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Question

Evaluate: limxπ2tan2xxπ2

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Solution

We have, limxπ2tan2xxπ2

At x=π2, the value of the given function takes the form 00

If, we put x=h+π2 such that xπ2 then h 0

Therefore limxπ2tan2xxπ2=limh 0tan2(h+π2)h

limh 0tan(π+2h)h{ tan(π+θ)=tanθ}

=limh 0tan 2hh×22

=2limh 0tan 2hh{limx 0tan xx+1}

=2× 1=2

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