If limx→0 2ax+(a−1)sin⁡xtan3⁡x=l, then a+l is equal to 

If limx02ax+(a1)sinxtan3x=l, then a+l is equal to 

  1. A

    23

  2. B

    13

  3. C

    29

  4. D

    49

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    Solution:

    We have

    limx02ax+(a1)sinxtan3x=l limx02a+(a1) sinxxx2tanxx3=l

    We observe that the numerator on LHS tends to 3a-1 and the denominator tends to 0. So, the limit will be a finite number I 
    only if 

    3a1=0 i.e., a=13

     limx03(1sinx)x2tanxx3=1

     limx0xsinxx3tanxx3=3l2

     16=3l2l=19        limx0xsinxx3=16

    Hence, a+l=13+19=49

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