$\lim\limits_{x \rightarrow \frac{\pi}{2}}\left(\lim\limits_{n \rightarrow \infty} \cos \frac{x}{2} \cos \frac{x}{2^2} \cos \frac{x}{2^3} ...... \frac{\cos x}{2^n}\right)$ equals to

1

$\lim\limits_{x \rightarrow \frac{\pi}{2}}\left(\lim\limits_{x \rightarrow \infty} \cos \frac{x}{2} . \cos \frac{x}{2^2} . \cos \frac{x}{2^3} .... \cos \frac{x}{2^n}\right)=2$

$=\lim\limits_{x \rightarrow \frac{\pi}{2}}\left(\lim\limits_{x \rightarrow \infty} \frac{\sin x}{2^n \sin \left(\frac{x}{2^n}\right)}\right)=2$

$=\lim\limits_{x \rightarrow \frac{\pi}{2}} \lim\limits_{n \rightarrow \infty}\left(\frac{\sin x}{x}\right)\left(\frac{x / 2^n}{\sin x / 2^n}\right)=1$

Hence (4) is the correct answer.