limx→1{−x−1x},where{.} denotes the fraction part function
A
is equal to 1
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B
is equal to 0
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C
Does not exist
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D
None of these
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Solution
The correct option is B is equal to 0 ∵limx→1{−x−1x} Substitute x=1+h ⇒RHL:limh→0+{−(1+h)−1(1+h)}={−2.0}=0 Similarly, ⇒LHL:limh→0−{−(1−h)−1(1−h)}={−2.0}=0 And, limx→1{−x−1x}={−1.0−11.0}={−2.0}=0 Hence option (B) is correct.