f(x)=1−x(1+|1−x|)|1−x|cos⁡11−x x≠1 . Then

We have, f(x)=1−x(1+1−x)1−xcos⁡11−x     for x<11−x(1−(1−x))−(1−x)cos⁡11−x     for x>1⇒ f(x)=1−x(2−x)1−xcos⁡11−x for x<11−x2−(1−x)cos⁡11−x for x>1⇒ f(x)=(1−x)cos⁡11−x for x<1(−1−x)cos⁡11−x for x>1∴ limx→1− f(x)=limx→1 (1−x)cos⁡11−x= 0×(An oscillating number between -1 and 1) =0limx→1+ f(x)=limx→1 −(1+x)cos⁡11−x=-2 (An oscillating number between -1 and 1)⇒ limx→I+ f(x) does not exist .