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Q.

Let P=limx→0+ 1+tan2⁡x12x then log⁡p is equals to

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a

12

b

14

c

2

d

1

answer is A.

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Detailed Solution

P=limx→01+tan2n12n=elimx→012x1+tan2x−11∞form=elimx→0tan2x2x=e12limx→0tan2xx2p=e12log⁡p=log⁡e12=12log⁡e=12
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