If limx→∞ x2+1x+1−ax−b=0 ,then

If limxx2+1x+1axb=0 ,then

  1. A

    a=1,b=1

  2. B

    a=1,b=1

  3. C

    a=1,b=1

  4. D

    a=1,b=1

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    Solution:

    We have, 

    limxx2+1x+1axb=0 

     limxx2(1a)x(a+b)+1bx+1=0

    Since the limit of the given expression is zero. Therefore, degree of the polynomial in numerator must be less than that of denominator. As the denominator is a first degree polynomial. So, numerator must be a constant i.e., a zero degree polynomial

     1a=0 and a+b=0a=1and b=1

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