Question

Use the drawing tools to sketch the graph of a rational function with a domain of x . Include one removable discontinuity and one nonremovable discontinuity. Label each discontinuity using the text tool.

Answer 1

If
`f(x)` has a removable discontinuity at
`x=a`, then the limit

`\displaystyle \lim_(x\to a) (f(x))/(x-a)`

exists and is finite.

A non-removable discontinuity at
`x=b` would entail a non-finite limit,

`\displaystyle \lim_(x\to b) (f(x))/(x-b) = \pm\infty`

or the limit does not exist (which could be due to the limits from either side of
`x=b` not matching or existing).

For a rational function, we want

`f(x) = (p(x))/(q(x))`

where
`p` and
`q` are polynomials in
`x`. To get a removable discontinuity at
`x=a`, both
`p` and
`q` must be divisible by
`x-a`, and the limit of their quotient after removing these factors still exists. That is,

`\displaystyle \lim_(x\to a) f(x) = \lim_(x\to a) (p(x))/(q(x)) = \lim_(x\to a)  ((x-a)p^*(x))/((x-a)q^*(x)) = \lim_(x\to a) (p^*(x))/(q^*(x)) = (p^*(a))/(q^*(a))`

On the flip side, we get a non-removable discontinuity
`x=b` if
`p` is not divisible by
`x-b`, in which case

`\displaystyle \lim_(x\to b) f(x) = \lim_(x\to b) (p(x))/(q(x)) = \lim_(x\to b)  (p(x))/((x-b)q^*(x)) = (p(b))/(0* q^*(b))`

and this is undefined.

Suppose
`f(x)` has a non-removable discontinuity at
`x=-5` and a removable one at
`x=4`. Then one such function could be

`f(x) = (x-4)/((x-4)(x+5)) = (x-4)/(x^2+x-20)`