3
$\begingroup$

In the third function from this image. The function when $x$ is not equal to $2$ can be transformed into $ x+1 $, resulting in having a value of $3$ at the point $ x = 2 $.

Why is this not considered a removable discontinuity?

enter image description here

$\endgroup$
2
  • $\begingroup$ Essentially what the answer states is that the hole has been filled in by the piecewise part. $\endgroup$ Sep 10, 2021 at 10:37
  • $\begingroup$ A removed discontinuity is not removable, because there's nothing more to remove :( $\endgroup$
    – Trebor
    Sep 10, 2021 at 11:02

1 Answer 1

4
$\begingroup$

It is because

$f(x)=\begin{cases}\frac{x^2-x-2}{x-2}; x\ne 2\\ 3; x=2\end{cases}$

is continuous at $x=2$ (Note that $\lim _{x\to 2} f(x)=\lim_{x\to 2}(x+1)=3$ and $f(2)=3$) and as per the given question, you are required to choose functions which have a removable discontinuity at $x=2$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .