# What is a removable discontinuity?

Apr 9, 2016

Please see the explanation section, below.

#### Explanation:

Recall that:

a function, $f$, is continuous at a number, $a$ if and only if ${\lim}_{x \rightarrow a} f \left(x\right) = f \left(a\right)$.

Furthermore,

${\lim}_{x \rightarrow a} f \left(x\right) = f \left(a\right)$.if and only if

(i) ${\lim}_{x \rightarrow a} f \left(x\right)$ exists,

(ii) $f \left(a\right)$ exists , and

(iii) the numbers in (i) and (ii) are equal.

$f$ has a removable discontinuity at $a$ if and only if ${\lim}_{x \rightarrow a} f \left(x\right)$ exists, but $f$ is not continuous at $a$.

This mean that ${\lim}_{x \rightarrow a} f \left(x\right)$ exists, but that $f \left(a\right)$ either does not exist or $f \left(a\right)$ is different from the limit.

Discontinuities in general

Many presentations of calculus do not give a precise definition of "$f$ has a discontinuity at $a$"
Mathematicians generally mean something like: $f$ is defined for some values near $a$ (in an open interval containing $a$) though possibly not at $a$ and $f$ is not continuous at $a$.

(At other times, mathematicians seems to mean $f$ is continuous near $a$ though possibly not at $a$ and $f$ is not continuous at $a$.)

For example, The square root function (as a function $\mathbb{R} \rightarrow \mathbb{R}$ ) is not continuous at $- 6$, but it's not even defined near $- 6$. So many mathematicians would not say "the square root function has a discontinuity at $- 6$".