Continuous Functions, calculus lesson

A thermometer reading changes smoothly through the day. It never leaps from 12 degrees to 15 degrees without passing every value in between. Continuity is the precise version of that idea. It says the value of a function at a point is exactly what the nearby values are heading toward, so small changes in the input never cause a sudden jump in the output.

Key terms

Interior points and endpoints

Definition Let $D$ be a set of real numbers. A point $c \in D$ is an interior point of $D$ if some open interval around it lies in $D$ , that is, if there is $r > 0$ with $(c - r, c + r) \subseteq D$ . A point $c \in D$ is a left endpoint of $D$ if there is $r > 0$ with $[c, c + r) \subseteq D$ and $(c - r, c) \cap D = \varnothing$ . It is a right endpoint of $D$ if there is $r > 0$ with $(c - r, c] \subseteq D$ and $(c, c + r) \cap D = \varnothing$ .

In plain words At an interior point you can move a little in both directions and stay in the domain. At an endpoint the domain stops on one side.

Example The domain of $f(x) = \sqrt{x}$ is $D = [0, \infty)$ . The point $0$ is a left endpoint, because $[0, 1) \subseteq D$ and no negative number is in $D$ . Every $c > 0$ is an interior point, since $(c - c/2, c + c/2) \subseteq D$ .

Continuity at a point

Definition Let $f$ be defined on an open interval containing $c$ . Then $f$ is continuous at $c$ if

\lim_{x \to c} f(x) = f(c).

This packs three conditions into one line. The value $f(c)$ is defined, the limit $\lim_{x \to c} f(x)$ exists as a finite number, and the two are equal. In terms of $\varepsilon$ and $\delta$ , it says that for every $\varepsilon > 0$ there is $\delta > 0$ such that

\lvert x - c \rvert < \delta \implies \lvert f(x) - f(c) \rvert < \varepsilon.

In plain words You can predict $f(c)$ from the values of $f$ near $c$ , and the prediction is right.

Example Let

f(x) = \begin{cases} \dfrac{x^2 - 4}{x - 2}, & x \neq 2, \\ 4, & x = 2. \end{cases}

For $x \neq 2$ the formula simplifies to $x + 2$ , so $\lim_{x \to 2} f(x) = 4$ . Since $f(2) = 4$ as well, $f$ is continuous at $2$ . If instead we had set $f(2) = 5$ , the limit would still be $4$ but it would no longer match the value, and $f$ would not be continuous at $2$ .

Left and right continuity

Definition Let $f$ be defined on an interval $(c - r, c]$ for some $r > 0$ . Then $f$ is continuous from the left at $c$ if

\lim_{x \to c^-} f(x) = f(c).

Let $f$ be defined on $[c, c + r)$ for some $r > 0$ . Then $f$ is continuous from the right at $c$ if

\lim_{x \to c^+} f(x) = f(c).

When $f$ is defined on an open interval around $c$ , it is continuous at $c$ exactly when it is continuous from both the left and the right at $c$ .

In plain words Left continuity only looks at inputs slightly smaller than $c$ . Right continuity only looks at inputs slightly larger.

Example The floor function $\lfloor x \rfloor$ gives the greatest integer less than or equal to $x$ . At $c = 1$ we have $\lfloor 1 \rfloor = 1$ . For $x$ just above $1$ the value is $1$ , so $\lim_{x \to 1^+} \lfloor x \rfloor = 1$ and the function is continuous from the right at $1$ . For $x$ just below $1$ the value is $0$ , so the left limit is $0 \neq 1$ and it is not continuous from the left there.

Continuity relative to the domain

Definition Let $f$ have domain $D$ and let $c \in D$ . Then $f$ is continuous at $c$ if for every $\varepsilon > 0$ there is $\delta > 0$ such that

x \in D \text{ and } \lvert x - c \rvert < \delta \implies \lvert f(x) - f(c) \rvert < \varepsilon.

At an interior point of $D$ this is the same as $\lim_{x \to c} f(x) = f(c)$ . At a left endpoint it is the same as continuity from the right, and at a right endpoint the same as continuity from the left. A function is called continuous, with no interval mentioned, when it is continuous at every point of its domain.

In plain words Only inputs that are actually in the domain count. Continuity is a question about points of the domain, so at a point where $f$ is not defined the question of continuity is never asked.

Example The function $g(x) = \frac{1}{x}$ has domain $D = \{x \mid x \neq 0\}$ , and every point $c$ of $D$ is interior with

\lim_{x \to c} \frac{1}{x} = \frac{1}{c} = g(c).

So $g$ is a continuous function, even though its graph breaks at $x = 0$ . The break happens at a point outside $D$ . The next lesson still calls $0$ a discontinuity of $g$ , because the word discontinuity is used for gaps in the domain as well. Both statements are true together, since being a continuous function only concerns points of $D$ . Likewise $\sqrt{x}$ is continuous at its left endpoint $0$ , because $\lim_{x \to 0^+} \sqrt{x} = 0 = \sqrt{0}$ .

Continuity on an interval

Definition Let $I$ be an interval contained in the domain of $f$ . Then $f$ is continuous on $I$ if it is continuous at every interior point of $I$ , continuous from the right at the left endpoint of $I$ if that endpoint belongs to $I$ , and continuous from the left at the right endpoint of $I$ if that endpoint belongs to $I$ .

In plain words Check every point of the interval. Inside, use the two-sided limit. At an endpoint that is included, only look from inside the interval.

Example The floor function is continuous on $[0, 1)$ , because it equals $0$ there. It is not continuous on $[0, 1]$ , because at the included right endpoint $1$ the left limit is $0$ while $\lfloor 1 \rfloor = 1$ . As a second example, $h(x) = \sqrt{1 - x^2}$ is continuous on $[-1, 1]$ . For $-1 < c < 1$ the limit laws for roots give

\lim_{x \to c} \sqrt{1 - x^2} = \sqrt{1 - c^2} = h(c),

and at the ends the one-sided limits match the values,

\lim_{x \to -1^+} h(x) = 0 = h(-1), \qquad \lim_{x \to 1^-} h(x) = 0 = h(1).

Check yourself

Find the value of $k$ that makes $f(x) = 3x - 1$ for $x < 2$ and $f(x) = kx^2$ for $x \ge 2$ continuous at $2$ .

Show the answer

The left limit is $3 \cdot 2 - 1 = 5$ . The right limit and the value are both $4k$ . Continuity needs $4k = 5$ , so $k = \frac{5}{4}$ .

Is $f(x) = \sqrt{x - 3}$ continuous at $x = 3$ ?

Show the answer

Yes. The domain is $[3, \infty)$ and $3$ is its left endpoint, so only right continuity is needed. Since $\lim_{x \to 3^+} \sqrt{x - 3} = 0 = f(3)$ , the function is continuous at $3$ .

On which of the intervals $[1, 2)$ and $[1, 2]$ is $\lfloor x \rfloor$ continuous?

Show the answer

It is continuous on $[1, 2)$ , where it is constantly $1$ . It is not continuous on $[1, 2]$ , because at $2$ the left limit is $1$ while $\lfloor 2 \rfloor = 2$ .

Continuous Functions

Theory, part 27

Practice, part 28

Key terms

Interior points and endpoints

Continuity at a point

Left and right continuity

Continuity relative to the domain

Continuity on an interval

Check yourself

How this lesson connects

Builds on

Used later in