limx→π/6 2sin2⁡x+sin⁡x−12sin2⁡x−3sin⁡x+1= 

limxπ/62sin2x+sinx12sin2x3sinx+1= 

  1. A

    3

  2. B

    -3

  3. C

    6

  4. D

    0

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    Solution:

    We have,

    limxπ/62sin2x+sinx12sin2x3sinx+1=limxπ/6(2sinx1)(sinx+1)(2sinx1)(sinx1)=limxπ/6sinx+1sinx1=3

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