limit of $\cos^n(\pi/n)$ as $n\to\infty$ (related to Malus's law)

Question

I am trying to find the limit of $\cos^n(\pi/n)$ as $n$ approaches infinity. Using a graphing calculator it appears that the answer is $1$, however I haven't been able to prove it.

The answer relates to a physics question of whether there is a bound to the portion of the intensity of light that can pass through two polarising filters that are at right angles, if you may place as many polarising filters as you would like between them. This is because that relationship follows Malus's Law for intensity of light between polarising lenses,

$$I = I_0 \cos^2(θ)$$

If anyone could show how to solve this (probably using the squeeze theorem) that would be great.

Answer 1

Because of the form of the Taylor series for $\cos(y)$ near $y = 0$, for a fixed value of $x$, if $n$ is large enough we have $$1 - {x^2 \over n^2} \leq \cos{x \over n} \leq 1 \tag{1}$$ This leads to $$\bigg(1 - {x^2 \over n^2}\bigg)^n \leq \bigg(\cos {x \over n}\bigg)^n \leq 1\tag{2}$$ But $$\lim_{x \rightarrow \infty} \bigg(1 - {x^2 \over n^2}\bigg)^n = \lim_{x \rightarrow \infty} \bigg(1 - {x \over n}\bigg)^n\lim_{x \rightarrow \infty} \bigg(1 + {x \over n}\bigg)^n\tag{3}$$ $$= e^{-x} e^x$$ $$= 1$$ Hence by the squeeze test applied to $(2)$, for any $x$ one has $$\lim_{x \rightarrow \infty}\bigg(\cos {x \over n}\bigg)^n = 1$$ In particular this holds for $x = \pi$, the case at hand.

Answer 2

Name the expression $$ f(n) = \left(\cos \frac{\pi}{n}\right)^n $$ As $n \to +\infty$, this is a $1^\infty$ pattern limit, so let's take a logarithm to get a product, then invert one term to get a quotient: $$ \ln f(n) = n \ln \left(\cos \frac{\pi}{n}\right) = \frac{\ln \left(\cos \frac{\pi}{n}\right)}{\frac 1n} $$

Or subsituting $n= \frac 1x$,

$$ \ln f\!\left( \frac 1x \right) = \frac{\ln(\cos \pi x)}{x} $$

This is a $\frac{\infty}{\infty}$ pattern limit, so we can try L'Hopital's rule. Supposing the limits exist,

$$ \lim_{x \to 0^+} \ln f\!\left( \frac 1x \right) = \lim_{x \to 0^+} \frac{\ln(\cos \pi x)}{x} = \lim_{x \to 0^+} \frac{-\pi \sin(\pi x)}{\cos(\pi x)} = 0 $$

So these limits do all exist. Relating this back to the original variable $n$,

$$ \lim_{n \to +\infty} \ln f(n) = \lim_{x \to 0^+} \ln f\!\left(\frac 1x\right) = 0 $$

Since $\ln$ is continuous,

$$ \ln \left(\lim_{n \to +\infty} f(n)\right) = \lim_{n \to +\infty} \ln f(n) = 0 $$

$$ \lim_{n \to +\infty} f(n) = e^0 = 1 $$

Answer 3

Consider $$a_n=\cos ^n\left(\frac{\pi }{n}\right)\quad \implies \log(a_n)=n\log \left(\cos \left(\frac{\pi }{n}\right)\right)$$

Using the usual Taylor series $$\cos \left(\frac{\pi }{n}\right)=1-\frac{\pi ^2}{2 n^2}+\frac{\pi ^4}{24 n^4}+O\left(\frac{1}{n^6}\right)$$ $$\log \left(\cos \left(\frac{\pi }{n}\right)\right)=-\frac{\pi ^2}{2 n^2}-\frac{\pi ^4}{12 n^4}+O\left(\frac{1}{n^6}\right)$$ $$\log(a_n)=-\frac{\pi ^2}{2 n}-\frac{\pi ^4}{12 n^3}++O\left(\frac{1}{n^5}\right)$$ $$a_n=e^{\log(a_n)}=1-\frac{\pi ^2}{2 n}+\frac{\pi ^4}{8 n^2}+O\left(\frac{1}{n^3}\right)$$