# calculating $lim_{n\rightarrow\infty} e^{-t\sqrt{n}}\left(1-\frac{t}{\sqrt{n}}\right)^{-n}$

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?

• 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 $\lim_{n \to \infty} \frac{n^2}{n} = n^2 \lim_{n \to \infty} \frac{1}{n} = 0$. – Viktor Glombik May 11 at 22:57

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