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Question

limx0+(4x(2x4x)(2x+1)x2).

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Solution

limx0+=4x(2x1)1(2x1)x2limx0+=4x(2x1)1(2x1)x2limx0+=limh0(40+h1)(20+h1)x2limx0+=limh0(40+h1)(20+h1)x2ThenweuseD.Lhospitalrule:limx0+=4xlog4(2x1)+(4x1)2xlog22xItisalso00formlimx0+=(23x4x)log4+(23x2x)log22xlimx0+=(23x4x)log4+(23x2x)log22x=(log8log4)log4+(log8log2)log2223x=(23)x=8x23x=(23)x=8x=2log2log42=log2log4

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