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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.

  • 5
    Is this question any different than one you asked about just a while ago and were pointed to [this answer](http://math.stackexchange.com/a/87067/15941)? If so, I think this question should be closed.2011-12-09
  • 1
    32 minutes. $ $2011-12-09

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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 $\left(\frac{1}{f(s)}\right)^s = e^{-s \ln f(s)} \to e^{t^2/2}$.