Evaluate `lim_(ntooo) {cos((x)/(2))cos((x)/(4))cos((x)/(8))...cos((x)/(2^(n)))}`.

Explanation

We know that
`costhetacos2thetacos4theta...cos2^(n-1)theta=(sin2^(n)theta)/(2^(n)sintheta)`
`:." "cos((x)/(2))cos((x)/(4))cos((x)/(8))...cos((x)/(2^(n)))=(sinx)/(2^(n)"sin"((x)/(2^(n))))`
`:." ""Given limits is "underset(ntooo)lim(sinx)/(2^(n)sin((x)/(2^(n))))=underset(ntooo)lim((sinx)/(x)((x)/(2^(n))))/(sin((x)/(2^(n))))`
`=(sinx)/(x)`