Why doesn't a removable discontinuity produce an asymptote on the graph of a polynomial function, even though it is excluded from the domain of the function?

Question

amistre64 · Answer

becasue an asymptote is defined more by the graph snuggling up close to it;

amistre64 · Answer

a hole is created when the badness of the bottom is controled by the top so that it simply vanishes at the hole

amistre64 · Answer

an asymptote has no controlling feature so that it slides up to it

amistre64 · Answer

consider the graphs of: 3/x and 3x/x

amistre64 · Answer

http://www.wolframalpha.com/input/?i=y%3D3%2Fx++and++y%3D3x%2Fx one provides control, the other not