REMOVABLE and NON-REMOVABLE DISCONTINUITY
if you have factors on the top and bottom that cancel, then this is a
removable discontinuity. You can find the y-value of the
discontinuity by cancelling the common factors, and then plugging
in the x-value of the hole to what's left
In other words, a removable discontinuity looks like 0/0
a non-removable discontinuity looks like 1 / 0
ex: f(x) = (x - 2)(x + 3) /[(x - 2) (x - 4)]
when x = 2, this becomes 0 / 0
x = 2 is a removable discontinuity
cancel the (x - 2) / (x - 2) to leave (x + 3) / (x - 4)
when x = 2, this is 5 / -2 =-5/2
so the removable discontinuity is the point (2 , -5/2) (that is, the
"hole" in the graph)
x = 4 is a non-removable discontinuity, and will be an asymptote of
the graph
ex: the graphs of y = x - 2 and y = (x^2 - 4) / (x + 2) are identical,
except for the removable discontinuity at x = -2
(this would be the point (-2 , -4) that's removed)
that's because (x^2 - 4) / (x + 2) = (x - 2)(x + 2) / (x + 2)
they look the same, but the (x + 2) / (x + 2) creates the hole that is
removed from the function