Why does $e^{-(x^2/2)} \approx \cos[\frac{x}{\sqrt{n}}]^n$ hold for large $n$? Why does this hold:
$$
e^{-x^2/2} = \lim_{n \to \infty} \cos^n \left( \frac{x}{\sqrt{n}} \right)
$$
I am not sure how to solve this using the limit theorem.

For sure, this is not as elegant as previous answers but I love too much Taylor expansions !
Starting with $$ \cos \left( \frac{x}{\sqrt{n}} \right)=1-\frac{x^2}{2 n}+\frac{x^4}{24 n^2}-\frac{x^6}{720 n^3}+\frac{x^8}{40320
   n^4}-\frac{x^{10}}{3628800 n^5}+O\left(x^{12}\right)$$ and raising to power $n$ (using binomial theorem) $$ \cos^n \left( \frac{x}{\sqrt{n}} \right)=1+\sum_{n=1}^\infty a_n x^{2n}$$ with $$a_1=-\frac 12$$ $$a_2=\frac{3n-2}{24n}\to \frac 18$$ $$a_3=-\frac{15 n^2-30 n+16}{720 n^2}\to -\frac 1{48} $$ $$a_4=\frac{105 n^3-420 n^2+588 n-272}{40320 n^3}\to  \frac 1{384}$$ $$a_5=-\frac{945 n^4-6300 n^3+16380 n^2-18960 n+7936}{3628800 n^4}\to -\frac 1{3840}$$ $$a_6=\frac{10395 n^5-103950 n^4+429660 n^3-893640 n^2+911328 n-353792}{479001600 n^5}\to \frac 1{46080}$$
