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.

$$
\begin{align}
\cos^n\!\left(\frac{x}{\sqrt{n}}\right)
&=\left(1-\sin^2\left(\frac{x}{\sqrt{n}}\right)\right)^{n/2}\\
&=\left(1-\frac{x^2}n\color{#C00000}{\left[\frac{\sin\left(\frac{x}{\sqrt{n}}\right)}{\frac{x}{\sqrt{n}}}\right]^2}\right)^{n/2}\\
\end{align}
$$
Since
$$
\lim_{n\to\infty}\frac{\sin\left(\frac{x}{\sqrt{n}}\right)}{\frac{x}{\sqrt{n}}}=1
$$
we can choose an $n$ large enough that the expression in red is as close to $1$ as we wish. Therefore, because $e^x$ is continuous for all $x$, we get
$$
\begin{align}
\lim_{n\to\infty}\cos^n\!\left(\frac{x}{\sqrt{n}}\right)
&=\lim_{n\to\infty}\left(1-\frac{x^2}n\right)^{n/2}\\[6pt]
&=e^{-x^2/2}
\end{align}
$$
