limx→a xa−axxx−aa is equal to

limxaxaaxxxaa is equal to

  1. A

    1+logea1logea

  2. B

    loge(e/a)loge(ae)

  3. C

    logc(a/e)loge(ae)

  4. D

    none of these

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    Solution:

    We have,

    limxaxaaxxxaa                   00 form

    =limxaaxa1axlogaxx(1+logx)0                       [Using L' Hospital's Rule]

    =aaaalogaaa(1+loga)=1loga1+loga=loge(e/a)loge(ae)

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