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Question

The value of the limit limxπ242sin3x+sinx2sin2xsin3x2+cos5x2-2+2cos2x+cos3x2 is _______________


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Solution

Find the value of the given limit

limxπ242sin3x+sinx2sin2xsin3x2+cos5x2-2+2cos2x+cos3x2=limxπ242sin3x+sinxcosx2-cos7x2+cos5x2-2+22cos2x-1+cos3x2[2sin(x)sin(y)=cos(x-y)-cos(x+y),cos(2x)=2cos2(x)-1]=limxπ242sin3x+sinxcosx2-cos3x2+cos5x2-cos7x2-22cos2x=limxπ2422sin2xcosx2sinxsinx2+2sin3xsinx2-22cos2x[cos(x)-cos(y)=2sin(y+x2)sin(y-x2),sin(x)+sin(y)=2sin(x+y2)cos(x-y2)]=limxπ282sin2xcosx2sinx2sinx+sin3x-22cos2x=limxπ282sin2xcosx2sinx22sin2xcosx-22cos2x[sin(x)+sin(y)=2sin(x+y2)cos(y-x2)]=limxπ2162sinxcos2x4sinx2cos2x2sinx-22cos2x[sin(2x)=2sin(x)cos(x)]=limxπ2162sinx8sinx2sinx-22=162812-22[sin(π2)=1,sin(π4)=12]=328-4=8Hence the value of the limit is 8


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