Explanation

(c)
We know that
$\cos{A\cos{2A\cos{4A}}}\ldots\cos 2^{n - 1}A = \frac{\sin{2^{n}A}}{2^{n}\sin A}$
$\lim_{n \rightarrow \infty}{\cos{\left( \frac{x}{2} \right)\cos{\left( \frac{x}{4} \right)\ldots\cos{\left( \frac{x}{2^{n - 1}} \right)\cos\left( \frac{x}{2^{n}} \right)}}}}$
$= \lim_{n \rightarrow \infty}\ \frac{\sin x}{2^{n}\sin{(x/2^{n})}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left\lbrack put\ A = \frac{x}{2^{n}} \right\rbrack$
$= \lim_{n \rightarrow \infty}\ \frac{\sin x}{x}.\frac{(x/2^{n})}{\sin{(x/2^{n})}}$
$= \frac{\sin x}{x}$