limx→0 sin2⁡x2−1+cos⁡x equals.

limx0sin2x21+cosx equals.

  1. A

    2

  2. B

    22

  3. C

    42

  4. D

    4

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    Solution:

    limx0sin2x21+cosx                          00 form 

    So, using L' Hospital' s rule, we get

    limx02sinxcosx121+cosx×(sinx)=limx02cosx×21+cosx=42

     

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