The value of limh→0 ln⁡(1+2h)−2ln⁡(1+h)h2, is

The value of limh0ln(1+2h)2ln(1+h)h2, is

  1. A

    1

  2. B

    -1

  3. C

    0

  4. D

    none of these

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    Solution:

    We have, 

    limh0log(1+2h)2log(1+h)h2=limh0log1+2h(1+h)2h2=limh0log(1+h)21+2hh2=limh0log1+h21+2hh21+2h(1+2h)=1=1.

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