The function #f(x)=(x+3)/((x-4)(x+3))# is undefined at #x=4# and #x=-3#, so it has discontinuities at those two values of #x#.

The one at #x=-3#, however, is "removable", because, after cancellation, #f(x)=1/(x-4)# whenever #x# is **not** equal to #-3#. The expression #1/(x-4)# is defined at #x=-3# and has a value of #1/(-3-4)=-1/7# there.

Because of this, #lim_{x->-3}f(x)=-1/7# and the graph of #f(x)=(x+3)/((x-4)(x+3))# only has a "hole" at #x=-3# (no vertical asymptote). This hole can be "filled in" by defining #f(-3)=-1/7# (in other words, make #f# a piecewise-defined function).

By filling in this hole in the graph of #f#, we have "removed" the original discontinuity there. That's why it's called a removable discontinuity.

The discontinuity at #x=4# is not removable. The graph of #f# has a vertical asymptote there. In fact, #lim_{x->4+}f(x)=+infty# (#x# approaches 4 from the right) and #lim_{x->4-}f(x)=-infty# (#x# approaches 4 from the left).

The graph of #f# is shown below. The hole is not seen because the computer is just connecting the dots. The vertical asymptote can be seen, however.

graph{(x+3)/((x-4)(x+3)) [-10.13, 10.145, -5.07, 5.065]}