For each t∈R, let [t] be the greatest integer less than or equal to t. Then, limx→1+(1−|x|+sin|1−x|)sin(π2[1−x])|1−x|[1−x]
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=limx→1+(1−|x|+sin|1−x|)sin(π2[1−x])|1−x|[1−x] ∵x→1+⇒x>1⇒[1−x]=−1,|1−x|=x−1 ⇒L=limx→1+(1−x+sin(x−1))sin(π2(−1))(x−1)(−1) =limx→1+(1−sin(x−1)x−1)(−1) =−1+1=0