a while ago and were pointed to this answer? If so, I think this question should be closed. – Dilip Sarwate Dec 9 '11 at 19:50
32 minutes.  – Did Dec 9 '11 at 20:31

## 1 Answer

Let $f(s) = (2 - e^{t/\sqrt{2s}}) e^{t/\sqrt{2s}}$. Then as $s \to \infty$, $e^{t/\sqrt{2s}} = 1 + \frac{t}{\sqrt{2s}} + \frac{t^2}{4s} + O(s^{-3/2})$, so $f(s) = 1 - \frac{t^2}{2s} + O(s^{-3/2})$, and $\ln f(s) = - \frac{t^2}{2s} + O(s^{-3/2})$. Therefore $\left(\frac{1}{f(s)}\right)^s = e^{-s \ln f(s)} \to e^{t^2/2}$.

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