The values of α and β such thatlimx→∞ x2+1x−1−αx−2β=32 are 

The values of α and β such that

limxx2+1x1αx2β=32 are 

  1. A

    α=1,β=34

  2. B

    α=1,β=14

  3. C

    α=1,β=54

  4. D

    α=1,β=34

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    Solution:

    limxx2+1x1αx2β

    =limxx2(1α)(2βα)x+1+2βx1

    Since the last exists so 1α=0α=1. In this
    case the last limit is equal to

    limx(2βα)+1+2βx11x=(2βα)=32

    2β=321=12β=14.

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