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i was trying to solve this limit: $\lim_{x\to\infty} \frac{e^{-it\sqrt{x}}}{(1-\frac{it}{\sqrt{x}})^x}$

I have tried to use the fact that $\lim_{x\to\infty} (1+1/x)^x =e$.

$\lim_{x\to\infty} \frac{e^{-it\sqrt{x}}}{(1-\frac{it}{\sqrt{x}})^x} = \lim_{x\to\infty} \frac{e^{-it\sqrt{x}}}{((1-\frac{it}{\sqrt{x}})^\sqrt{x})^\sqrt{x}} = \lim_{x\to\infty} \frac{e^{-it\sqrt{x}}}{e^{-it\sqrt{x}}}=1 $

I am doing something wrong, probably abusing the fact that $\lim_{x\to\infty} (1-it/\sqrt{x})^\sqrt{x}=e^{-it}$ in this situation.

The limit shoud be $e^{-t^2/2}$, but how can i prove it?

Thank you

1 Answers 1

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HINT:

$$\left(1-\frac{it}{\sqrt x}\right)^{-x}=e^{-x\log\left(1-\frac{it}{\sqrt x}\right)}=e^{x\left(\frac{it}{\sqrt x}+\frac12 \left(\frac{it}{\sqrt x}\right)^2+O\left(\frac{it}{\sqrt x}\right)^3\right)}$$

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    Thank you! I thought that the Taylor series would have been useful, but did not know how to use them – 2017-02-15
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    You're welcome! My pleasure. -Mark – 2017-02-15
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    And feel free to up vote and best vote as you see fit of course. ;-)) – 2017-02-15