Question

Does the limit exist for (x^2-1)/(abs(x-1)) as x goes to 1?

Answer 1

The equivalent function is
f1(x)=(x^2-1)/(x-1) when x>1
and
f2(x)=-(x^2-1)/(x-1) when x<1
f(x) is undefined when x=1.

To find the limit of
f(x)=(x^2-1)/abs(x-1) when x->1
we need to find the limit from both sides, i.e. limit x->1+ f1(x) and limit x->1- f2(x).
If they both exist and the values are identical, then the limit x->1 for f(x) exists.
If these two limits exist, but are different, then the required limit does not exist.