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Question:

Let f(x) = 1−x(1+|1−x|)/|1−x|cos(1/(1−x)) for x≠1. Then

limx→1−f(x) does not exist

limx→1−f(x) = 0

limx→1+f(x) does not exist

limx→1+f(x) = 0

Solution:

f(x) = 1−x(1+|1−x|)/|1−x|cos(1/(1−x))
For x<1 ⇒ (1−x)>0 ⇒ |1−x| = 1−x
Let h = 1−x
limx→1−f(x) = limh→01−(1−h)(1+h)/(h)cos(1/h) = limh→01−(1−h2)/(h)cos(1/h) = limh→0h2/hcos(1/h) = 0
∴ limx→1−f(x) = 0
For x>1 ⇒ 1−x<0 ⇒ |1−x| = −(1−x)
Let h = 1−x
limx→1+f(x) = limh→01−(1+h)(1+h)/(h)cos(1−h) = limh→01−(1+2h+h2)/(h)cos(1/h) = limh→0(−h)cos(1/h)
∴ limx→1+f(x) does not exist as the value of cos(1/(1−x)) is not fixed.
Hence, B and C are correct.