Question

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?

Answer 1

Hmm kind of a neat question :)

Answer 2

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.

Answer 3

Thank you so much! That really helped a lot :)