can fx have one vertical asymptote and one removable discontinuity

Question

I don't understand 

Answer 1

by removable discontinuity, we mean there's a missing point, or hole.

 

Consider: f(x) = (x-3)/[(x-3)(x+2)]

 

Domain: all Reals except 3 and -2. This is based on the ORIGINAL writing of the function.

 

NOW the factor (x-3) cancels, and f(x) simplifies to 1/(x+2); VA is x=-2.