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Question

limx0{sin(α+β)x+sin(αβ)x+sin2αx}cos2βxcos2αx.

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Solution

sin(α+β)x+sin(αβ)x=2sin(2αx2)cos(2βx2)
=2sin(αx)cos(βx)
and sin2αx=2sinαxcosαx
and cos2βxcos2αx=(cosβx+cosαx)(cosβxcosαx)1+1 as x0
=2(2sin(βα2)xsin(β+α2)x)
L=limx02sinαx4sin(βα2)xsin(β+α2)x
As limθ0sinθθ=1
limθ0sinθ=θ
L=limx02αx4(βα2)(β+α2)x2=2α(αβ)(α+β)x= Not defined.

1167262_1246275_ans_ba19fcca628b42dc809739cde981d2c1.jpg

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