$\underset{n\to \mathrm{\infty }}{lim} \prod _{r=1}^n \cos \left(\frac{x}{2^r}\right)=$

$L=\underset{n\to \mathrm{\infty }}{lim} \cos \frac{x}{2}\cdot \cos \frac{x}{2^2}\dots \cos \frac{x}{2^n}$

By using repeatedly $2\sin \theta \cos \theta = \sin 2\theta$  starting with $\theta =\frac{x}{2^n}$

$L=\underset{n\to \mathrm{\infty }}{lim} \frac{\sin x}{2^n\sin \frac{x}{2^n}}=\sin x\underset{\theta \to 0}{lim} \frac{\theta }{\sin x\theta },\text{ where }\theta =\frac{1}{2^n}$

$=\sin x\times \frac{1}{x}=\frac{\sin x}{x}$