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

Question

lgbasallote · Answer

because removable discontinuities make the graph go to infinity...that a valid reason?

anonymous · Answer

review your limits lessons. If you evaluate the limit of a removable discontinuity it will not go to infinity.

lgbasallote · Answer

it wont? lol..im getting so much wrong today @_@ last time ill answer a calculus question hahahaha =))))

lgbasallote · Answer

sorry asker!

anonymous · Answer

It's ok! lol :)

anonymous · Answer

I agree with @rlsadiz , but to put in simpler terms, the hole is basically where a common factor from the numerator and denominator cancelled to make the function look like another function but with the domain restriction of the original function...  wait... did that make sense?

anonymous · Answer

it did :) thanks!