If limx→0 ax−e4x−1axe4x−1=b exists finitely then 2(a+b)=

If limx0axe4x1axe4x1=b exists finitely then 2(a+b)=

  1. A

    -1

  2. B

    -7

  3. C

    1

  4. D

    7

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    Solution:

        limx0axe4x1axe4x1=b exists finitely

     limx0a4e4x14xae4x1=b  exists finitely

    The numerator becomes a-4 when x0 and the denominator tends to zero. Therefore, if a -40, LHS whereas RHS is finite. Therefore, a -4 = 0 i.e. a = 4.

    When a=4

     limx0a4e4x4x1ae4x1=b limx01e4x14xe4x1=b

      limx011+(4x)2!+(4x)23!+4x+(4x)22!+(4x)33!+=b12=b

     2(a+b)=2412=7       

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