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?