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Question

Evaluate the limit:
limx08x4x2x+1x2

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Solution

Given:
limx08x4x2x+1x2

=limx023x22x2x+1x2

=limx022x(2x1)1(2x1)x2

=limx0(22x1)(2x1)x2

=limx02×(22x1)2x×(2x1)x

=2limx0(22x1)2x×limx0(2x1)x

=2log2×log2
[limx0(ax1x)=loga]

=(log22)(log2)
=(log4)(log2)

Therefore,
limx08x4x2x+1x2=(log4)(log2)

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