solve for limits of 1+ and 1- for this function:
(x^2 + abs (x-1) - 1)/ abs (x-1)
i don't know how to simplify abs. value expression times polynomial. can someone please help.
2 answers
Help with this one ?
as x->1+, |x-1| = x-1, so
(x^2+|x-1|-1)/|x-1|
= (x^2+x-2)/(x-1)
= (x+2)(x-1)/(x-1)
= x+2
so limit = 1+2 = 3
as x->1-, |x-1| = -(x-1), so
(x^2+|x-1|-1)/|x-1|
= (x^2-x)/-(x-1)
= x(x-1)/-(x-1)
= -x
so limit = -1
(x^2+|x-1|-1)/|x-1|
= (x^2+x-2)/(x-1)
= (x+2)(x-1)/(x-1)
= x+2
so limit = 1+2 = 3
as x->1-, |x-1| = -(x-1), so
(x^2+|x-1|-1)/|x-1|
= (x^2-x)/-(x-1)
= x(x-1)/-(x-1)
= -x
so limit = -1