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.
 A: In a neighbourhood of the origin,
$$\log\cos z = -\frac{z^2}{2}\left(1+O(z^2)\right)\tag{1} $$
hence for any $x$ and for any $n$ big enough:
$$ \log\left(\cos^n\frac{x}{\sqrt{n}}\right)=-\frac{x^2}{2}\left(1+O\left(\frac{1}{n}\right)\right)\tag{2}$$
and the claim follows by exponentiating $(2)$:
$$ \cos^n\frac{x}{\sqrt{n}} = e^{-x^2/2}\cdot\left(1+O\left(\frac{1}{n}\right)\right).\tag{3}$$
A: $$
\begin{align}
\cos^n\!\left(\frac{x}{\sqrt{n}}\right)
&=\left(1-\sin^2\left(\frac{x}{\sqrt{n}}\right)\right)^{n/2}\\
&=\left(1-\frac{x^2}n\color{#C00000}{\left[\frac{\sin\left(\frac{x}{\sqrt{n}}\right)}{\frac{x}{\sqrt{n}}}\right]^2}\right)^{n/2}\\
\end{align}
$$
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
$$
\begin{align}
\lim_{n\to\infty}\cos^n\!\left(\frac{x}{\sqrt{n}}\right)
&=\lim_{n\to\infty}\left(1-\frac{x^2}n\right)^{n/2}\\[6pt]
&=e^{-x^2/2}
\end{align}
$$
A: $\cos(x)$ is the characteristic function of a signed Bernoulli.
A: Recall that if $t\in[0,\frac\pi2)$ then
$$ \sin t \le t \le \tan t $$
Apply $\int_0^x \cdot \,dt$ to obtain that if $x\in[0,\frac\pi2)$ then
$$ 1 - \cos x \le \frac{x^2}{2} \le \ln\sec x $$
Rearranging yields that if $x\in[0,\frac\pi2)$ then
$$ 1 - \frac{x^2}{2} \le \cos x \le e^{-x^2/2} $$
Everything is even, so in fact this holds for $x\in(-\frac\pi2,\frac\pi2)$.  For general $x$, we have $\frac{x}{\sqrt n}\in (-\frac\pi2,\frac\pi2)$ for all sufficiently large $n$, and so
$$ 1 - \frac{x^2}{2n} \le \cos\Big(\frac{x}{\sqrt n}\Big) \le e^{-x^2/2n} $$
for sufficiently large $n$.  Raising both sides to the power $n$ and applying the squeeze theorem finishes the job.
A: Not as elegant as Jack's answer, but rewrite the limit as 
$$\lim_{n\rightarrow \infty} e^{\dfrac{\ln(\cos(x/\sqrt{n}))}{1/n}}$$
and use the continuity of $e^x$ to perform l'Hospital's rule twice. 
A: Hint: Use the Taylor series approximation for $~\cos t\simeq1-\dfrac{t^2}2~$ when $t\to0$, in conjunction with the limit definition of $~e=\displaystyle\lim_{u\to\infty}\bigg(1+\frac1u\bigg)^u$
A: 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 $$a_1=-\frac 12$$ $$a_2=\frac{3n-2}{24n}\to \frac 18$$ $$a_3=-\frac{15 n^2-30 n+16}{720 n^2}\to -\frac 1{48} $$ $$a_4=\frac{105 n^3-420 n^2+588 n-272}{40320 n^3}\to  \frac 1{384}$$ $$a_5=-\frac{945 n^4-6300 n^3+16380 n^2-18960 n+7936}{3628800 n^4}\to -\frac 1{3840}$$ $$a_6=\frac{10395 n^5-103950 n^4+429660 n^3-893640 n^2+911328 n-353792}{479001600 n^5}\to \frac 1{46080}$$
