The value of limx→1−[sin sin−1x] is
1
does not exist
π/2
0
limx→1−[sin sin−1x]=limx→1[sin sin−1x]=1
(Hence x→1 means x→1-, ∵ sin sin-1 x is not defined for x>1 )
Hence (a) is the correct answer
If f(x) = {x,when 0≤ x ≥ 1 2−x,2-x when 1 ≤ x ≥ 2 then limx→1 f(x) =