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Question

limxπ41tan xxπ4

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Solution

limxπ41tan xxπ4

Let x=π4+y

y=xπ4asxπ4=y0

=limy01tan(y+π4)y

=limy01(tan y+tanπ41tan y tanπ4)y

=limy01tan y+11tan yy

=limy0(1tan ytan y1)y(1tan y)

[tan π4=1]

=limy0(2 tan y)y(1tan y)

=2limy0tan yy×1limy0(1tan y)

=2×11(10)=2


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