2. Limits

# 2.4 Continuity

### 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 conditions:

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,

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:

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

### Definition

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

- is defined
- 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

- Check to see if is defined. If is undefined, we need go no further. The function is not continuous at . If is defined, continue to step 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 and the problem is solved. If exists, then continue to step 3.
- Compare and . If , then the function is not continuous at . If , then the function is continuous at .

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.

### 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).

### 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).

### 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 .

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.

#### Solution

is not continuous at 1 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 (Figure) to determine the points at which a given rational function is continuous.

### Continuity of a Rational Function

For what values of is continuous?

#### Solution

The rational function is continuous for every value of except .

# Types of Discontinuities

As we have seen in (Figure) and (Figure), 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.

These three discontinuities are formally defined as follows:

### Definition

If is discontinuous at , then

- has a
**removable discontinuity**at if exists. (Note: When we state that exists, we mean that , where is a real number.) - has a
**jump discontinuity**at 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 .) - has an
**infinite discontinuity**at if or .

### Classifying a Discontinuity

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

#### Solution

To classify the discontinuity at 2 we must evaluate :

Since is discontinuous at 2 and exists, has a removable discontinuity at .

### Classifying a Discontinuity

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

#### Solution

Earlier, we showed that is discontinuous at 3 because does not exist. However, since and both exist, we conclude that the function has a jump discontinuity at 3.

### Classifying a Discontinuity

Determine whether is continuous at −1. If the function is discontinuous at −1, classify the discontinuity as removable, jump, or infinite.

#### Solution

The function value is undefined. Therefore, the function is not continuous at −1. To determine the type of discontinuity, we must determine the limit at −1. We see that and . Therefore, the function has an infinite discontinuity at −1.

For , decide whether is continuous at 1. If is not continuous at 1, classify the discontinuity as removable, jump, or infinite.

#### Solution

Discontinuous at 1; removable

#### Hint

Follow the steps in (Figure). If the function is discontinuous at 1, look at and use the definition to determine the type of discontinuity.

# 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 if .

A function is said to be** continuous from the left** at 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 and is continuous from the left at . Analogously, a function is continuous over an interval of the form if it is continuous over and is continuous from the left at . 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 .

### 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 .

### 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 in . We also know that exists and exists. Therefore, is continuous over the interval .

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

#### Solution

#### Hint

Use (Figure) as a guide for solving.

The (Figure) 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 (Figure), recall that earlier, in the section on limit laws, we showed . Consequently, we know that is continuous at 0. In (Figure) we see how to combine this result with the composite function theorem.

### Limit of a Composite Cosine Function

Evaluate .

#### Solution

The given function is a composite of and . Since and is continuous at 0, we may apply the composite function theorem. Thus,

The proof of the next theorem uses the composite function theorem as well as the continuity of and at the point 0 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 .

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 and 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 in satisfying . (See (Figure)).

### 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.

### 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 .

### 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.

Show that has a zero over the interval .

#### Solution

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

#### Hint

Find and . Apply the Intermediate Value Theorem.

### 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 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.

For the following exercises, determine the point(s), if any, at which each function is discontinuous. Classify any discontinuity as jump, removable, infinite, or other.

**1. **

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The function is defined for all in the interval .

**2. **

**3. **

#### Solution

Removable discontinuity at ; infinite discontinuity at

**4. **

**5. **

#### Solution

Infinite discontinuity at

**6. **

**7. **

#### Solution

Infinite discontinuities at , for

**8. **

For the following exercises, decide if the function continuous at the given point. If it is discontinuous, what type of discontinuity is it?

**9. ** at

#### Solution

No. It is a removable discontinuity.

**10. ** at

**11. ** at

#### Solution

Yes. It is continuous.

**12. **, at

**13. ** at

#### Solution

Yes. It is continuous.

**14. ** at

In the following exercises, find the value(s) of that makes each function continuous over the given interval.

**15. **

#### Solution

**16. **

**17. **

#### Solution

**18. **

**19. **

#### Solution

In the following exercises, use the Intermediate Value Theorem (IVT).

**20. **Let Over the interval , there is no value of such that , although and . Explain why this does not contradict the IVT.

**21. **A particle moving along a line has at each time a position function , which is continuous. Assume and . Another particle moves such that its position is given by . Explain why there must be a value for such that .

#### Solution

Since both and are continuous everywhere, then is continuous everywhere and, in particular, it is continuous over the closed interval . Also, and . Therefore, by the IVT, there is a value such that .

**22. [T]** Use the statement “The cosine of is equal to cubed.”

- Write the statement as a mathematical equation.
- Prove that the equation in part a. has at least one real solution.
- Use a calculator to find an interval of length 0.01 that contains a solution of the equation.

**23. **Apply the IVT to determine whether has a solution in one of the intervals or . Briefly explain your response for each interval.

#### Solution

The function is continuous over the interval and has opposite signs at the endpoints.

**24. **Consider the graph of the function shown in the following graph.

- Find all values for which the function is discontinuous.
- For each value in part a., use the formal definition of continuity to explain why the function is discontinuous at that value.
- Classify each discontinuity as either jump, removable, or infinite.

**25. **Let

- Sketch the graph of .
- Is it possible to find a value such that , which makes continuous for all real numbers? Briefly explain.

#### Solution

a.

1. It beings with an open circle at (1,3).”>

b. It is not possible to redefine since the discontinuity is a jump discontinuity.

**26. **Let for .

- Sketch the graph of .
- Is it possible to find values and such that and , and that makes continuous for all real numbers? Briefly explain.

**27. **Sketch the graph of the function with properties i. through vii.

- The domain of is .
- has an infinite discontinuity at .
- is left continuous but not right continuous at .
- and

#### Solution

Answers may vary; see the following example:

3.”>

**28. **Sketch the graph of the function with properties i. through iv.

- The domain of is .
- and exist and are equal.
- is left continuous but not continuous at , and right continuous but not continuous at .
- has a removable discontinuity at , a jump discontinuity at , and the following limits hold: and .

In the following exercises, suppose is defined for all . For each description, sketch a graph with the indicated property.

**29. **Discontinuous at with and

#### Solution

Answers may vary; see the following example:

1. It begins at (1,3).”>

**30. **Discontinuous at but continuous elsewhere with

Determine whether each of the given statements is true. Justify your response with an explanation or counterexample.

**31. ** is continuous everywhere.

#### Solution

False. It is continuous over .

**32. **If the left- and right-hand limits of as exist and are equal, then cannot be discontinuous at .

**33. **If a function is not continuous at a point, then it is not defined at that point.

#### Solution

False. Consider

**34. **According to the IVT, has a solution over the interval .

**35. **If is continuous such that and have opposite signs, then has exactly one solution in .

#### Solution

False. Consider on .

**36. **The function is continuous over the interval .

**37. **If is continuous everywhere and , then there is no root of in the interval .

#### Solution

False. The IVT does *not* work in reverse! Consider over the interval .

The following problems consider the scalar form of Coulomb’s law, which describes the electrostatic force between two point charges, such as electrons. It is given by the equation , where is Coulomb’s constant, are the magnitudes of the charges of the two particles, and is the distance between the two particles.

**38. [T]** To simplify the calculation of a model with many interacting particles, after some threshold value , we approximate as zero.

- Explain the physical reasoning behind this assumption.
- What is the force equation?
- Evaluate the force using both Coulomb’s law and our approximation, assuming two protons with a charge magnitude of , and the Coulomb constant are 1 m apart. Also, assume . How much inaccuracy does our approximation generate? Is our approximation reasonable?
- Is there any finite value of for which this system remains continuous at ?

**39. [T] **Instead of making the force 0 at , instead we let the force be for . Assume two protons, which have a magnitude of charge , and the Coulomb constant . Is there a value that can make this system continuous? If so, find it.

#### Solution

Recall the discussion on spacecraft from the chapter opener. The following problems consider a rocket launch from Earth’s surface. The force of gravity on the rocket is given by , where is the mass of the rocket, is the distance of the rocket from the center of Earth, and is a constant.

**40. [T]** Determine the value and units of given that the mass of the rocket on Earth is 3 million kg. (*Hint*: The distance from the center of Earth to its surface is 6378 km.)

**41. [T]** After a certain distance has passed, the gravitational effect of Earth becomes quite negligible, so we can approximate the force function by Find the necessary condition such that the force function remains continuous.

#### Solution

km

**42. **As the rocket travels away from Earth’s surface, there is a distance where the rocket sheds some of its mass, since it no longer needs the excess fuel storage. We can write this function as Is there a value such that this function is continuous, assuming ?

Prove the following functions are continuous everywhere

**43. **

#### Solution

For all values of is defined, exists, and . Therefore, is continuous everywhere.

**44. **

**45. **Where is continuous?

#### Solution

Nowhere

## Glossary

- continuity 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, and (3)

- continuity from the left
- A function is continuous from the left at if

- continuity from the right
- A function is continuous from the right at 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 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

## Hint

Check each condition of the definition.