# How do you find the points of continuity of a function?

Jul 18, 2015

For functions we deal with in lower level Calculus classes, it is easier to find the points of discontinuity. Then the points of continuity are the points left in the domain after removing points of discontinuity

#### Explanation:

A function cannot be continuous at a point outside its domain, so, for example:

$f \left(x\right) = {x}^{2} / \left({x}^{2} - 3 x\right)$ cannot be continuous at $0$, nor at $3$.

It is worth learning that rational functions are continuous on their domains.

This brings up a general principle: a function that has a denominator is not defined (and hence, not continuous) at points where the denominator is $0$.
This include "hidden" denominators as we have in $\tan x$, for example. We don't see the denominator $\cos x$, but we know it's there.

For functions defined piecewise, we must check the partition number, the points where the rules change. The function may or may not be continuous at those points.

Recall that in order for $f$ to be continuous at $c$, we must have:

$f \left(c\right)$ exists ($c$ is in the domain of $f$)

${\lim}_{x \rightarrow c} f \left(x\right)$ exists

and also these two numbers are equal (a strange phrase, but it is common enough -- I mean these two descriptions pick out the same number.)

It is important and relevant for piecewise function, to remember that in order for ${\lim}_{x \rightarrow c} f \left(x\right)$ to exist, both the left and right limits must exist and they must be equal.

Example:

$f \left(x\right) = \left\{\begin{matrix}3 x - 1 & 0 < x < 2 \\ 8 & x = 2 \\ {x}^{2} + 1 & 2 < x < 7 \\ x - 4 & 7 \le x < 9 \\ \frac{5}{x - 8} & 9 \le x\end{matrix}\right.$

The domain of $f$ is $\left(0 , \infty\right)$ so $f$ is continuous on some subset of that domain.

Each of the 4 functions is continuous on the interval on which it is used: $3 x - 1$, ${x}^{2} + 1$, and $x - 4$ are polynomials, hence continuous everywhere.
$\frac{5}{x - 2}$ is discontinuous at $2$, but it is not used near $2$, so that is not a problem.

We need to check for continuity at the numbers 2, 7, and 9. Note first (if you have not already done so) that each is in the domain of $f$. (If the last rule were for $9 < x$, then $9$ would not be in the domain of the function.)

$f \left(x\right) = \left\{\begin{matrix}3 x - 1 & 0 < x < 2 \\ 8 & x = 2 \\ {x}^{2} + 1 & 2 < x < 7 \\ x - 4 & 7 \le x < 9 \\ \frac{5}{x - 8} & 9 \le x\end{matrix}\right.$

Checking at $2$

${\lim}_{x \rightarrow {2}^{-}} f \left(x\right) = {\lim}_{x \rightarrow {2}^{-}} \left(3 x - 1\right) = 5$

${\lim}_{x \rightarrow {2}^{+}} f \left(x\right) = {\lim}_{x \rightarrow {2}^{+}} \left({x}^{2} + 1\right) = 5$

Because these are equal, we have
${\lim}_{x \rightarrow 2} f \left(x\right) = 5$

However, $f \left(2\right) = 8$, so the function is discontinuous at $2$

Checking at $7$, the left and right limits are: $50$ and $3$, respectively. The limit does not exist and the function is discontinuous at $7$

At $9$ , we have equal one-sided limits: $5$, so ${\lim}_{x \rightarrow 9} f \left(x\right) = 5$ and $f \left(9\right) = 5$ so $f$ is continuous at $9$

$f$ is continuous at every $x$ greater than zero except $2$ and $7$.

In interval notation:

$f$ is continuous on $\left(0 , 2\right) \cup \left(2 , 7\right) \cup \left(7 , \infty\right)$