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Mathematics LibreTexts

14.2: Limits and Continuity

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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 Section 2.5 that the definition of a limit of a function of one variable:

Let f(x) be defined for all xa in an open interval containing a. Let L be a real number. Then

limxaf(x)=L

if for every ε>0, there exists a δ>0, such that if 0<|xa|<δ for all x in the domain of f, then

|f(x)L|<ε.

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.

Definition: δ Disks

Consider a point (a,b)R2. A δ disk centered at point (a,b) is defined to be an open disk of radius δ centered at point (a,b) —that is,

{(x,y)R2(xa)2+(yb)2<δ2}

as shown in Figure 14.2.1.

On the xy plane, the point (2, 1) is shown, which is the center of a circle of radius δ.
Figure 14.2.1: A δ disk centered around the point (2,1).

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

aδ<x<a+δ.

In more than one dimension, we use a δ disk.

Definition: limit of a function of two variables

Let f be a function of two variables, x and y. The limit of f(x,y) as (x,y) approaches (a,b) is L, written

lim(x,y)(a,b)f(x,y)=L

if for each ε>0 there exists a small enough δ>0 such that for all points (x,y) in a δ disk around (a,b), except possibly for (a,b) itself, the value of f(x,y) is no more than ε away from L (Figure 14.2.2).

Using symbols, we write the following: For any ε>0, there exists a number δ>0 such that

|f(x,y)L|<ε

whenever

0<(xa)2+(yb)2<δ.

In xyz space, a function is drawn with point L. This point L is the center of a circle of radius ॉ, with points L ± ॉ marked. On the xy plane, there is a point (a, b) drawn with a circle of radius δ around it. This is denoted the δ-disk. There are dashed lines up from the δ-disk to make a disk on the function, which is called the image of delta disk. Then there are dashed lines from this disk to the circle around the point L, which is called the ॉ-neighborhood of L.
Figure 14.2.2: The limit of a function involving two variables requires that f(x,y) be within ε of L whenever (x,y) is within δ of (a,b). The smaller the value of ε, the smaller the value of δ.

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 section titled The Limit Laws in single-variable calculus.

Theorem 14.2.1: Limit Laws for Functions of Two Variables

Let f(x,y) and g(x,y) be defined for all (x,y)(a,b) in a neighborhood around (a,b), and assume the neighborhood is contained completely inside the domain of f. Assume that L and M are real numbers such that

lim(x,y)(a,b)f(x,y)=L

and

lim(x,y)(a,b)g(x,y)=M,

and let c be a constant. Then each of the following statements holds:

Constant Law:

lim(x,y)(a,b)c=c

Identity Laws:

lim(x,y)(a,b)x=a

lim(x,y)(a,b)y=b

Sum Law:

lim(x,y)(a,b)(f(x,y)+g(x,y))=L+M

Difference Law:

lim(x,y)(a,b)(f(x,y)g(x,y))=LM

Constant Multiple Law:

lim(x,y)(a,b)(cf(x,y))=cL

Product Law:

lim(x,y)(a,b)(f(x,y)g(x,y))=LM

Quotient Law:

lim(x,y)(a,b)f(x,y)g(x,y)=LM for M0

Power Law:

lim(x,y)(a,b)(f(x,y))n=Ln

for any positive integer n.

Root Law:

lim(x,y)(a,b)nf(x,y)=nL

for all L if n is odd and positive, and for L0 if n 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.

Example 14.2.1: Finding the Limit of a Function of Two Variables

Find each of the following limits:

  1. lim(x,y)(2,1)(x22xy+3y24x+3y6)
  2. lim(x,y)(2,1)2x+3y4x3y
Solution

a. First use the sum and difference laws to separate the terms:

lim(x,y)(2,1)(x22xy+3y24x+3y6)=(lim(x,y)(2,1)x2)(lim(x,y)(2,1)2xy)+(lim(x,y)(2,1)3y2)(lim(x,y)(2,1)4x)+(lim(x,y)(2,1)3y)(lim(x,y)(2,1)6).

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

=(lim(x,y)(2,1)x2)2(lim(x,y)(2,1)xy)+3(lim(x,y)(2,1)y2)4(lim(x,y)(2,1)x)+3(lim(x,y)(2,1)y)lim(x,y)(2,1)6.

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

(lim(x,y)(2,1)x)22(lim(x,y)(2,1)x)(lim(x,y)(2,1)y)+3(lim(x,y)(2,1)y)24(lim(x,y)(2,1)x)+3(lim(x,y)(2,1)y)lim(x,y)(2,1)6.

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

lim(x,y)(2,1)(x22xy+3y24x+3y6)=(2)22(2)(1)+3(1)24(2)+3(1)6=6.

b. 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,

lim(x,y)(2,1)(4x3y)=lim(x,y)(2,1)4xlim(x,y)(2,1)3y=4(lim(x,y)(2,1)x)3(lim(x,y)(2,1)y)=4(2)3(1)=11.

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:

lim(x,y)(2,1)(2x+3y)=lim(x,y)(2,1)2x+lim(x,y)(2,1)3y=2(lim(x,y)(2,1)x)+3(lim(x,y)(2,1)y)=2(2)+3(1)=1.

Therefore, according to the quotient law we have

lim(x,y)(2,1)2x+3y4x3y=lim(x,y)(2,1)(2x+3y)lim(x,y)(2,1)(4x3y)=111.

Exercise 14.2.1:

Evaluate the following limit:

lim(x,y)(5,2)3x2yy2+x1.

Hint

Use the limit laws.

Answer

lim(x,y)(5,2)3x2yy2+x1=32

Since we are taking the limit of a function of two variables, the point (a,b) is in R2, 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 (a,b). If this is the case, then the limit fails to exist. In other words, the limit must be unique, regardless of path taken.

Example 14.2.2: Limits That Fail to Exist

Show that neither of the following limits exist:

  1. lim(x,y)(0,0)2xy3x2+y2
  2. lim(x,y)(0,0)4xy2x2+3y4
Solution

a. The domain of the function f(x,y)=2xy3x2+y2 consists of all points in the xy-plane except for the point (0,0) (Figure 14.2.3). To show that the limit does not exist as (x,y) approaches (0,0), 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 (0,0). First, consider the line y=0 in the xy-plane. Substituting y=0 into f(x,y) gives

f(x,0)=2x(0)3x2+02=0

for any value of x. Therefore the value of f remains constant for any point on the x-axis, and as y approaches zero, the function remains fixed at zero.

Next, consider the line y=x. Substituting y=x into f(x,y) gives

f(x,x)=2x(x)3x2+x2=2x24x2=12.

This is true for any point on the line y=x. If we let x approach zero while staying on this line, the value of the function remains fixed at 12, regardless of how small x is.

Choose a value for ε that is less than 1/2—say, 1/4. Then, no matter how small a δ disk we draw around (0,0), the values of f(x,y) for points inside that δ disk will include both 0 and 12. Therefore, the definition of limit at a point is never satisfied and the limit fails to exist.

Figure 14.2.3: Graph of the function f(x,y)=2xy3x2+y2. Along the line y=0, the function is equal to zero; along the line y=x, the function is equal to 12.

b. In a similar fashion to a., we can approach the origin along any straight line passing through the origin. If we try the x-axis (i.e., y=0), then the function remains fixed at zero. The same is true for the y-axis. Suppose we approach the origin along a straight line of slope k. The equation of this line is y=kx. Then the limit becomes

lim(x,y)(0,0)4xy2x2+3y4=lim(x,y)(0,0)4x(kx)2x2+3(kx)4=lim(x,y)(0,0)4k2x3x2+3k4x4=lim(x,y)(0,0)4k2x1+3k4x2=lim(x,y)(0,0)(4k2x)lim(x,y)(0,0)(1+3k4x2)=0.

regardless of the value of k. 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 x=y2. Substituting y2 in place of x in f(x,y) gives

lim(x,y)(0,0)4xy2x2+3y4=lim(x,y)(0,0)4(y2)y2(y2)2+3y4=lim(x,y)(0,0)4y4y4+3y4=lim(x,y)(0,0)1=1.

By the same logic in part a, it is impossible to find a δ disk around the origin that satisfies the definition of the limit for any value of ε<1. Therefore,

lim(x,y)(0,0)4xy2x2+3y4

does not exist.

Exercise 14.2.2:

Show that

lim(x,y)(2,1)(x2)(y1)(x2)2+(y1)2

does not exist.

Hint

Pick a line with slope k passing through point (2,1).

Answer

If y=k(x2)+1, then lim(x,y)(2,1)(x2)(y1)(x2)2+(y1)2=k1+k2. Since the answer depends on k, the limit fails to exist.

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.

Definition: interior and boundary points

Let S be a subset of R2 (Figure 14.2.4).

  1. A point P0 is called an interior point of S if there is a δ disk centered around P0 contained completely in S.
  2. A point P0 is called a boundary point of S if every δ disk centered around P0 contains points both inside and outside S.
On the xy plane, a closed shape is drawn. There is a point (–1, 1) drawn on the inside of the shape, and there is a point (2, 3) drawn on the boundary. Both of these points are the centers of small circles.
Figure 14.2.4: In the set S shown, (1,1) is an interior point and (2,3) is a boundary point.
Definition: Open and closed sets

Let S be a subset of R2 (Figure 14.2.4).

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

An example of an open set is a δ 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 δ disk but not the other half, then the set is neither open nor closed.

Definition: connected sets and Regions

Let S be a subset of R2 (Figure 14.2.4).

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

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

Definition

Let f be a function of two variables, x and y, and suppose (a,b) is on the boundary of the domain of f. Then, the limit of f(x,y) as (x,y) approaches (a,b) is L, written

lim(x,y)(a,b)f(x,y)=L,

if for any ε>0, there exists a number δ>0 such that for any point (x,y) inside the domain of f and within a suitably small distance positive δ of (a,b), the value of f(x,y) is no more than ε away from L (Figure 14.2.2). Using symbols, we can write: For any ε>0, there exists a number δ>0 such that

|f(x,y)L|<εwhenever0<(xa)2+(yb)2<δ.

Example 14.2.3: Limit of a Function at a Boundary Point

Prove

lim(x,y)(4,3)25x2y2=0.

Solution

The domain of the function f(x,y)=25x2y2 is {(x,y)R2x2+y225}, which is a circle of radius 5 centered at the origin, along with its interior as shown in Figure 14.2.5.

A circle with radius 5 centered at the origin with its interior filled in.
Figure 14.2.5: Domain of the function f(x,y)=25x2y2.

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

lim(x,y)(4,3)25x2y2=lim(x,y)(4,3)(25x2y2)=lim(x,y)(4,3)25lim(x,y)(4,3)x2lim(x,y)(4,3)y2=254232=0

See the following graph.

The upper hemisphere in xyz space with radius 5 and center the origin.
Figure 14.2.6: Graph of the function f(x,y)=25x2y2.
Exercise 14.2.3

Evaluate the following limit:

lim(x,y)(5,2)29x2y2.

Hint

Determine the domain of f(x,y)=29x2y2.

Answer

lim(x,y)(5,2)29x2y2=2952(2)2=29254=0=0

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 f(x) to be continuous at point x=a

  1. f(a) exists.
  2. limxaf(x) exists.
  3. limxaf(x)=f(a).

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

Definition: continuous Functions

A function f(x,y) is continuous at a point (a,b) in its domain if the following conditions are satisfied:

  1. f(a,b) exists.
  2. lim(x,y)(a,b)f(x,y) exists.
  3. lim(x,y)(a,b)f(x,y)=f(a,b).
Example 14.2.4: Demonstrating Continuity for a Function of Two Variables

Show that the function

f(x,y)=3x+2yx+y+1

is continuous at point (5,3).

Solution

There are three conditions to be satisfied, per the definition of continuity. In this example, a=5 and b=3.

1. f(a,b) exists. This is true because the domain of the function f consists of those ordered pairs for which the denominator is nonzero (i.e., x+y+10). Point (5,3) satisfies this condition. Furthermore,

f(a,b)=f(5,3)=3(5)+2(3)5+(3)+1=1562+1=3.

2. lim(x,y)(a,b)f(x,y) exists. This is also true:

lim(x,y)(a,b)f(x,y)=lim(x,y)(5,3)3x+2yx+y+1=lim(x,y)(5,3)(3x+2y)lim(x,y)(5,3)(x+y+1)=15653+1=3.

3. lim(x,y)(a,b)f(x,y)=f(a,b). This is true because we have just shown that both sides of this equation equal three.

Exercise 14.2.4

Show that the function

f(x,y)=262x2y2

is continuous at point (2,3).

Hint

Use the three-part definition of continuity.

Answer
  1. The domain of f contains the ordered pair (2,3) because f(a,b)=f(2,3)=162(2)2(3)2=3
  2. lim(x,y)(a,b)f(x,y)=3
  3. lim(x,y)(a,b)f(x,y)=f(a,b)=3

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 (x0,y0) in its domain if for every ε>0 there exists a δ>0 such that, whenever (xx0)2+(yy0)2<δ it is true, |f(x,y)f(a,b)|<ε. 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:

Theorem 14.2.2: The Sum of Continuous Functions is Continuous

If f(x,y) is continuous at (x0,y0), and g(x,y) is continuous at (x0,y0), then f(x,y)+g(x,y) is continuous at (x0,y0).

Theorem 14.2.3: The Product of Continuous Functions is Continuous

If g(x) is continuous at x0 and h(y) is continuous at y0, then f(x,y)=g(x)h(y) is continuous at (x0,y0).

Theorem 14.2.4: The Composition of Continuous Functions is Continuous

Let g be a function of two variables from a domain DR2 to a range RR. Suppose g is continuous at some point (x0,y0)D and define z0=g(x0,y0). Let f be a function that maps R to R such that z0 is in the domain of f. Last, assume f is continuous at z0. Then fg is continuous at (x0,y0) as shown in Figure 14.2.7.

A shape is shown labeled the domain of g with point (x, y) inside of it. From the domain of g there is an arrow marked g pointing to the range of g, which is a straight line with point z on it. The range of g is also marked the domain of f. Then there is another arrow marked f from this shape to a line marked range of f.
Figure 14.2.7: The composition of two continuous functions is continuous.

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

Example 14.2.5: More Examples of Continuity of a Function of Two Variables

Show that the functions f(x,y)=4x3y2 and g(x,y)=cos(4x3y2) are continuous everywhere.

Solution

The polynomials g(x)=4x3 and h(y)=y2 are continuous at every real number, and therefore by the product of continuous functions theorem, f(x,y)=4x3y2 is continuous at every point (x,y) in the xy-plane. Since f(x,y)=4x3y2 is continuous at every point (x,y) in the xy-plane and g(x)=cosx is continuous at every real number x, the continuity of the composition of functions tells us that g(x,y)=cos(4x3y2) is continuous at every point (x,y) in the xy-plane.

Exercise 14.2.5

Show that the functions f(x,y)=2x2y3+3 and g(x,y)=(2x2y3+3)4 are continuous everywhere.

Hint

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

Answer

The polynomials g(x)=2x2 and h(y)=y3 are continuous at every real number; therefore, by the product of continuous functions theorem, f(x,y)=2x2y3 is continuous at every point (x,y) in the xy-plane. Furthermore, any constant function is continuous everywhere, so g(x,y)=3 is continuous at every point (x,y) in the xy-plane. Therefore, f(x,y)=2x2y3+3 is continuous at every point (x,y) in the xy-plane. Last, h(x)=x4 is continuous at every real number x, so by the continuity of composite functions theorem g(x,y)=(2x2y3+3)4 is continuous at every point (x,y) in the xy-plane.

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 f(x,y,z) that gives the temperature at a physical location (x,y,z) in three dimensions. Or perhaps a function g(x,y,z,t) can indicate air pressure at a location (x,y,z) at time t. How can we take a limit at a point in R3? 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 δ 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.

Definition: δ-balls

Let (x0,y0,z0) be a point in R3. Then, a δ-ball in three dimensions consists of all points in R3 lying at a distance of less than δ from (x0,y0,z0) —that is,

{(x,y,z)R3(xx0)2+(yy0)2+(zz0)2<δ}.

To define a δ-ball in higher dimensions, add additional terms under the radical to correspond to each additional dimension. For example, given a point P=(w0,x0,y0,z0) in R4, a δ ball around P can be described by

{(w,x,y,z)R4(ww0)2+(xx0)2+(yy0)2+(zz0)2<δ}.

To show that a limit of a function of three variables exists at a point (x0,y0,z0), it suffices to show that for any point in a δ ball centered at (x0,y0,z0), 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.

Example 14.2.6: Finding the Limit of a Function of Three Variables

Find

lim(x,y,z)(4,1,3)x2y3z2x+5yz.

Solution

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,

lim(x,y,z)(4,1,3)(2x+5yz)=2(lim(x,y,z)(4,1,3)x)+5(lim(x,y,z)(4,1,3)y)(lim(x,y,z)(4,1,3)z)=2(4)+5(1)(3)=16.

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

lim(x,y,z)(4,1,3)(x2y3z)=(lim(x,y,z)(4,1,3)x)2(lim(x,y,z)(4,1,3)y)3lim(x,y,z)(4,1,3)z=(42)(1)3(3)=16+9=25

Last, applying the quotient law:

lim(x,y,z)(4,1,3)x2y3z2x+5yz=lim(x,y,z)(4,1,3)(x2y3z)lim(x,y,z)(4,1,3)(2x+5yz)=2516

Exercise 14.2.6

Find

lim(x,y,z)(4,1,3)13x22y2+z2

Hint

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

Answer

lim(x,y,z)(4,1,3)13x22y2+z2=2

Key Concepts

  • To study limits and continuity for functions of two variables, we use a δ disk centered around a given point.
  • A function of several variables has a limit if for any point in a δ ball centered at a point P, 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.

Glossary

boundary point
a point P0 of R is a boundary point if every δ disk centered around P0 contains points both inside and outside R
closed set
a set S that contains all its boundary points
connected set
an open set S that cannot be represented as the union of two or more disjoint, nonempty open subsets
δ disk
an open disk of radius δ centered at point (a,b)
δ ball
all points in R3 lying at a distance of less than δ from (x0,y0,z0)
interior point
a point P0 of R is a boundary point if there is a δ disk centered around P0 contained completely in R
open set
a set S that contains none of its boundary points
region
an open, connected, nonempty subset of R2

14.2: Limits and Continuity is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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