The value of limn→∞ cos⁡x2cos⁡x4cos⁡x8…cos⁡x2n, is

The value of limncosx2cosx4cosx8cosx2n, is

  1. A

    1

  2. B

    sinxx

  3. C

    xsinx

  4. D

    none of these

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    Solution:

    We know that 

    cosAcos2Acos4Acos2n1A=sin2nA2nsinA

    Taking A=x2n, we get 

    cosx2ncosx2n1cosx4cosx2=sinx2nsinx2n limncosx2cosx4cosx2n1cosx2n=limnsinx2nsinx2n=limnsinxxx/2nsinx/2n=sinxx

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