Differentiation of Functions of Several Variables

# 22 Limits and Continuity

### Learning Objectives

- Calculate the limit of a function of two variables.
- Learn how a function of two variables can approach different values at a boundary point, depending on the path of approach.
- State the conditions for continuity of a function of two variables.
- Verify the continuity of a function of two variables at a point.
- Calculate the limit of a function of three or more variables and verify the continuity of the function at a point.

We have now examined functions of more than one variable and seen how to graph them. In this section, we see how to take the limit of a function of more than one variable, and what it means for a function of more than one variable to be continuous at a point in its domain. It turns out these concepts have aspects that just don’t occur with functions of one variable.

### Limit of a Function of Two Variables

Recall from The Limit of a Function the definition of a limit of a function of one variable:

Let be defined for all in an open interval containing Let be a real number. Then

if for every there exists a such that if for all in the domain of then

Before we can adapt this definition to define a limit of a function of two variables, we first need to see how to extend the idea of an open interval in one variable to an open interval in two variables.

Consider a point A disk centered at point is defined to be an open disk of radius centered at point —that is,

as shown in the following graph.

The idea of a disk appears in the definition of the limit of a function of two variables. If is small, then all the points in the disk are close to This is completely analogous to being close to in the definition of a limit of a function of one variable. In one dimension, we express this restriction as

In more than one dimension, we use a disk.

Let be a function of two variables, and The limit of as approaches is written

if for each there exists a small enough such that for all points in a disk around except possibly for itself, the value of is no more than away from ((Figure)). Using symbols, we write the following: For any there exists a number such that

Proving that a limit exists using the definition of a limit of a function of two variables can be challenging. Instead, we use the following theorem, which gives us shortcuts to finding limits. The formulas in this theorem are an extension of the formulas in the limit laws theorem in The Limit Laws.

Let and be defined for all in a neighborhood around and assume the neighborhood is contained completely inside the domain of Assume that and are real numbers such that and and let be a constant. Then each of the following statements holds:

**Constant Law:**

**Identity Laws:**

**Sum Law:**

**Difference Law:**

**Constant Multiple Law:**

**Product Law:**

**Quotient Law:**

**Power Law:**

for any positive integer

**Root Law:**

for all if is odd and positive, and for if is even and positive.

The proofs of these properties are similar to those for the limits of functions of one variable. We can apply these laws to finding limits of various functions.

Find each of the following limits:

- First use the sum and difference laws to separate the terms:

*** QuickLaTeX cannot compile formula: \begin{array}{}\\ \underset{\left(x,y\right)\to \left(2,-1\right)}{\text{lim}}\left({x}^{2}-2xy+3{y}^{2}-4x+3y-6\right)\hfill \\ =\left(\underset{\left(x,y\right)\to \left(2,-1\right)}{\text{lim}}{x}^{2}\right)-\left(\underset{\left(x,y\right)\to \left(2,-1\right)}{\text{lim}}2xy\right)+\left(\underset{\left(x,y\right)\to \left(2,-1\right)}{\text{lim}}3{y}^{2}\right)-\left(\underset{\left(x,y\right)\to \left(2,-1\right)}{\text{lim}}4x\right)\hfill \\ \phantom{\rule{0.8em}{0ex}}+\left(\underset{\left(x,y\right)\to \left(2,-1\right)}{\text{lim}}3y\right)-\left(\underset{\left(x,y\right)\to \left(2,-1\right)}{\text{lim}}6\right).\hfill \end{array} *** Error message: Missing # inserted in alignment preamble. leading text: $\begin{array}{} Missing $ inserted. leading text: ...,y\right)\to \left(2,-1\right)}{\text{lim}} Extra }, or forgotten $. leading text: ...,y\right)\to \left(2,-1\right)}{\text{lim}} Missing } inserted. leading text: ...^{2}-2xy+3{y}^{2}-4x+3y-6\right)\hfill \\ = Extra }, or forgotten $. leading text: ...^{2}-2xy+3{y}^{2}-4x+3y-6\right)\hfill \\ = Missing } inserted. leading text: ...^{2}-2xy+3{y}^{2}-4x+3y-6\right)\hfill \\ = Extra }, or forgotten $. leading text: ...^{2}-2xy+3{y}^{2}-4x+3y-6\right)\hfill \\ = Missing } inserted. leading text: ...^{2}-2xy+3{y}^{2}-4x+3y-6\right)\hfill \\ = Extra }, or forgotten $. leading text: ...^{2}-2xy+3{y}^{2}-4x+3y-6\right)\hfill \\ = Missing } inserted.

Next, use the constant multiple law on the second, third, fourth, and fifth limits:

Now, use the power law on the first and third limits, and the product law on the second limit:

*** QuickLaTeX cannot compile formula: \begin{array}{}\\ \\ \\ ={\left(\underset{\left(x,y\right)\to \left(2,-1\right)}{\text{lim}}x\right)}^{2}-2\left(\underset{\left(x,y\right)\to \left(2,-1\right)}{\text{lim}}x\right)\left(\underset{\left(x,y\right)\to \left(2,-1\right)}{\text{lim}}y\right)+3{\left(\underset{\left(x,y\right)\to \left(2,-1\right)}{\text{lim}}y\right)}^{2}\hfill \\ \phantom{\rule{0.5em}{0ex}}-4\left(\underset{\left(x,y\right)\to \left(2,-1\right)}{\text{lim}}x\right)+3\left(\underset{\left(x,y\right)\to \left(2,-1\right)}{\text{lim}}y\right)-\underset{\left(x,y\right)\to \left(2,-1\right)}{\text{lim}}6.\hfill \end{array} *** Error message: Missing # inserted in alignment preamble. leading text: $\begin{array}{} Missing $ inserted. leading text: $\begin{array}{}\\ \\ \\ ={\left Extra }, or forgotten $. leading text: ...\to \left(2,-1\right)}{\text{lim}}x\right)} Missing } inserted. leading text: ...{\text{lim}}y\right)}^{2}\hfill \\ \phantom Extra }, or forgotten $. leading text: ...{\text{lim}}y\right)}^{2}\hfill \\ \phantom Missing } inserted. leading text: ...{\text{lim}}y\right)}^{2}\hfill \\ \phantom Extra }, or forgotten $. leading text: ...{\text{lim}}y\right)}^{2}\hfill \\ \phantom Missing } inserted. leading text: ...{\text{lim}}y\right)}^{2}\hfill \\ \phantom Extra }, or forgotten $. leading text: ...{\text{lim}}y\right)}^{2}\hfill \\ \phantom Missing } inserted. leading text: ...{\text{lim}}y\right)}^{2}\hfill \\ \phantom

Last, use the identity laws on the first six limits and the constant law on the last limit:

- Before applying the quotient law, we need to verify that the limit of the denominator is nonzero. Using the difference law, constant multiple law, and identity law,

Since the limit of the denominator is nonzero, the quotient law applies. We now calculate the limit of the numerator using the difference law, constant multiple law, and identity law:

Therefore, according to the quotient law we have

Evaluate the following limit:

Use the limit laws.

Since we are taking the limit of a function of two variables, the point is in and it is possible to approach this point from an infinite number of directions. Sometimes when calculating a limit, the answer varies depending on the path taken toward If this is the case, then the limit fails to exist. In other words, the limit must be unique, regardless of path taken.

Show that neither of the following limits exist:

- The domain of the function consists of all points in the except for the point ((Figure)). To show that the limit does not exist as approaches we note that it is impossible to satisfy the definition of a limit of a function of two variables because of the fact that the function takes different values along different lines passing through point First, consider the line in the Substituting into gives

for any value of Therefore the value of remains constant for any point on the and as approaches zero, the function remains fixed at zero.

Next, consider the line Substituting into gives

This is true for any point on the line If we let approach zero while staying on this line, the value of the function remains fixed at regardless of how small is.

Choose a value for that is less than —say, Then, no matter how small a disk we draw around the values of for points inside that disk will include both and Therefore, the definition of limit at a point is never satisfied and the limit fails to exist.

In a similar fashion to a., we can approach the origin along any straight line passing through the origin. If we try the (i.e., then the function remains fixed at zero. The same is true for the Suppose we approach the origin along a straight line of slope The equation of this line is Then the limit becomes

regardless of the value of It would seem that the limit is equal to zero. What if we chose a curve passing through the origin instead? For example, we can consider the parabola given by the equation Substituting in place of in gives

By the same logic in a., it is impossible to find a disk around the origin that satisfies the definition of the limit for any value of Therefore, does not exist.

Show that

does not exist.

If then Since the answer depends on the limit fails to exist.

Pick a line with slope passing through point

### Interior Points and Boundary Points

To study continuity and differentiability of a function of two or more variables, we first need to learn some new terminology.

Let *S* be a subset of ((Figure)).

- A point is called an interior point of if there is a disk centered around contained completely in
- A point is called a boundary point of if every disk centered around contains points both inside and outside

Let *S* be a subset of ((Figure)).

- is called an open set if every point of is an interior point.
- is called a closed set if it contains all its boundary points.

An example of an open set is a disk. If we include the boundary of the disk, then it becomes a closed set. A set that contains some, but not all, of its boundary points is neither open nor closed. For example if we include half the boundary of a disk but not the other half, then the set is neither open nor closed.

Let *S* be a subset of ((Figure)).

- An open set is a connected set if it cannot be represented as the union of two or more disjoint, nonempty open subsets.
- A set is a region if it is open, connected, and nonempty.

The definition of a limit of a function of two variables requires the disk to be contained inside the domain of the function. However, if we wish to find the limit of a function at a boundary point of the domain, the is not contained inside the domain. By definition, some of the points of the are inside the domain and some are outside. Therefore, we need only consider points that are inside both the disk and the domain of the function. This leads to the definition of the limit of a function at a boundary point.

Let be a function of two variables, and and suppose is on the boundary of the domain of Then, the limit of as approaches is written

if for any there exists a number such that for any point inside the domain of and within a suitably small distance positive of the value of is no more than away from ((Figure)). Using symbols, we can write: For any there exists a number such that

Prove

The domain of the function is which is a circle of radius centered at the origin, along with its interior as shown in the following graph.

We can use the limit laws, which apply to limits at the boundary of domains as well as interior points:

See the following graph.

Evaluate the following limit:

Determine the domain of

### Continuity of Functions of Two Variables

In Continuity, we defined the continuity of a function of one variable and saw how it relied on the limit of a function of one variable. In particular, three conditions are necessary for to be continuous at point

- exists.
- exists.

These three conditions are necessary for continuity of a function of two variables as well.

A function is continuous at a point in its domain if the following conditions are satisfied:

- exists.
- exists.

Show that the function is continuous at point

There are three conditions to be satisfied, per the definition of continuity. In this example, and

- exists. This is true because the domain of the function consists of those ordered pairs for which the denominator is nonzero (i.e., Point satisfies this condition. Furthermore,

- exists. This is also true:

- This is true because we have just shown that both sides of this equation equal three.

Show that the function is continuous at point

- The domain of contains the ordered pair because

Use the three-part definition of continuity.

Continuity of a function of any number of variables can also be defined in terms of delta and epsilon. A function of two variables is continuous at a point in its domain if for every there exists a such that, whenever it is true, This definition can be combined with the formal definition (that is, the *epsilon–delta definition*) of continuity of a function of one variable to prove the following theorems:

If is continuous at and is continuous at then is continuous at

If is continuous at and is continuous at then is continuous at

Let be a function of two variables from a domain to a range Suppose is continuous at some point and define Let be a function that maps to such that is in the domain of Last, assume is continuous at Then is continuous at as shown in the following figure.

Let’s now use the previous theorems to show continuity of functions in the following examples.

Show that the functions and are continuous everywhere.

The polynomials and are continuous at every real number, and therefore by the product of continuous functions theorem, is continuous at every point in the Since is continuous at every point in the and is continuous at every real number the continuity of the composition of functions tells us that is continuous at every point in the

Show that the functions and are continuous everywhere.

The polynomials and are continuous at every real number; therefore, by the product of continuous functions theorem, is continuous at every point in the Furthermore, any constant function is continuous everywhere, so is continuous at every point in the Therefore, is continuous at every point in the Last, is continuous at every real number so by the continuity of composite functions theorem is continuous at every point in the

Use the continuity of the sum, product, and composition of two functions.

### Functions of Three or More Variables

The limit of a function of three or more variables occurs readily in applications. For example, suppose we have a function that gives the temperature at a physical location in three dimensions. Or perhaps a function can indicate air pressure at a location at time How can we take a limit at a point in What does it mean to be continuous at a point in four dimensions?

The answers to these questions rely on extending the concept of a disk into more than two dimensions. Then, the ideas of the limit of a function of three or more variables and the continuity of a function of three or more variables are very similar to the definitions given earlier for a function of two variables.

Let be a point in Then, a ball in three dimensions consists of all points in lying at a distance of less than from —that is,

To define a ball in higher dimensions, add additional terms under the radical to correspond to each additional dimension. For example, given a point in a ball around can be described by

To show that a limit of a function of three variables exists at a point it suffices to show that for any point in a ball centered at the value of the function at that point is arbitrarily close to a fixed value (the limit value). All the limit laws for functions of two variables hold for functions of more than two variables as well.

Find

Before we can apply the quotient law, we need to verify that the limit of the denominator is nonzero. Using the difference law, the identity law, and the constant law,

Since this is nonzero, we next find the limit of the numerator. Using the product law, difference law, constant multiple law, and identity law,

Last, applying the quotient law:

Find

Use the limit laws and the continuity of the composition of functions.

### Key Concepts

- To study limits and continuity for functions of two variables, we use a disk centered around a given point.
- A function of several variables has a limit if for any point in a ball centered at a point the value of the function at that point is arbitrarily close to a fixed value (the limit value).
- The limit laws established for a function of one variable have natural extensions to functions of more than one variable.
- A function of two variables is continuous at a point if the limit exists at that point, the function exists at that point, and the limit and function are equal at that point.

For the following exercises, find the limit of the function.

2.0

Show that the limit exists and is the same along the paths: and and along

For the following exercises, evaluate the limits at the indicated values of If the limit does not exist, state this and explain why the limit does not exist.

The limit does not exist because when and both approach zero, the function approaches which is undefined (approaches negative infinity).

For the following exercises, complete the statement.

A point in a plane region is an interior point of if _________________.

A point in a plane region is called a boundary point of if ___________.

every open disk centered at contains points inside and outside

For the following exercises, use algebraic techniques to evaluate the limit.

For the following exercises, evaluate the limits of the functions of three variables.

The limit does not exist.

For the following exercises, evaluate the limit of the function by determining the value the function approaches along the indicated paths. If the limit does not exist, explain why not.

- Along the
- Along the
- Along the path

Evaluate using the results of previous problem.

The limit does not exist. The function approaches two different values along different paths.

- Along the
*x*-axis - Along the
*y*-axis - Along the path

Evaluate using the results of previous problem.

The limit does not exist because the function approaches two different values along the paths.

Discuss the continuity of the following functions. Find the largest region in the in which the following functions are continuous.

The function is continuous in the region

The function is continuous at all points in the except at

For the following exercises, determine the region in which the function is continuous. Explain your answer.

(*Hint*: Show that the function approaches different values along two different paths.)

The function is continuous at since the limit of the function at is the same value of

Determine whether is continuous at

The function is discontinuous at The limit at fails to exist and does not exist.

Create a plot using graphing software to determine where the limit does not exist. Determine the region of the coordinate plane in which is continuous.

Determine the region of the in which the composite function is continuous. Use technology to support your conclusion.

Since the function is continuous over is continuous where is continuous. The inner function is continuous on all points of the except where Thus, is continuous on all points of the coordinate plane *except* at points at which

Determine the region of the in which is continuous. Use technology to support your conclusion. (*Hint*: Choose the range of values for carefully!)

At what points in space is continuous?

All points in space

At what points in space is continuous?

Show that does not exist at by plotting the graph of the function.

The graph increases without bound as both approach zero.

**[T]** Evaluate by plotting the function using a CAS. Determine analytically the limit along the path

**[T]**

- Use a CAS to draw a contour map of
- What is the name of the geometric shape of the level curves?
- Give the general equation of the level curves.
- What is the maximum value of
- What is the domain of the function?
- What is the range of the function?

a.

b. The level curves are circles centered at with radius c. d. e. f.

*True or False*: If we evaluate along several paths and each time the limit is we can conclude that

Use polar coordinates to find You can also find the limit using L’Hôpital’s rule.

Use polar coordinates to find

Discuss the continuity of where and

is continuous at all points that are not on the line

Given find

Given find

### Glossary

- boundary point
- a point of is a boundary point if every disk centered around contains points both inside and outside

- closed set
- a set that contains all its boundary points

- connected set
- an open set that cannot be represented as the union of two or more disjoint, nonempty open subsets

- disk
- an open disk of radius centered at point

- ball
- all points in lying at a distance of less than from

- interior point
- a point of is a boundary point if there is a disk centered around contained completely in

- open set
- a set that contains none of its boundary points

- region
- an open, connected, nonempty subset of