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Question

Evaluate the following limits:

limx0sinα+βx+sinα-βx+sin2αxcos2βx-cos2αx

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Solution


limx0sinα+βx+sinα-βx+sin2αxcos2βx-cos2αx=limx02sinα+βx+α-βx2cosα+βx-α-βx2+2sinαx cosαx1-sin2βx-1-sin2αx=limx02sinαx cosβx+2sinαx cosαxsin2αx-sin2βx=limx02sinαxcosβx+cosαxsin2αx-sin2βx
=limx02αx×sinαxαx×cosβx+cosαxα2x2×sin2αxα2x2-β2x2×sin2βxβ2x2=2α×limx0sinαxαx×limx0cosβx+cosαxα2×limx0sinαxαx2-β2×limx0sinβxβx2×limx0xx2=2α×1×1+1α2×1-β2×1×limx01x=4αα2-β2×=

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