$\lim_{n\rightarrow \infty} \frac{1}{2^n}\left\{(1 + \cos x)(1 + \cos \frac{x}{2})(1 + \cos \frac{x}{4})\mbox{.....}(1+ \cos \frac{x}{2^{n - 1}})\right\}$
$\lim_{n\rightarrow \infty} \frac{1}{2^n}\left\{(1 + \cos x)(1 + \cos \frac{x}{2})(1 + \cos \frac{x}{4})\mbox{.....}(1+ \cos \frac{x}{2^{n - 1}})\right\}$
if $x=0,$ the limit is clearly 1. for $x \neq 0$ we have $f_n(x)=\frac{1}{2^n}\prod_{j=0}^{n-1} \left[1 + \cos \left(\frac{x}{2^j} \right) \right]=\prod_{j=1}^n \cos^2 \left(\frac{x}{2^j} \right)=\frac{\sin^2 x}{4^n \sin^2 (\frac{x}{2^n})}$ and thus $\lim_{n\to\infty} f_n(x)=\frac{\sin^2 x}{x^2}.$