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Question

The value of the limit limx(x2+xx) is equal to

A
limxx2+x+1x2+1
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B
limxxxx22x
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C
limx0(cos2x)log(2x)x2cos2x
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D
limx1(x+14)tan(π2x)
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Solution

The correct option is C limx0(cos2x)log(2x)x2cos2x
limx(x2+xx)=limxx(x2+x+x)=limx1(1+1x+1)=12

limxx2+x+1x2+1=limxxx2+x+1+x2+1=limxxx1+1x+1x2+x1+1x2=12

limxxx+x22x
=limxxx+x 12x=limx11+12x=12

limx0(cos2x)log(2x)x2cos2x=elimx0log(2x)x2cos2x(2sin2x)=e(2)log(2)=(2)2=12

limx1(x+14)tan(π2x)=limx1(t4)tan(π2(t1))
By replacing x+1=t such that x1t0 So, we have
14limt0π2(t)tanπ2(t)2π=142π=12π

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