Question
limx→0(1+2x1+3x)1x2.e1x=
Answer: Option A
:
A
limx→0e1x2(log(1+2x)−log(1+3x))+1xelimx→0(log(1+2x)−log(1+3x))+xx2
Using the expansion we get -
=e52
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:
A
limx→0e1x2(log(1+2x)−log(1+3x))+1xelimx→0(log(1+2x)−log(1+3x))+xx2
Using the expansion we get -
=e52
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