I always have issues dealing with asymptotic notation...
I am trying to verify the following step:
$$\left(1-\frac{t^2}{2n} + o(1/n)\right)^n \to e^{-t^2/2}.$$
To change this into $(1-t^2/(2n))^n \to e^{-t^2/2}$, it suffices to show that $$\left(1+\frac{o(1/n)}{1-t^2/(2n)}\right)^n \to 1,$$ but I don't see where to go from here. Any hints would be appreciated!
Also apparently the same result holds when replacing $o(1/n)$ with $O(1/n^2)$...
Since $$ \left(1-\frac{t^2}{2n}+o\left(\frac1n\right)\right)=\left(1-\frac{t^2}{2n}\right)\left(1+o\left(\frac1n\right)\right) $$ we can raise both to the $n^{\text{th}}$ power: $$ \begin{align} \lim_{n\to\infty}\left(1-\frac{t^2}{2n}+o\left(\frac1n\right)\right)^n &=\lim_{n\to\infty}\left(1-\frac{t^2}{2n}\right)^n\lim_{n\to\infty}\left(1+o\left(\frac1n\right)\right)^n\\ &=e^{-t^2/2}\lim_{n\to\infty}e^{o(1)}\\ &=e^{-t^2/2}\,e^0\\[6pt] &=e^{-t^2/2} \end{align} $$
If there is doubt that $\lim\limits_{n\to\infty}\left(1+o\left(\frac1n\right)\right)^n=1$, note that since $o\left(\frac1n\right)=\frac{o(1)}n$, we have that for any $\varepsilon\gt0$, $$ e^{-\varepsilon}=\lim_{n\to\infty}\left(1-\frac\varepsilon n\right)^n \le\lim_{n\to\infty}\left(1+o\left(\frac1n\right)\right)^n \le\lim_{n\to\infty}\left(1+\frac\varepsilon n\right)^n=e^\varepsilon $$ Finish with the Squeeze Theorem.