Question about limit $\lim\limits_{n\rightarrow\infty}\cos^n\left(\frac{\omega}{\sqrt{n}}\right)$

Question

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?

Answer 1

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 $(1+\frac{C_2}{n^2})^n\to 1$.

Answer 2

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

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

Answer 3

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.

Answer 4

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

Answer 5

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.