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2.5: Continuity

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    Nov 2, 2021
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Learning Objectives

  • Explain the three conditions for continuity at a point.
  • Describe three kinds of discontinuities.
  • Define continuity on an interval.
  • State the theorem for limits of composite functions.
  • Provide an example of the intermediate value theorem.

Many functions have the property that their graphs can be traced with a pencil without lifting the pencil from the page. Such functions are called continuous. Other functions have points at which a break in the graph occurs, but satisfy this property over intervals contained in their domains. They are continuous on these intervals and are said to have a discontinuity at a point where a break occurs.

We begin our investigation of continuity by exploring what it means for a function to have continuity at a point. Intuitively, a function is continuous at a particular point if there is no break in its graph at that point.

Continuity at a Point

Before we look at a formal definition of what it means for a function to be continuous at a point, let’s consider various functions that fail to meet our intuitive notion of what it means to be continuous at a point. We then create a list of conditions that prevent such failures.

Our first function of interest is shown in Figure . We see that the graph of has a hole at . In fact, is undefined. At the very least, for to be continuous at , we need the following condition:

i. is defined

A graph of an increasing linear function f(x) which crosses the x axis from quadrant three to quadrant two and which crosses the y axis from quadrant two to quadrant one. A point a greater than zero is marked on the x axis. The point on the function f(x) above a is an open circle; the function is not defined at a.
Figure : The function is not continuous at a because is undefined.

However, as we see in Figure , this condition alone is insufficient to guarantee continuity at the point . Although is defined, the function has a gap at . In this example, the gap exists because does not exist. We must add another condition for continuity at —namely,

ii. exists

The graph of a piecewise function f(x) with two parts. The first part is an increasing linear function that crosses from quadrant three to quadrant one at the origin. A point a greater than zero is marked on the x axis. At fa. on this segment, there is a solid circle. The other segment is also an increasing linear function. It exists in quadrant one for values of x greater than a. At x=a, this segment has an open circle.
Figure : The function is not continuous at a because does not exist.

However, as we see in Figure , these two conditions by themselves do not guarantee continuity at a point. The function in this figure satisfies both of our first two conditions, but is still not continuous at . We must add a third condition to our list:

iii.

The graph of a piecewise function with two parts. The first part is an increasing linear function that crosses the x axis from quadrant three to quadrant two and which crosses the y axis from quadrant two to quadrant one. A point a greater than zero is marked on the x axis. At this point, there is an open circle on the linear function. The second part is a point at x=a above the line.
Figure : The function is not continuous at a because .

Now we put our list of conditions together and form a definition of continuity at a point.

Definition: Continuous at a Point

A function is continuous at a point if and only if the following three conditions are satisfied:

  1. is defined
  2. exists

A function is discontinuous at a point if it fails to be continuous at .

The following procedure can be used to analyze the continuity of a function at a point using this definition.

Problem-Solving Strategy: Determining Continuity at a Point

  1. Check to see if is defined. If is undefined, we need go no further. The function is not continuous at a. If is defined, continue to step 2.
  2. Compute . In some cases, we may need to do this by first computing and . If does not exist (that is, it is not a real number), then the function is not continuous at a and the problem is solved. If exists, then continue to step 3.
  3. Compare and . If , then the function is not continuous at a. If , then the function is continuous at a.

The next three examples demonstrate how to apply this definition to determine whether a function is continuous at a given point. These examples illustrate situations in which each of the conditions for continuity in the definition succeed or fail.

Example : Determining Continuity at a Point, Condition 1

Using the definition, determine whether the function is continuous at . Justify the conclusion.

Solution

Let’s begin by trying to calculate . We can see that , which is undefined. Therefore, is discontinuous at 2 because is undefined. The graph of is shown in Figure .

A graph of the given function. There is a line which crosses the x axis from quadrant three to quadrant two and which crosses the y axis from quadrant two to quadrant one. At a point in quadrant one, there is an open circle where the function is not defined.
Figure : The function is discontinuous at because is undefined.

Example : Determining Continuity at a Point, Condition 2

Using the definition, determine whether the function is continuous at . Justify the conclusion.

Solution

Let’s begin by trying to calculate .

.

Thus, is defined. Next, we calculate . To do this, we must compute and :

and

.

Therefore, does not exist. Thus, is not continuous at 3. The graph of is shown in Figure .

A graph of the given piecewise function, which has two parts. The first is a downward opening parabola which is symmetric about the y axis. Its vertex is on the y axis, greater than zero. There is a closed circle on the parabola for x=3. The second part is an increasing linear function in the first quadrant, which exists for values of x > 3. There is an open circle at the end of the line where x would be 3.
Figure : The function is not continuous at 3 because does not exist.

Example : Determining Continuity at a Point, Condition 3

Using the definition, determine whether the function is continuous at .

Solution

First, observe that

Next,

.

Last, compare and . We see that

.

Since all three of the conditions in the definition of continuity are satisfied, is continuous at .

Exercise

Using the definition, determine whether the function is continuous at . If the function is not continuous at 1, indicate the condition for continuity at a point that fails to hold.

Hint

Check each condition of the definition.

Answer

is not continuous at because .

By applying the definition of continuity and previously established theorems concerning the evaluation of limits, we can state the following theorem.

Continuity of Polynomials and Rational Functions

Polynomials and rational functions are continuous at every point in their domains.

Proof

Previously, we showed that if and are polynomials, for every polynomial and as long as . Therefore, polynomials and rational functions are continuous on their domains.

□

We now apply Note to determine the points at which a given rational function is continuous.

Example :Continuity of a Rational Function

For what values of x is continuous?

Solution

The rational function is continuous for every value of except .

Exercise

For what values of is continuous?

Hint

Use the Continuity of Polynomials and Rational Functions stated above.

Answer

is continuous at every real number.

Types of Discontinuities

As we have seen in Example and Example, discontinuities take on several different appearances. We classify the types of discontinuities we have seen thus far as removable discontinuities, infinite discontinuities, or jump discontinuities. Intuitively, a removable discontinuity is a discontinuity for which there is a hole in the graph, a jump discontinuity is a noninfinite discontinuity for which the sections of the function do not meet up, and an infinite discontinuity is a discontinuity located at a vertical asymptote. Figure illustrates the differences in these types of discontinuities. Although these terms provide a handy way of describing three common types of discontinuities, keep in mind that not all discontinuities fit neatly into these categories.

Three graphs, each showing a different discontinuity. The first is removable discontinuity. Here, the given function is a line with positive slope. At a point x=a, where a>0, there is an open circle on the line and a closed circle a few units above the line. The second is a jump discontinuity. Here, there are two lines with positive slope. The first line exists for x<=a, and the second exists for x>a, where a>0. The first line ends at a solid circle where x=a, and the second begins a few units up with an open circle at x=a. The third discontinuity type is infinite discontinuity. Here, the function has two parts separated by an asymptote x=a. The first segment is a curve stretching along the x axis to 0 as x goes to negative infinity and along the y axis to infinity as x goes to zero. The second segment is a curve stretching along the y axis to negative infinity as x goes to zero and along the x axis to 0 as x goes to infinity.
Figure : Discontinuities are classified as (a) removable, (b) jump, or (c) infinite.

These three discontinuities are formally defined as follows:

Definition

If is discontinuous at a, then

1. has a removable discontinuity at a if exists. (Note: When we state that exists, we mean that , where L is a real number.)

2. has a jump discontinuity at a if and both exist, but . (Note: When we state that and both exist, we mean that both are real-valued and that neither take on the values .)

3. has an infinite discontinuity at a if or .

Example : Classifying a Discontinuity

In Example, we showed that is discontinuous at . Classify this discontinuity as removable, jump, or infinite.

Solution

To classify the discontinuity at 2 we must evaluate :

Erroneous nesting of equation structures\displaystyle \begin{align*} \lim_{x→2}f(x) &=\lim_{x→2}\frac{x^2−4}{x−2}\\[4pt] &= \lim_{x→2}\frac{(x−2)(x+2)}{x−2}\\[4pt] &= \lim_{x→2}(x+2)\\[4pt] &=4.\end{align*}

Since is discontinuous at and exists, f has a removable discontinuity at .

Example : Classifying a Discontinuity

In Example, we showed that is discontinuous at . Classify this discontinuity as removable, jump, or infinite.

Solution

Earlier, we showed that is discontinuous at because does not exist. However, since and both exist, we conclude that the function has a jump discontinuity at .

Example : Classifying a Discontinuity

Determine whether is continuous at . If the function is discontinuous at , classify the discontinuity as removable, jump, or infinite.

Solution

The function value is undefined. Therefore, the function is not continuous at . To determine the type of discontinuity, we must determine the limit at . We see that and . Therefore, the function has an infinite discontinuity at .

Types of Discontinuities in the Real World

Decide whether the given real world example includes a Removable Discontinuity, a Jump Discontinuity, an Infinite Discontinuity, or is Continuous.

 

    Removable Discontinuity
    Jump Discontinuity
    Infinite Discontinuity
    Continuous

f(t) is the function that represents the total mass of the universe over time, t, measured in years since the big bang. t goes from -1,000,000 to 1,000,000. Assume that before the big bang, no matter was present and at the big bang, the universe was suddenly created.

Click on the type of discontinuity (or continuous) that the given function has.

 

 

Exercise

For , decide whether is continuous at . If is not continuous at , classify the discontinuity as removable, jump, or infinite.

Hint

Consider the definitions of the various kinds of discontinuity stated above. If the function is discontinuous at , look at

Answer

Discontinuous at ; removable

Continuity over an Interval

Now that we have explored the concept of continuity at a point, we extend that idea to continuity over an interval. As we develop this idea for different types of intervals, it may be useful to keep in mind the intuitive idea that a function is continuous over an interval if we can use a pencil to trace the function between any two points in the interval without lifting the pencil from the paper. In preparation for defining continuity on an interval, we begin by looking at the definition of what it means for a function to be continuous from the right at a point and continuous from the left at a point.

Continuity from the Right and from the Left

A function is said to be continuous from the right at a if .

A function is said to be continuous from the left at a if

A function is continuous over an open interval if it is continuous at every point in the interval. A function is continuous over a closed interval of the form if it is continuous at every point in and is continuous from the right at a and is continuous from the left at b. Analogously, a function is continuous over an interval of the form if it is continuous over and is continuous from the left at b. Continuity over other types of intervals are defined in a similar fashion.

Requiring that and ensures that we can trace the graph of the function from the point to the point without lifting the pencil. If, for example, , we would need to lift our pencil to jump from to the graph of the rest of the function over .

Example : Continuity on an Interval

State the interval(s) over which the function is continuous.

Solution

Since is a rational function, it is continuous at every point in its domain. The domain of is the set . Thus, is continuous over each of the intervals , and .

Example : Continuity over an Interval

State the interval(s) over which the function is continuous.

Solution

From the limit laws, we know that for all values of a in . We also know that exists and exists. Therefore, is continuous over the interval .

Exercise

State the interval(s) over which the function is continuous.

Hint

Use Example as a guide.

Answer

The Note allows us to expand our ability to compute limits. In particular, this theorem ultimately allows us to demonstrate that trigonometric functions are continuous over their domains.

Composite Function Theorem

If is continuous at and , then

Before we move on to Example, recall that earlier, in the section on limit laws, we showed . Consequently, we know that is continuous at . In Example we see how to combine this result with the composite function theorem.

Example : Limit of a Composite Cosine Function

Evaluate .

Solution

The given function is a composite of and . Since and is continuous at , we may apply the composite function theorem. Thus,

Exercise :

Evaluate .

Hint

is continuous at . Use Example as a guide.

Answer

The proof of the next theorem uses the composite function theorem as well as the continuity of and at the point to show that trigonometric functions are continuous over their entire domains.

Continuity of Trigonometric Functions

Trigonometric functions are continuous over their entire domains.

Proof

We begin by demonstrating that is continuous at every real number. To do this, we must show that for all values of .

Erroneous nesting of equation structures\displaystyle \begin{align*} \lim_{x→a}\cos x &=\lim_{x→a}\cos((x−a)+a) & & \text{Rewrite }x=x−a+a.\\[4pt] &= \lim_{x→a}(\cos(x−a)\cos a−\sin(x−a)\sin a) & & \text{Apply the identity for the cosine of the sum of two angles.}\\[4pt] &= \cos(\lim_{x→a}(x−a))\cos a−\sin(\lim_{x→a}(x−a))\sin a & &\text{Since }\lim_{x→a}(x−a)=0,\text{ and }\sin x\text{ and }\cos x\text{ are continuous at }0.\\[4pt] &= \cos(0)\cos a−\sin(0)\sin a & & \text{Evaluate }\cos(0)\text{ and }\sin(0)\text{ and simplify.}\\[4pt] &= 1⋅\cos a−0⋅\sin a=\cos a. \end{align*}

The proof that is continuous at every real number is analogous. Because the remaining trigonometric functions may be expressed in terms of and , their continuity follows from the quotient limit law.

□

As you can see, the composite function theorem is invaluable in demonstrating the continuity of trigonometric functions. As we continue our study of calculus, we revisit this theorem many times.

The Intermediate Value Theorem

Functions that are continuous over intervals of the form , where a and b are real numbers, exhibit many useful properties. Throughout our study of calculus, we will encounter many powerful theorems concerning such functions. The first of these theorems is the Intermediate Value Theorem.

The Intermediate Value Theorem

Let be continuous over a closed, bounded interval . If is any real number between and , then there is a number c in satisfying in Figure .

A diagram illustrating the intermediate value theorem. There is a generic continuous curved function shown over the interval [a,b]. The points fa. and fb. are marked, and dotted lines are drawn from a, b, fa., and fb. to the points (a, fa.) and (b, fb.). A third point, c, is plotted between a and b. Since the function is continuous, there is a value for fc. along the curve, and a line is drawn from c to (c, fc.) and from (c, fc.) to fc., which is labeled as z on the y axis.
Figure : There is a number that satisfies .

Example : Application of the Intermediate Value Theorem

Show that has at least one zero.

Solution

Since is continuous over , it is continuous over any closed interval of the form . If you can find an interval such that and have opposite signs, you can use the Intermediate Value Theorem to conclude there must be a real number in that satisfies . Note that

and

.

Using the Intermediate Value Theorem, we can see that there must be a real number in that satisfies . Therefore, has at least one zero.

Example : When Can You Apply the Intermediate Value Theorem?

If is continuous over and , can we use the Intermediate Value Theorem to conclude that has no zeros in the interval ? Explain.

Solution

No. The Intermediate Value Theorem only allows us to conclude that we can find a value between and ; it doesn’t allow us to conclude that we can’t find other values. To see this more clearly, consider the function . It satisfies , and .

Example : When Can You Apply the Intermediate Value Theorem?

For and . Can we conclude that has a zero in the interval ?

Solution

No. The function is not continuous over . The Intermediate Value Theorem does not apply here.

Exercise

Show that has a zero over the interval .

Hint

Find and . Apply the Intermediate Value Theorem.

Answer

is continuous over . It must have a zero on this interval.

Key Concepts

  • For a function to be continuous at a point, it must be defined at that point, its limit must exist at the point, and the value of the function at that point must equal the value of the limit at that point.
  • Discontinuities may be classified as removable, jump, or infinite.
  • A function is continuous over an open interval if it is continuous at every point in the interval. It is continuous over a closed interval if it is continuous at every point in its interior and is continuous at its endpoints.
  • The composite function theorem states: If is continuous at L and , then .
  • The Intermediate Value Theorem guarantees that if a function is continuous over a closed interval, then the function takes on every value between the values at its endpoints.

Glossary

continuity at a point
A function is continuous at a point a if and only if the following three conditions are satisfied: (1) is defined, (2) exists, and (3)
continuity from the left
A function is continuous from the left at b if
continuity from the right
A function is continuous from the right at a if
continuity over an interval
a function that can be traced with a pencil without lifting the pencil; a function is continuous over an open interval if it is continuous at every point in the interval; a function is continuous over a closed interval of the form [] if it is continuous at every point in (), and it is continuous from the right at and from the left at
discontinuity at a point
A function is discontinuous at a point or has a discontinuity at a point if it is not continuous at the point
infinite discontinuity
An infinite discontinuity occurs at a point if or
Intermediate Value Theorem
Let be continuous over a closed bounded interval [] if is any real number between and , then there is a number c in [] satisfying
jump discontinuity
A jump discontinuity occurs at a point if and both exist, but
removable discontinuity
A removable discontinuity occurs at a point if is discontinuous at , but exists

Contributors and Attributions

  • Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.

This page titled 2.5: Continuity is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform.

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