Question

hey calculus anyone?

Answer 1

How do i test if a discontinuity is removable or not?

Answer 2

does a removable discontinuity always give you indeterminate form?

Answer 3

Removable discontinuities often occur with rational functions like this,
\[\frac{x^2-1}{x-1}\]
The numerator can be factored, and the \(x-1\) factors in numerator/denominator cancel out, thus making \(x=1\) the removable discontinuity.

However, just because \(\dfrac{x^2-1}{x-1}\) can be changed to \(x+1\), this does not make them the same function. Plugging in \(x=1\) for the first gives you an undefined \(\dfrac{0}{0}\), whereas the other gives you \(2\).

Answer 4

how about when I have a trinomial such as x^2+4x-45? I got -9 and 5 as the discontinuities but they both equal 0 when plugged in. How do I know whether it's removable?

Answer 5

sorry i just realized how dumb a mistake i just made. I didnt copy down the numerator

Answer 6

Do you mean you're given \(x^2+4x-45\), or \(\dfrac{1}{x^2+4x-45}\)?

If you're just given a polynomial like the first one, you don't have to worry about discontinuities. Polynomials are defined for all values \(x\).

In the rational case, \(x=-9,5\) are not removable discontinuities because, well, you can't "remove" them.

Answer 7

Thank you I found my mistake!

Answer 8

yw