$\cos A\cos 2A\cos 4A....\cos {2^{n - 1}}A = \frac{{\sin {2^n}A}}{{{2^n}\sin A}}$
Taking $A = \frac{x}{{{2^n}}},$ we get
$\cos \,\left( {\frac{x}{{{2^n}}}} \right)\,\cos \,\left( {\frac{x}{{{2^{n - 1}}}}} \right)\,...\cos \left( {\frac{x}{4}} \right)\cos \,\left( {\frac{x}{2}} \right) = \frac{{\sin x}}{{{2^n}\sin \left( {\frac{x}{{{2^n}}}} \right)}}$
$\therefore \,\,\,\mathop {\lim }\limits_{n \to \infty } \,\,\cos \,\left( {\frac{x}{2}} \right)\cos \,\left( {\frac{x}{4}} \right)...\cos \,\left( {\frac{x}{{{2^{n - 1}}}}} \right)\,\cos \,\left( {\frac{x}{{{2^n}}}} \right)$
$ = \mathop {\lim }\limits_{n \to \infty } \,\frac{{\sin x}}{{{2^n}\sin \,\left( {\frac{x}{{{2^n}}}} \right)}} = \mathop {\lim }\limits_{n \to \infty } \,\frac{{\sin x}}{x}\frac{{(x/{2^n})}}{{\sin \,(x/{2^n})}} = \frac{{\sin x}}{x}$.
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