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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?

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    $\begingroup$ It would be better to write $\cos(\frac{\omega}{\sqrt{n}}) = 1 - \frac{\omega^2}{2n} + o(\frac{1}{n})$ and then try to evaluate the limit $\endgroup$ – Dominik Kutek Aug 6 at 15:47
  • $\begingroup$ Similar question $\endgroup$ – A.Γ. Aug 6 at 16:23
  • $\begingroup$ Yes,Dominik Kutek, that would be a more precise mathematical expression . $\endgroup$ – IEDC PHY Aug 6 at 16:30
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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.

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  • $\begingroup$ I could write $\cos\left(\frac{\omega}{\sqrt{n}}\right)=1-\frac{\omega^2}{2n}+O\left(\frac{1}{n^2}\right)$ as Dominik Kutek suggested in the comments above? $\endgroup$ – IEDC PHY Aug 6 at 16:35
  • $\begingroup$ Yes, that is what I had in mind. $\endgroup$ – Mohammad Riazi-Kermani Aug 6 at 16:38
  • $\begingroup$ Thanks. Mohammad $\endgroup$ – IEDC PHY Aug 6 at 16:40
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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 $(1+\frac{C_2}{n^2})^n\to 1$.

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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: $$\cos^2x=1-\sin^2x\\ \lim_\limits{x\to 0} \frac{\sin x}{x}=1.$$

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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.

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