anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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lgbasallote
  • lgbasallote
because removable discontinuities make the graph go to infinity...that a valid reason?
anonymous
  • anonymous
review your limits lessons. If you evaluate the limit of a removable discontinuity it will not go to infinity.
lgbasallote
  • lgbasallote
it wont? lol..im getting so much wrong today @_@ last time ill answer a calculus question hahahaha =))))

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lgbasallote
  • lgbasallote
sorry asker!
anonymous
  • anonymous
It's ok! lol :)
anonymous
  • anonymous
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
  • anonymous
it did :) thanks!

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