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Question

limx0{sin(α+β)x+sin(αβ)x+sin 2αx}cos2βxcos2αx


Solution

limx0{sin(α+β)x+sin(αβ)x+sin 2αx}cos2βxcos2αx

=limx02sin[(α+β)x+(αβ)x2]cos[(α+β)x(αβ)x2]+2sin αx cos αx(1sin2βx)(1sin2αx)

=limx02sin αx cos βx+2sin αx cos αxsin2 αxsin2 βx

=limx02sin αx(cos βx+cos αx)sin2 αxsin2 βx

=limx02αx×sinαxαx×(cos βx+cos αx)α2x2×sinαxα2x2β2x2×sin2βxβ2x2

=limx02α×limx0(sinαxαx)×limx0(cos βx+cos αx))α2×(limx0sinαxαx)2β2×(limx0sinβxβx)2×limx0xx2

=2α×1×(1+1)α2×1β2×1×limx01x

=4αα2β2×=


Mathematics
RD Sharma
Standard XI

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