The value of limx→01n(1+{x}){x} is (where {x} denotes the fractional part of x)
Does not exist
LHL = limx→0−f(x)
= limh→0f(0−h)
= limh→0In(1+{0−h}){0−h}
= limh→0In(2−h)(1−h) = In 2
RHL = limx→0+f(x) = limh→0f(0+h)
= limh→0In(1+{h}){h} = In(1+h)h = 1
LHL ≠ RHL