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I understand

$\lim_{n\to\infty}$$e^{-\sqrt{n}x}$$(1-\frac{x}{\sqrt{n}})^{\sqrt{n}x-n}$ = $e^{\frac{-x^2}{2}}$

But wanted to see the explicit steps for evaluating the limit. Any help would be really appreciated!

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    Here is my try. I don't get the factor 1/2 in the exponent. And I'm not very sure about how "fair" is the move of taking limits "by steps". $$\lim_{n\to\infty}e^{-\sqrt{n}x}\left(\left(1-\frac{x}{\sqrt{n}}\right)^{-\frac{\sqrt n}{x}}\right)^{-\frac{x(\sqrt{n}x-n)}{\sqrt n}}=$$ $$=\lim_{n\to\infty}e^{-\sqrt{n}x}\left(\left(1-\frac{x}{\sqrt{n}}\right)^{-\frac{\sqrt n}{x}}\right)^{-x^2+\sqrt nx}=$$ $$=\lim_{n\to\infty}e^{-\sqrt{n}x}e^{-x^2+\sqrt nx}=e^{-x^2}$$ – 2017-01-28

2 Answers 2

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Hint

Consider $$A=e^{-\sqrt{n}x}(1-\frac{x}{\sqrt{n}})^{\sqrt{n}x-n}$$ Take logarithms $$\log(A)={-\sqrt{n}x}+({\sqrt{n}x-n})\log\left(1-\frac{x}{\sqrt{n}} \right)$$ Now, for large values of $n$, by Taylor expansion or using equivalents $$\log\left(1-\frac{x}{\sqrt{n}} \right)\approx -\frac{x}{\sqrt{n}}$$ Replace and simplify.

Edit

As Rafa Budría commented, he/she and I missed the factor $\frac 12$ which make our answers wrong.

To get the right answer, we need to write $$\log\left(1-\frac{x}{\sqrt{n}} \right)=-\frac x {\sqrt n} -\frac{x^2}{2 n}+O\left(\frac{1}{n^{3/2}}\right)$$ which makes, after simplifications, to $$\log(A)=-\frac{x^2}{2}-\frac{x^3}{{6} \sqrt{n}}-\frac{x^4}{12 n}+O\left(\frac{1}{n^{3/2}}\right)$$ and Taylor again $$A=e^{\log(A)}=e^{-\frac{x^2}{2}}\left(1-\frac{x^3}{{6} \sqrt{n}}+O\left(\frac{1}{n}\right) \right)$$

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    It gets the same: doesn't appear the 1/2 factor in the exponent. – 2017-01-28
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    @RafaBudría. I agree with you. We miise the 1/2 factor. – 2017-01-29
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Let us do it in this way!! we can write $$(1-\frac{x}{\sqrt{n}})^{\sqrt{n}x-n}=e^{(\sqrt{n}x-n)\log(1-\frac{x}{\sqrt{n}})}.$$ So \begin{equation} \lim_{n\to\infty}e^{-\sqrt{n}x}(1-\frac{x}{\sqrt{n}})^{\sqrt{n}x-n}=e^{\lim_{n\to\infty}(-\sqrt{n}x)+\lim_{n\to\infty}\left((\sqrt{n}x-n)\log(1-\frac{x}{\sqrt{n}})\right)} \end{equation} Now \begin{align*} \lim_{n\to\infty}\left((\sqrt{n}x-n)\log(1-\frac{x}{\sqrt{n}})\right)&=\lim_{n\to\infty}\frac{\log(1-\frac{x}{\sqrt{n}})}{\frac{1}{(\sqrt{n}x-n)}}~~~~~~~~~(\frac{0}{0} ~~\text{form})\\\\ &=\lim_{n\to\infty}\dfrac{\frac{1}{\left(1-\frac{x}{\sqrt{n}}\right)}\frac{x}{2\sqrt{n}}}{\frac{-\left(\frac{x}{2\sqrt{n}}-1\right)}{\left(\sqrt{n}x-n\right)^2}}\\\\ &=\lim_{n\to\infty}\frac{x}{2}\left(\dfrac{n-\sqrt{n}x}{\sqrt{n}-\frac{x}{2}}\right)~~~~~~~~~(\frac{\infty}{\infty} ~~\text{form})\\ \\ &=\lim_{n\to\infty}\frac{x}{2}~~\dfrac{1-\frac{x}{2\sqrt{n}}}{\frac{1}{2\sqrt{n}}}\\\\ &=\lim_{n\to\infty}\left(x\sqrt{n}-\frac{x^2}{2}\right) \end{align*} Hence \begin{align*} \lim_{n\to\infty}e^{-\sqrt{n}x}(1-\frac{x}{\sqrt{n}})^{\sqrt{n}x-n}&=e^{\lim_{n\to\infty}(-\sqrt{n}x)+\lim_{n\to\infty}\left((\sqrt{n}x-n)\log(1-\frac{x}{\sqrt{n}})\right)}\\ \\ &=e^{\lim_{n\to\infty}(-\sqrt{n}x)+\lim_{n\to\infty}\left(x\sqrt{n}-\frac{x^2}{2}\right)}\\\\ &=e^{-\frac{x^2}{2}} \end{align*}