We turn now to an idea closely related to limits, and a concept that we’ll use often and need to define: “continuity.” As you’ll see, whether we can apply certain tools to a function at a given point requires that we know that the function is “continuous” there, so we need to understand what continuity *is!*

Your everyday sense of “continuity” is quite useful for thinking about continuity in mathematics and science: time sweeps along continuously; river water flows continuously; “I biked continuously for twenty miles without stopping.” The key idea is that there is no interruption in something that is continuous — no gaps, leaps, or sudden breaks.

If you can draw the curve without lifting your pencil from the page, the function is continuous

We can translate these everyday ideas to a simple way to determine if a function is continuous by considering its graph: if you can draw a function’s curve without lifting your pencil from the page, then that function is continuous. For example, each of the functions in the following figure is continuous:By contrast, a function is

A more mathematical way to think about continuity is to notice that for a continuous function, if we make a small change in the function’s input (imagine moving your hand a bit to the right as you draw the graph), the function’s output changes by a small amount (so you don’t have to suddenly move where your pencil is to draw the next bit). In fact,

By contrast, if the function is discontinuous at a point (like at $x=2$ in each of the Discontinuous Example graphs above), then you *cannot* make the function’s change in output as small as you’d like. Instead, the function’s value on the left-side of the point is simply disconnected from its value on the right-side of the point.

The words, “make the change-in-output as small as you’d like by making the change-in-input sufficiently small” probably remind you of limits, as they should: the mathematical definition of “continuity” had to wait for the notion of “limit” to be developed, and relies heavily on that notion. Here is the official definition of Continuity at a Point:

A function

\[\lim_{x \to a}f(x) = f(a)\]

First, for a function to be continuous at $x=a$ it must be defined there; otherwise there’s automatically a “gap” of some sort, and so the function is discontinuous. But being *defined* at $x=a$ is not sufficient! For example in the bottom graph above, the function *is* defined at $x=2,$ but is clearly discontinuous at that point.

The bottom graph thus highlights the importance of the limit in the definition of continuity. Applying our understanding of limits from earlier, in words the definition says: a function is continuous at $x =a$ *if* the function is defined at $x=a,$ *and* if we can make the the function’s output $f(x)$ as close as we’d like to its value at $x=a,$ $f(a),$ by making *x* sufficiently close to *a.* That is, using our $\epsilon$ and $\delta$ notation, we *must* be able to be within $\pm \epsilon$ of $f(a)$ (“as close as we’d like”) by being within $\pm \delta$ of $a$ (“sufficiently close”). Kinda cool: see how important your understanding of the concept of “limits” is?

Memorize these

As a practical matter, the continuity definition has three requirements, which you should memorize since a question that asks you to prove continuity at the point $x=a$ means that you must verify each:1. $f(a)$ exists: *a* is in the domain of *f*.

2. $\displaystyle{\lim_{x \to a}f(x)}$ exists.

3. $\displaystyle{\lim_{x \to a}f(x) = f(a)}$: the limit equals the function’s value at $x=a.$

2. $\displaystyle{\lim_{x \to a}f(x)}$ exists.

3. $\displaystyle{\lim_{x \to a}f(x) = f(a)}$: the limit equals the function’s value at $x=a.$

You can see how each function in the Discontinuous Examples figure above fails one of the requirements: In the top two figures, the function is not defined at $x=2,$ and so we know immediately that the function is discontinuous there. (On an exam that’s all you’d have to write; no need to proceed to the other items.) In the bottom graph, $f(2)$ is defined so requirement #1 is fulfilled, but then $\displaystyle{\lim_{x \to 2}f(x)}$ does not exist and so the function is discontinuous at $x=2$ by requirement #2. We’ll of course let you practice showing that a function is continuous, or not, in problems below.

Before then, let’s introduce the idea of **one-sided continuity**, which is of course tied closely to the idea of one-sided limits. The figure to the right shows one of the example continuous functions from above, $f(x) = \sqrt{x}.$ As we know, this function defined only for $x \ge 0$, and simply doesn’t exist for $x \lt 0.$ Its limit as $x \to 0$ thus doesn’t exist, and so we can’t talk about its (general) continuity at $x=0.$ It is, however, “continuous from the right,” meaning it satisfies the one-sided limit condition $\displaystyle{\lim_{x \to 0^+} = f(0)}.$

More generally, one-sided continuity at a point is defined by

\begin{align*}

\text{Continuous from the right at }x=a: &\quad \lim_{x \to a^+} = f(a) \\[8px] \text{Continuous from the left at }x=a: &\quad \lim_{x \to a^-} = f(a) \\[8px] \end{align*}

Let’s consider an example to show how this all works in practice.

Consider the function $f(x) = \sqrt{1-x^2},$ which has the domain $[-1, 1].$ Show that *f* is

(i) continuous from the right at the endpoint $x=-1,$ and

(ii) continuous from the left at $x=1.$

*Solution.*

We use the Three Requirements for Continuity, now as applied for one-sided continuity:

(i) At $x = -1$:

1. $x=-1$ is in the domain of the function.

2. $\displaystyle{\lim_{x \to -1^+}\sqrt{1 – x^2} = \sqrt{\lim_{x \to -1^+}1 – \lim_{x \to -1^+}x^2} = \sqrt{1-1}=0 }$

and so exists.

3. Since $f(-1) = 0,$ $\displaystyle{\overbrace{\lim_{x \to -1^+}\sqrt{1 – x^2} = 0}^{\text{from 2.}} = f(-1).}$

2. $\displaystyle{\lim_{x \to -1^+}\sqrt{1 – x^2} = \sqrt{\lim_{x \to -1^+}1 – \lim_{x \to -1^+}x^2} = \sqrt{1-1}=0 }$

and so exists.

3. Since $f(-1) = 0,$ $\displaystyle{\overbrace{\lim_{x \to -1^+}\sqrt{1 – x^2} = 0}^{\text{from 2.}} = f(-1).}$

Hence

(i) At $x = 1$:

1. $x=1$ is in the domain of the function.

2. $\displaystyle{\lim_{x \to 1^-}\sqrt{1 – x^2} = \sqrt{\lim_{x \to 1^-}1 – \lim_{x \to 1^-}x^2} = \sqrt{1-1}=0 }$

and so exists.

3. Since $f(1) = 0,$ $\displaystyle{\overbrace{\lim_{x \to 1^-}\sqrt{1 – x^2} = 0}^{\text{from 2.}} = f(1).}$

2. $\displaystyle{\lim_{x \to 1^-}\sqrt{1 – x^2} = \sqrt{\lim_{x \to 1^-}1 – \lim_{x \to 1^-}x^2} = \sqrt{1-1}=0 }$

and so exists.

3. Since $f(1) = 0,$ $\displaystyle{\overbrace{\lim_{x \to 1^-}\sqrt{1 – x^2} = 0}^{\text{from 2.}} = f(1).}$

Hence

Quick aside about Substitution and Continuity

The preceding example actually contains a subtle point: do you remember when we first introduced Substitution as a tactic for finding limits we said, rather informally, that we were looking at “a function that is defined at the point of interest and that behaves smoothly near the point of interest, meaning there are no jumps or gaps there”?

Open for reminder of discussion about Substitution to find a limit

When finding the limit $\displaystyle{\lim_{x \to a}f(x) },$ if you substitute $x=a$ into $f(x)$ and obtain a number for $f(a)$ (and

\[\lim_{x \to a} f(x) = f(a)\]

As you saw in the examples above, when Substitution works you can find the limit easily, often in one line, which makes these problems quick. And we don’t want to overcomplicate that. At the same time, we want to emphasize again that $\displaystyle{\lim_{x \to a}f(x) }$ and $f(a)$ are *two distinct quantities.*

Recall our working definition of limits:

\[\lim_{x \to a} f(x) = L\]

is a number *L* (if one exists) such that $f(x)$ is as close to *L* as we want

whenever *x* is sufficiently close to *a.*

We saw earlier various cases where $\displaystyle{\lim_{x \to a} f(x) \ne f(a) }$ (figure (a), below). By contrast, for these “nice, smooth functions” the limit does simply equal $f(a)$ (figure (b) below).

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We see now that what we were actually saying is that

Continuity is crucial: we can only use certain tools where a function is continuous

Indeed, very soon we’ll see that we can only apply certain Calculus tools (like substitution, and later differentiation and integration) Everything above has focused on a function’s continuity at a

Continuity on open intervals

No surprise: If the function *f* is continuous at every number in the open interval $(a, b),$ then *f* is continuous on the interval $(a, b)$.

Similarly, if it is continuous at every number in the infinite interval $(-\infty, b),$ $(a, \infty),$ or $(-\infty, \infty),$ then it is continuous on that interval.

Continuity on closed intervals

We can extend these ideas to apply to continuity on a closed interval, simply by adding in the requirement that the function also be one-sided continuous at both endpoints of the interval:

If a function

\[\lim_{x \to a^+}f(x) = f(a) \quad \text{and} \quad \lim_{x \to b^-}f(x) = f(b)\]

Example 1 (continued): Semicircle continuous over its entire domain

Consider again the function $f(x) = \sqrt{1-x^2},$ which has the domain $[-1, 1].$

Show that *f* is continuous on its domain.

*Solution.*

To show that *f* is continuous on the open interval $(1,1)$:

1. *f* is defined on this interval.

2. Consider an input-value $x=c$ in the interval $-1 \lt c \lt 1.$ Then

\[\lim_{x \to c}f(x) = \sqrt{\lim_{x \to c}1- \lim_{x \to c}x^2} = \sqrt{1-c^2} \] and so exists.

3. Still thinking about $x=c$ in the same interval, since $f(c) = \sqrt{1-c^2},$

\[\overbrace{\lim_{x \to c}f(x) = \sqrt{1-c^2}}^{\text{from 2.}} = f(c)\]

2. Consider an input-value $x=c$ in the interval $-1 \lt c \lt 1.$ Then

\[\lim_{x \to c}f(x) = \sqrt{\lim_{x \to c}1- \lim_{x \to c}x^2} = \sqrt{1-c^2} \] and so exists.

3. Still thinking about $x=c$ in the same interval, since $f(c) = \sqrt{1-c^2},$

\[\overbrace{\lim_{x \to c}f(x) = \sqrt{1-c^2}}^{\text{from 2.}} = f(c)\]

Hence

We already showed above in Example 1 that *f* is one-sided continuous at its endpoints $x= -1$ and $x = 1,$ and won’t repeat the arguments here. Summarizing the results there, we have

\[\lim_{x \to 1^+}f(x) = 0 = f(-1) \; \blacktriangleleft \quad \text{and} \quad \lim_{x \to 1^-}f(x) = 0 = f(1) \; \blacktriangleleft\]
Hence *f* is continuous on its entire domain $[-1, 1]. \; \cmark$

Time for some practice problems so you can consolidate the ideas here for yourself.

Most students find problems that ask you to that a function is continuous a little awkward at first, but — as with most things — become less so with practice.

If you are going to be tested on continuity, make sure you know how to complete Problems #4 and #5, which are *extremely* common on exams.

Practice Problem #1

Given the statement, "The function *f* is continuous at $x=5,$" which of the following must be true?

I. $\displaystyle{\lim_{x \to 5} f(x)}$ exists.

II. $\displaystyle{\lim_{x \to 5}f(x) = f(5)}$.

III. $f(5)$ exists.

IV. $f(5.001)$ exists.

\begin{array}{lllll} \text{(A) I, II, III, & IV} && \text{(B) II only} && \text{(C) I & II only} && \text{(D) I, II & III only} && \text{(E) none of these} \end{array}II. $\displaystyle{\lim_{x \to 5}f(x) = f(5)}$.

III. $f(5)$ exists.

IV. $f(5.001)$ exists.

Practice Problem #2

Let *f* be the function defined by
\[f(x) =
\begin{cases}
\sqrt{x+2} & \text{for } -2 \le x \le 7 \\[8px]
x -4 & \text{for } 7 \lt x
\end{cases} \]
Prove that *f* is, or is not, continuous at $x=7.$

Practice Problem #3

Show that $f(x) = \dfrac{1}{x-1}$ is continuous on the interval $[2, 5].$

The next few problems are typical exam questions.

Practice Problem #4

Let *f* be the function defined by
\[f(x) =
\begin{cases}
x^2 & \text{for } x \le 2 \\[8px]
-x + k & \text{for } 2 \lt x
\end{cases} \]
Find the value of *k* such that *f* is continuous at $x = 2.$

Open for interactive graph of the function, with a slider for $k$

You can use the slider beneath the graph to vary the value of *k* and see visually how setting it to a particular value makes the function continuous at $x = 2.$ This is just a helpful visual that you wouldn’t have on an exam; you still need to use the tools we’ve developed to find the answer and show your work as you would need to on a test.

Use the slider to change the value of *k*:

Currently*k* = 0

Currently

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Practice Problem #5

Find the values of *b* and *c* that make *f* continuous for all real values of *x.* \[f(x) =
\begin{cases}
\dfrac{x^2 - 9}{x-3} & \text{for } x \le 3 \\[8px]
x^2 + bx + c & \text{for } 3 \lt x \le 4 \\[8px]
2x + c & \text{for } 4 \le x
\end{cases} \]

On the next screen we’ll give names to the different

For now, what questions or comments do you have about the material on this screen, or any other Calculus concepts? Please join the discussion over on the Forum!

- Informally, a function is continuous if you can draw its curve without lifting your pencil. Formally, a function
*f*is continuous at $x = a$ if

\[\lim_{x \to a}f(x) = f(a)\] - To prove that a function is continuous at $x=a,$ we must show three things (which you should memorize):

1. $f(a)$ exists:*a*is in the domain of*f*.

2. $\displaystyle{\lim_{x \to a}f(x)}$ exists.

3. $\displaystyle{\lim_{x \to a}f(x) = f(a)}$: the limit equals the function’s value at $x=a.$ - For one-sided continuity, we consider only the limit from the left or from the right:

\begin{align*}

\text{Continuous from the right at }x=a: &\quad \lim_{x \to a^+} = f(a) \\[8px] \text{Continuous from the left at }x=a: &\quad \lim_{x \to a^-} = f(a) \\[8px] \end{align*} - A function is continuous on the open interval $(a, b)$ if it is continuous at every point in that interval. It is continuous on the closed interval $[a, b]$ if it is continuous on the open interval $(a, b)$ and also at the endpoints, such that

\[\lim_{x \to a^+}f(x) = f(a) \quad \text{and} \quad \lim_{x \to b^-}f(x) = f(b)\]