$\lim \limits_{n \to \infty}$ $\prod_{r=1}^{n} \cos(\frac{x}{2^r})$ $\lim \limits_{n \to \infty}$ $\prod_{r=1}^{n} \cos(\frac{x}{2^r})$  
How do I simplify this limit?  
I tried multiplying dividing $\sin(\frac{x}{2^r})$ to use half angle formula but it doesnt give a telescopic which would have simplified it.

Hint: Using the identity $\sin 2t = 2\sin t \cos t$, we have
$$\prod_{r = 1}^n \cos \frac{x}{2^r} = \prod_{r = 1}^n \frac{\sin \frac{x}{2^{r-1}}}{2\sin \frac{x}{2^r}} = \frac{\sin x}{2^n \sin \frac{x}{2^n}}$$
Now use the fact that $(\sin t)/t \to 1$ as $t\to 0$ to show that the limit of the above expression is $(\sin x)/x$.
