How do you find the x values at which #f(x)=abs(x3)/(x3)# is not continuous, which of the discontinuities are removable?
2 Answers
Refer to the Discussion given in the Explanation Section below.
Explanation:
Observe that
continuous at that point.
For
This means that
In other words,
We conclude that
Explanation:
The function is discontinuous for
Then we analyze separately:

For
#x in (3,+oo)# , we have:
#(x3) > 0 => abs(x3)= (x3) => f(x) = abs(x3)/(x3)= (x3)/(x3) = 1# 
For
#x in (oo,3)# , we have:
#(x3) < 0 => abs(x3)= (x3) => f(x) = abs(x3)/(x3)= (x3)/(x3) = 1#
Thus,
The discontinuity cannot be removed, as clearly:
As the limit from the right and from the left are different, then the function does not have a limit for
This is the graph of
graph{(x3)/(x3) [2, 5, 3, 3]}