hugy0212
  • hugy0212
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
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SOLVED
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chestercat
  • chestercat
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zepdrix
  • zepdrix
Hmm kind of a neat question :)
zepdrix
  • zepdrix
I guess I would say because of the fact that it's a `removable discontinuity`. That point of discontinuity has been `removed`. So any asymptotic behavior that might have existed around the point has been completely removed. Do you know what a removable discontinuity looks like in a function? Here is an example: \(\large\rm \dfrac{x(x-2)}{(x-2)}\) It looks like the straight line with a hole at =2. Here is an example of a function with an asymptote: \(\large\rm \dfrac{1}{x-2}\) We get the asymptotic behavior because the discontinuity doesn't cancel out with anything else.
hugy0212
  • hugy0212
Thank you so much! That really helped a lot :)

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