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Question

Solve limxπ2[tanxln(sinx)]

A
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1
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Solution

The correct option is A 0
limxπ2[tanxln(sinx)]
limxπ2[ln(sinx)cotx]
Applying L'Hopital's rule, we get:
limxπ2[(cosxsinx)cosec2x]
limxπ2[cotxcosec2x]
limxπ2[cosxcosec3x].
Direct substitution now, yields: 01, which equals 0.
Therefore,
limxπ2[tanxln(sinx)]=0

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