1.The value of limn→∞(cosx2cosx4cosx8…cosx2n+1)\displaystyle\lim_{n\to\infty}\left(\cos\frac{x}{2}\cos\frac{x}{4}\cos\frac{x}{8}\ldots\cos\frac{x}{2^{n+1}}\right)n→∞lim(cos2xcos4xcos8x…cos2n+1x) is equal toa.x/sinxx/\sin xx/sinxb.sinx/x\sin x/xsinx/xc.000d.None of theseLogin to continueOnly logged in users canattempt or see the solution.
