finite mathematics & applied calculus topic summary:
Introduction to the derivative: Part 2 of 3: Continuity
Tools:

Continuous functions

Let $f$ be a function, and let $a$ be a number in its domain. Then $f$ is continuous at the point $a$ if
a. $\displaystyle \lim_{x \to a} f(x)$ exists, and

b. $\displaystyle \lim_{x \to a} f(x) = f(a).$
Notes
1. If $a$ is an endpoint of the domain of $f$, then $\lim_{x \to a} f(x)$ is, as usual, understood to be the appropriate one-sided limit.
2. If $a$ is an isolated point in the domain of $f$; that is, $f$ is defined at $a$ but at no other points within some distance of $a$ (for example, $f(x)=\sqrt{-x^2}$ is defined only at $x = 0$) then, even though there can be no limit at $a,$ we regard $f$ as continuous at $a.$

The function $f$ is said to be continuous on its domain if it is continuous at every point in its domain. If $f$ is not continuous at a particular point, $a$, in its domain, we say that $f$ is discontinuous at $x = a$ or that $f$ has a discontinuity at $x = a$.

Continuity of closed form functions

We saw in Part 1 that, if $f$ is a closed form function, then for every $a$ in its domain, $\lim_{x \to a}f(x)$ exists and equals $f(a)$. Thus closed form functions are continuous on their domains.

Examples: Continuous functions

Consider the function $f$ defined by the following graph (each grid step represents one unit):
 Checking continuity at $\bold{x = 1:}$ (red circle) • $\displaystyle \lim_{x \to 1} f(x)$ exists, and equals $0.5$. ✓ • $f(1) = 0.5$ as well ✓ Therefore, $f$ is continuous at $x = 1.$ Checking continuity at $\bold{x = 0:}$ (blue circle) • $\displaystyle \lim_{x \to 0} f(x)$ does not exist. ✗ Therefore, $f$ is not continuous at $x = 0.$ Checking continuity at $\bold{x = -1:}$ (green circle) • $\displaystyle \lim_{x \to -1} f(x)$ exists, and equals $1$. ✓ • $f(-1) = 0,$ which does not equal the limit. ✗ Therefore, $f$ is not continuous at $x = -1.$

The function $f(x) = 3x^2-4x+2$ is a closed form function, and therefore is continuous on its domain (the set of all real numbers).

The function $\displaystyle g(x) = \frac{4x^2+1}{x-3}$ is also a closed form function, and therefore is continuous on its domain (the set of all real numbers except 3).

The point $x = 3$ is not a discontinuity (as $x = 3$ is not in its domain) but is called a singularity of $g$ (see the next subtopic).

Practice:

Practice:

Singularities

If $f$ is a function such that $f(a)$ is not defined but $f(x)$ is defined for (at least some) values of $x$ arbitrarily close to $a,$ we say that $f$ has a singularity at $\bold{a,}$ or that $\bold{a}$ is a singular point of $\bold{f.}$

 Singularity at $\bold{a}$ Discontinuity at $\bold{a}$
Examples: Singularities

The functions $\displaystyle f(x) = \dfrac{1}{x}$ and $f(x) = \dfrac{1}{x^2}$ are not defined at $x = 0$ but are defined for points aribtrarily close to $x=0$, and so both these functions have singularities at $x = 0$.

The function $f(x) = \begin{cases} \dfrac{1}{x} & \text{if } x \ne 0 \\3 & \text{if } x = 0 \end{cases} \quad$ is defined at $x = 0$ and so is not singular there. It is, however, discontinuous at $x=0$.

Practice:

Removable and essential singularities

If $f$ has a singular point at $a$, we can sometimes remove the singularity by properly defining $f(a)$ (see the left-hand graph below). This occurs when $\displaystyle \lim_{x \to a}f(x)$ exists (and is finite); we just define $f(a)$ to be that limit. In such cases, we say that $f$ has a removable singularity at $\bold{a}$. If the limit does not exist (see the right-hand graph), then no matter how we define $f(a)$, the resulting function will be discontinuaous at $a$, so we say that $f$ has an essential singularity at $\bold{a}$.

 Removable singularity at $\bold{a}$ Essential singularity at $\bold{a}$ $\displaystyle \lim_{x \to a}f(x) = 1$ $\displaystyle \lim_{x \to a}g(x)$ dos not exist. $f$ can be made continuous by defining $f(a) = 1$. $g$ cannot be made continuous.
Examples: Removable and essential singularities

The functions $\displaystyle f(x) = \dfrac{1}{x}$ and $f(x) = \dfrac{1}{x^2}$ have essential singularities at $x = 0$ as, in the first case the limit at zero does not exist, and in the second, the limit is not finite.

$\displaystyle f(x) = \frac{x^2-9}{x-3}$ has a singular point at $x = 3$. In Part 1 of this summary we saw that
$\displaystyle \lim_{x \to 3} f(x) = \lim_{x \to 3} \frac{x^2-9}{x-3} = 6.$
Thus the singularity is removable, and we can remove the singularity and make the function continuous at $x = 3$ by canceling the offending $x-3$ in the denominator:
$\displaystyle \frac{x^2-9}{x-3} = \frac{(x-3)(x+3)}{x-3} = x+3$,
which is now continuous at $x=3$.

Practice: