The value of limx→1 log⁡xsin⁡πx, is 

The value of limx1logxsinπx, is 

  1. A

    1π

  2. B

    π

  3. C

    π

  4. D

    -1π

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    Solution:

    We have,

    limx1logxsinπx=limx1log[1+(x1)}sin(ππx)=limx1log{1+(x1)}sinπ(1x)=limx1log{1+(x1)}x1×x1sinπ(1x)=1πlimx1log{1+(x1)}x1×π(1x)sinπ(1x)=1π

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