If f(x) = {x,when 0≤ x ≥ 1 2−x,2-x when 1 ≤ x ≥ 2 then limx→1 f(x) =
1
2
0
Does not exist
and limx→1+ f(x) = limh→0 2-(1+h) = 1 Hence limit of the function is 1
If the function f(x) satisfies limx→1f(x)−2x2−1=π,then limx→1f(x)=
limx→∞|x|x is equal to