First slide
Introduction to limits
Question

The value of limxπ/422(cosx+sinx)31sin2x,is 

Moderate
Solution

We have,

limxπ/422(cosx+sinx)31sin2x=limxπ/423/2(cosx+sinx)23/22(1+sin2x)=limxπ/423/2(1+sin2x)3/22(1+sin2x) 

=limy2y3/223/2y2,where y=1+sin2x

=32(2)3/21=32×2=32

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