f(x)=|x(x−1)(x−2)|x(x−1)(x−2)=|x|x|x−1|x−1|x−2|x−2
Consider the points x=0,1,2;
We redefine the function,
f(x)=⎧⎪
⎪
⎪
⎪⎨⎪
⎪
⎪
⎪⎩−1x<0(−1)(−1)(−1)=−110<x<1(1)(−1)(−1)=1−11<x<2(1)(1)(−1)=−11x>2(1)(1)(1)=1
limf(x) as x→0,1,2 does not exist as both R.H.L. and L.H.L. will be differnet.
Hence 0, 1, 2. are points of discontinuity.