limx→2 x−2|x−2| is equal to

∵|x−2|=x−2,     if x≥2−(x−2),     if x<2Now, RHL =limx→2+ x−2x−2=limx→2+ 1=1and LHL =limx→2− x−2−(x−2)              =−limx→2− x−2x−2=−1since limx→2+ f(x)≠limx→2− f(x)Therefore, the limit does not exist.