# What are the the three things a continuous function can't have?

Nov 23, 2016

There are more than three, but I'll try: In order for a function to be continuous on $\left(- \infty , \infty\right)$ it can have no holes, jumps and vertical asymptotes.

#### Explanation:

I am taking as a definiton:

$f$ is a continuous function if and only if , for every real number, $a$, $f$ is continuous at $a$

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

In order for $f$ to be continuous at $a$,

$f \left(a\right)$ must exist, so there can be no hole at $x = a$

${\lim}_{x \rightarrow a} f \left(x\right)$ must exist which means no jump (different one-sided limits) and no vertical asymptote at $x = a$.

Those are the first 3 I thought of, but the list in not complete. A continuous function also cannot have the kind of "infinite discontinuity" that

$f \left(x\right) = \left\{\begin{matrix}\sin \left(\frac{1}{x}\right) & \mathmr{if} & x \ne 0 \\ 0 & \mathmr{if} & x = 0\end{matrix}\right.$

has at $0$. The limit does not exist, but there is no jump and no vertical asymptote.

Here is most of the graph of $f \left(x\right) = \left\{\begin{matrix}\sin \left(\frac{1}{x}\right) & \mathmr{if} & x \ne 0 \\ 0 & \mathmr{if} & x = 0\end{matrix}\right.$.

(I can't get Socratic's graphing utility to put $\left(0 , 0\right)$ on the graph.)

graph{(y-sin(1/x))=0 [-0.743, 0.756, -1.2, 1.2]}