Question about limit $\lim\limits_{n\rightarrow\infty}\cos^n\left(\frac{\omega}{\sqrt{n}}\right)$ My question is how to calculate this limit.
$$\lim_{n\rightarrow\infty}\cos^n\left(\frac{\omega}{\sqrt{n}}\right)$$
I know that the answer is $e^{-\frac{\omega^2}{2}}$. 
Attempts:
I tried to reduce the limit to the known limit
$$\lim_{n\rightarrow \infty}\left(1+\frac{a}{n}\right)^n=e^{a}$$
So, I wrote $$\cos\left(\frac{\omega}{\sqrt{n}}\right)\approx1-\frac{\omega^2}{2n}$$
using the cosine Taylor series, and stop there because $\frac{\omega}{\sqrt{n}}$ gets very small as $n\rightarrow \infty$.
Then, the limit is
$$ \lim_{n\rightarrow\infty}\cos^n\left(\frac{\omega}{\sqrt{n}}\right)=\lim_{n\rightarrow \infty}\left(1-\frac{\omega^2}{2n}\right)^n=e^{-\frac{\omega^2}{2}}$$
I also tried using $\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$ and then using the binomial theorem with no success.
Is this answer correct?
 A: For $n>\omega^2$, we have
$$
1-\frac{\omega^2}{n}+\frac{\omega^4}{24n^2}>\cos(\omega/\sqrt{n})>1-\frac{\omega^2}{2n}
$$
hence
$$
e^{-\omega^2/2}\leftarrow\left(1-\frac{\omega^2}{2n}+\frac{\omega^4}{24n^2}\right)^n>\cos^n(\omega/\sqrt{n})
>\left(1-\frac{\omega^2}{2n}\right)^n\to e^{-\omega^2/2}
$$
The limit on the left is obtained by
$$
\frac{\left(1-\frac{\omega^2}{2n}+\frac{\omega^4}{24n^2}\right)^n}{\left(1-\frac{\omega^2}{2n}\right)^n}=\left(1+\frac{1}{n^2}\cdot\frac{\omega^4}{24(1-\frac{\omega^2}{2n})}\right)^n=a_n^n
$$
where
$$
1+\frac{C_1}{n^2}<a_n<1+\frac{C_2}{n^2}
$$
for suitable $C_1,\,C_2>0$. But $(1+\frac{C_2}{n^2})^n\to 1$.
A: $$\lim_{n\rightarrow\infty}\cos^n\left(\frac{\omega}{\sqrt{n}}\right)$$
$$=\lim_{n\rightarrow\infty}\left(\cos\left(\frac{\omega}{\sqrt{n}}\right)\right)^{n}$$$$=\lim_{n\rightarrow\infty}\left(1+\left(\cos\left(\frac{\omega}{\sqrt{n}}\right)-1\right)\right)^{^{n}}$$$$=\exp\left(\lim_{n\rightarrow\infty}\frac{\cos\left(\frac{\omega}{\sqrt{n}}\right)-1}{\frac{1}{n}}\right)$$$$=\exp\left(-\frac{\omega^{2}}{2}\left(\lim_{n\rightarrow\infty}\frac{\sin\left(\frac{\omega}{\sqrt{n}}\right)}{\frac{\omega}{\left(\sqrt{n}\right)}}\right)^{2}\right)$$$$=\exp\left(-\frac{\omega^{2}}{2}\right)$$
Here I used the  $\lim_{n\rightarrow\infty}\frac{\sin\left(n\right)}{n}=1$
A: Yes, your approach is valid and your answer is correct.
However  you may enhance your proof by a more precise 
 explanation about the validity of $$\cos\left(\frac{\omega}{\sqrt{n}}\right)\approx1-\frac{\omega^2}{2n}$$
in your proof.
A: Alternatively:
$$\lim_{n\rightarrow\infty}\cos^n\left(\frac{\omega}{\sqrt{n}}\right)=\lim_{2m\rightarrow\infty}\cos^{2m}\left(\frac{\omega}{\sqrt{2m}}\right)=\lim_{m\rightarrow\infty}\left(1-\sin^2 \frac{\omega}{\sqrt{2m}}\right)^{m}=\\
\lim_{m\rightarrow\infty}\left[\left(1-\sin^2 \frac{\omega}{\sqrt{2m}}\right)^{-\frac{1}{\sin^2 \frac{\omega}{\sqrt{2m}}}}\right]^{\frac{-\sin^2 \frac{\omega}{\sqrt{2m}}}{\frac{\omega^2}{(\sqrt{2m})^2}}\cdot \frac{\omega^2}2}=e^{-\frac{\omega^2}{2}}.$$
Note that the following was used: 
$$\cos^2x=1-\sin^2x\\
\lim_\limits{x\to 0} \frac{\sin x}{x}=1.$$
A: You can use the identity
$$
\cos^n\frac{\omega}{\sqrt{n}} = \prod_{k = 1}^\infty\left(1 - \frac{\omega^2}{n\pi^2\left(k - \frac{1}{2}\right)^2}\right)^n
$$
that yields
$$
\lim_{n\rightarrow\infty}\cos^n\frac{\omega}{\sqrt{n}}=\prod_{k = 1}^\infty e^{-\frac{\omega^2}{\pi^2\left(k - \frac{1}{2}\right)^2}}=e^{-\sum_{k=1}^\infty\frac{\omega^2}{\pi^2\left(k - \frac{1}{2}\right)^2}}.
$$
Now, by noting that
$$
\sum_{k=1}^\infty\frac{1}{\pi^2\left(k - \frac{1}{2}\right)^2}=\frac{1}{2}
$$
you get the limit.
