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a
Limx→1− fx=0
b
Limx→1+ fx=0
c
Limx→1+f(x) does not exist
d
Limx→1-f(x) does not exist
answer is A.
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Detailed Solution
x→1−⇒1−x2−x1−x.cos11−x since x<1⇒1-x=1-xx<1⇒1−x→0+⇒1−xcos11−x→0 by sadwitch theoremx→1+⇒1−xxx−1cos11−x since x>1⇒1-x=-1-xx>1⇒−x+1cos11−x does not exist