Find the limit of the cosine sequence.

Question

The question is to find $\lim\limits_{n\to\infty} \cos \frac{x}{2} \cdot \cos \frac{x}{2^2} \cdot \cos \frac{x}{2^3}\cdot\cdot\cdot \cos \frac{x}{2^n}$.

I first thought of several facts:
1). $-1 \leq \cos(a) \leq 1$ (From this I know the sequence is also bounded by -1 and 1)
2). $\cos(0) = 1$ (This is what the items in the sequence tend to, as n becomes bigger)
3). When $x=0$, the limit is 1.

Then I failed to think more and the above do not get me anywhere close to finding the limit...

Can anybody teach me how to find the limit of this sequence? Thanks a lot.

Answer 1

May be, you could start using $$\sin(x)=2\sin(\frac x2)\cos(\frac x2)=2\cos(\frac x2)\Big(2\sin(\frac x4)\cos(\frac x4)\Big)=2^2\cos(\frac x2)\cos(\frac x4)\sin(\frac x4)=$$ $$2^2\cos(\frac x2)\cos(\frac x4)\Big(2\sin(\frac x8)\cos(\frac x8)\Big)=2^3\cos(\frac x2)\cos(\frac x4)\cos(\frac x8)\sin(\frac x8)$$ and continue to get $$\sin(x)=2^n \cos(\frac x2)\cos(\frac x4)\cos(\frac x8) \cdots\cos(\frac x {2^n})\sin(\frac x {2^n})$$ When $n$ becomes very large $\sin(\frac x {2^n})\approx \frac x {2^n}$ which makes the product of the cosines becomes closer and closer to $\frac{\sin(x)} x$.

Answer 2

If $x \ne 0$, then \begin{align} \lim_{n\to\infty}\cos \frac{x}{2}\cos\frac{x}{2^2}\cos\frac{x}{2^3}\cdots \cos \frac{x}{2^n}&=\lim_{n\to\infty}\frac{\cos \frac{x}{2}\cos\frac{x}{2^2}\cos\frac{x}{2^3}\cdots \cos \frac{x}{2^n}\sin\frac{x}{2^n}}{\sin\frac{x}{2^n}}\\ &=\lim_{n\to\infty}\frac{\cos \frac{x}{2}\cos\frac{x}{2^2}\cos\frac{x}{2^3}\cdots \cos \frac{x}{2^{n-1}}\sin\frac{x}{2^{n-1}}}{2\sin\frac{x}{2^n}}\\ &=\lim_{n\to\infty}\frac{\cos \frac{x}{2}\cos\frac{x}{2^2}\cos\frac{x}{2^3}\cdots \cos \frac{x}{2^{n-2}}\sin\frac{x}{2^{n-2}}}{2^2\sin\frac{x}{2^n}}\\ &=\cdots\\ &=\lim_{n\to\infty}\frac{\sin x}{2^n \sin\frac{x}{2^n}}\\ &=\frac{\sin x}{x}. \end{align} If $x=0$, then given limit goes to $1$.