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For my probability homework I have to show that a certain limit exists and equals $e^{\frac{1}{2}t^2}$.

The limit in question is $\lim_{n\rightarrow\infty} e^{-t\sqrt{n}}\left(1-\frac{t}{\sqrt{n}}\right)^{-n}$.

I have tried the following simplifications: \begin{align*} &\quad\ \text{substitute $m = \sqrt{n}$}\\ &= \lim_{m\rightarrow\infty} e^{-tm}\left(1+\frac{-t}{m}\right)^{-m^2}\\ &= \lim_{m\rightarrow\infty} e^{-tm}\left(\left(1+\frac{-t}{m}\right)^m\right)^{-m}\\ &= \lim_{m\rightarrow\infty} e^{-tm}\left(e^{-t}\right)^{-m}\\ &= \lim_{m\rightarrow\infty} e^{-tm+tm}\\ &= 1 \end{align*}

But according to wolfram alpha during the third equality the outcome changes.

Can anyone help me on how to properly calculate this limit?

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    $\begingroup$ You can't just calculate the limit for $m \to \infty$ for some of the "inner" terms and leave the rest unchanged. That is like saying $\lim_{n \to \infty} \frac{n^2}{n} = n^2 \lim_{n \to \infty} \frac{1}{n} = 0$. $\endgroup$ – Viktor Glombik May 11 at 22:57
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Consider the logarithm which equals $$ -t\sqrt{n}-n\log\left(1-\frac{t}{\sqrt{n}}\right)=-t\sqrt{n}+n\left( \frac{t}{\sqrt{n}}+\frac{t^2}{2n}+o(n^{-1}) \right)=\frac{t^2}{2}+o(1)\to t^2/2 $$ as $n\to \infty$ where we used the taylor expansion of $-\log(1-x)=x+\frac{x^2}{2}+o(x^2)$ in the first equality.

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$-t\sqrt n-n\log(1-\frac t {\sqrt n})=-t\sqrt n-n(-\frac t {\sqrt n}-(\frac t {\sqrt n})^{2}/2-... \to t^{2} /2$ so the given limit is $e^{t^{2}/2}$.

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