Question

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

Answer 1

if you take the limit of a function as it approaches a removable discontinuity you will get the value it should have been if it was there. If you take the limit of a value at an asymptote you usually get infinity or 0, sometimes DNE

please correct me if i'm wrong, i hated algebra 2

Answer 2

A removable discontinuity is just a single point that has been excluded from the graph of a function.

An asymptote refers to a situation where you approach, but never reach, a particular value.

Although these situations are similar in a way, a removable discontinuity is not an asymptote, because the graph actually HAS approached this point - it's just been excluded from the domain.

Answer 3

And exclusion from the domain does not automatically produce an asymptote.

Answer 4

why is a discontinuity removable? because the limit of common factors is 1. common factors can be removed.

\[\lim_{x\to a}\frac{(x-a)(x-b)}{(x-a)(x-k)}\]
\[\lim_{x\to a}\frac{(x-a)}{(x-a)}~\lim_{x\to a}\frac{(x-b)}{(x-k)}\]
\[1~\lim_{x\to a}\frac{(x-b)}{(x-k)}\]
\[\frac{a-b}{a-k}\]

x=a is still not a part of the domain since the original setup may model a real life event.

A removable discontinuity is like a crack in the sidewalk, we can just step over it an move on.

A vertical asymptote is a brick wall that blocks our way, and we walk up and down it in a vain attempt to get around it.