What does it mean when it says: "non-removable discontinuity at x=4"?

Oct 22, 2015

See the explanation.

Explanation:

Here is the definition I am familiar with.

Function $f$ has a discontinuity at $a$ if $f$ is defined in an open interval containing $a$ except possibly at $x = a$, and $f$ is not continuous at $a$.

(This allows us to avoid saying, for example that $\sqrt{x}$ has a discontinuity at $x = - 5$, which is good. But, the function $f \left(x\right) = 1$ with domain all irrational numbers is not continuous at $3$, but is also not discontinuous at $3$, because is it not defined on an open interval containing $3$.)

$f$ has a removable discontinuity at $a$ if ${\lim}_{x \rightarrow a} f \left(x\right)$ exists. (Remember that writing ${\lim}_{x \rightarrow a} f \left(x\right) = \infty$ is a way of explaining why the limit Does Not Exist.)

$f$ has a non-removable discontinuity at $a$ if $f$ has a discontinuity at $a$ and ${\lim}_{x \rightarrow a} f \left(x\right)$ does not exist.

In the most familiar functions:

Rational and trigonometric functions have non-removable discontinuities at their vertical asymptotes. (Holes are removable.)

Piecewise-defined functions can have jump discontinuities, which are non-removable. (Holes are removable.)

The Greatest integer function (a.k.a. the Floor function) has a non-removable discontinuity at every integer.