# 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.

• 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 – Dominik Kutek Aug 6 '19 at 15:47
• Similar question – A.Γ. Aug 6 '19 at 16:23
• Yes,Dominik Kutek, that would be a more precise mathematical expression . – もっと酒 Aug 6 '19 at 16:30

However you may enhance your proof by a more precise explanation about the validity of $$\cos\left(\frac{\omega}{\sqrt{n}}\right)\approx1-\frac{\omega^2}{2n}$$

• 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? – もっと酒 Aug 6 '19 at 16:35
• Yes, that is what I had in mind. – Mohammad Riazi-Kermani Aug 6 '19 at 16:38
• Thanks. Mohammad – もっと酒 Aug 6 '19 at 16:40

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 $$1+\frac{C_1}{n^2} for suitable $$C_1,\,C_2>0$$. But $$(1+\frac{C_2}{n^2})^n\to 1$$.

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

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.

$$\lim_{n\rightarrow\infty}\cos^n\left(\frac{\omega}{\sqrt{n}}\right)$$ $$=\lim_{n\rightarrow\infty}\left(\cos\left(\frac{\omega}{\sqrt{n}}\right)\right)^{n}$$$$=\lim_{n\rightarrow\infty}\left(1+\left(\cos\left(\frac{\omega}{\sqrt{n}}\right)-1\right)\right)^{^{n}}$$$$=\exp\left(\lim_{n\rightarrow\infty}\frac{\cos\left(\frac{\omega}{\sqrt{n}}\right)-1}{\frac{1}{n}}\right)$$$$=\exp\left(-\frac{\omega^{2}}{2}\left(\lim_{n\rightarrow\infty}\frac{\sin\left(\frac{\omega}{\sqrt{n}}\right)}{\frac{\omega}{\left(\sqrt{n}\right)}}\right)^{2}\right)$$$$=\exp\left(-\frac{\omega^{2}}{2}\right)$$

Here I used the $$\lim_{n\rightarrow\infty}\frac{\sin\left(n\right)}{n}=1$$