$\lim \limits_{n \to \infty}$ $\prod_{r=1}^{n} \cos(\frac{x}{2^r})$ $\lim \limits_{n \to \infty}$ $\prod_{r=1}^{n} \cos(\frac{x}{2^r})$  
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.

$$\prod_{r=1}^{\infty} \frac{\cos (x/2^r) \sin (x/2^r)}{\sin (x/2^r)} = \prod_{r = 1}^{\infty} \frac{\sin (x/2^{r-1})}{2\sin(x/2^r)} = \lim_{r \to \infty} \frac{\sin x}{2^r \sin (x/2^r)}$$
Can you take it from here? Recall that $\lim_{t \to 0} \frac{\sin 
 \alpha t}{t} = \alpha$.
