2.4: Continuity

Continuity

A function that is "friendly" and doesn’t have any breaks or jumps in it is called continuous. More formally,

Continuity at a Point

A function $\bf f$ is continuous at $\bf x = a$ if and only if $\lim\limits_{x\to a} \mathbf{f(x)} = \mathbf{f(a)}$.

The graph below illustrates some of the different ways a function can behave at and near a point, and the table contains some numerical information about the function and its behavior.

$a$	$f(a)$	$\lim\limits_{x\to a} f(x)$
1	2	2
2	1	2
3	2	Does not exist (DNE)
4	Undefined	2

Based on the information in the table, we can conclude that $f$ is continuous at 1 since $\lim\limits_{x\to 1} f(x) = 2 = f(1)$.

We can also conclude from the information in the table that $f$ is not continuous at 2 or 3 or 4, because $\lim\limits_{x\to 2} f(x) \neq f(2)$, $\lim\limits_{x\to 3} f(x) \neq f(3)$, and $\lim\limits_{x\to 4} f(x) \neq f(4)$.

The behaviors at $x = 2$ and $x = 4$ exhibit a hole in the graph, sometimes called a removable discontinuity, since the graph could be made continuous by changing the value of a single point. The behavior at $x = 3$ is called a jump discontinuity, since the graph jumps between two values.

So which functions are continuous? It turns out pretty much every function you’ve studied is continuous where it is defined: polynomial, radical, rational, exponential, and logarithmic functions are all continuous where they are defined. Moreover, any combination of continuous functions is also continuous.

This is helpful, because the definition of continuity says that for a continuous function, $\lim\limits_{x\to a} f(x) = f(a)$. That means for a continuous function, we can find the limit by direct substitution (evaluating the function) if the function is continuous at $a$.

Example $\PageIndex{5}$

Evaluate using continuity, if possible:

1. $\lim\limits_{x\to 2} x^3-4x$
2. $\lim\limits_{x\to 2} \dfrac{x-4}{x+3}$
3. $\lim\limits_{x\to 2} \dfrac{x-4}{x-2}$

Solution

1. The given function is polynomial, and is defined for all values of $x$, so we can find the limit by direct substitution: $$\lim\limits_{x\to 2} x^3-4x = 2^3-4(2) = 0. \nonumber$$
2. The given function is rational. It is not defined at $x = -3$, but we are taking the limit as $x$ approaches 2, and the function is defined at that point, so we can use direct substitution: $$\lim\limits_{x\to 2} \dfrac{x-4}{x+3} = \dfrac{2-4}{2+3}= -\dfrac{2}{5}. \nonumber$$
3. This function is not defined at $x = 2$, and so is not continuous at $x = 2$. We cannot use direct substitution.