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Question

For each tR, let [t] be the greatest integer less than or equal to t. Then, limx1+(1|x|+sin|1x|)sin(π2[1x])|1x|[1x]

A
equals 0
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B
equals 1
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C
equals 1
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D
does not exist
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Solution

The correct option is A equals 0
L=limx1+(1|x|+sin|1x|)sin(π2[1x])|1x|[1x]
x1+x>1[1x]=1,|1x|=x1
L=limx1+(1x+sin(x1))sin(π2(1))(x1)(1)
=limx1+(1sin(x1)x1)(1)
=1+1=0

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