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Question

limxπ21sin x(π2x)2

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Solution

limxπ21sin x(π2x)2

Let x=π2+y

y=xπ2

as xπ2,y0

=limy01sin(π2y)y2

=limy01cos yy2=limy02 sin2y2y2

=2(limy0siny2y2)2×14[limy0sin xx=1]

=2×1×14=12


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