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Calculus Volume 3

4.2 Limits and Continuity

Calculus Volume 34.2 Limits and Continuity
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  1. Preface
  2. 1 Parametric Equations and Polar Coordinates
    1. Introduction
    2. 1.1 Parametric Equations
    3. 1.2 Calculus of Parametric Curves
    4. 1.3 Polar Coordinates
    5. 1.4 Area and Arc Length in Polar Coordinates
    6. 1.5 Conic Sections
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  3. 2 Vectors in Space
    1. Introduction
    2. 2.1 Vectors in the Plane
    3. 2.2 Vectors in Three Dimensions
    4. 2.3 The Dot Product
    5. 2.4 The Cross Product
    6. 2.5 Equations of Lines and Planes in Space
    7. 2.6 Quadric Surfaces
    8. 2.7 Cylindrical and Spherical Coordinates
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  4. 3 Vector-Valued Functions
    1. Introduction
    2. 3.1 Vector-Valued Functions and Space Curves
    3. 3.2 Calculus of Vector-Valued Functions
    4. 3.3 Arc Length and Curvature
    5. 3.4 Motion in Space
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  5. 4 Differentiation of Functions of Several Variables
    1. Introduction
    2. 4.1 Functions of Several Variables
    3. 4.2 Limits and Continuity
    4. 4.3 Partial Derivatives
    5. 4.4 Tangent Planes and Linear Approximations
    6. 4.5 The Chain Rule
    7. 4.6 Directional Derivatives and the Gradient
    8. 4.7 Maxima/Minima Problems
    9. 4.8 Lagrange Multipliers
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Chapter Review Exercises
  6. 5 Multiple Integration
    1. Introduction
    2. 5.1 Double Integrals over Rectangular Regions
    3. 5.2 Double Integrals over General Regions
    4. 5.3 Double Integrals in Polar Coordinates
    5. 5.4 Triple Integrals
    6. 5.5 Triple Integrals in Cylindrical and Spherical Coordinates
    7. 5.6 Calculating Centers of Mass and Moments of Inertia
    8. 5.7 Change of Variables in Multiple Integrals
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  7. 6 Vector Calculus
    1. Introduction
    2. 6.1 Vector Fields
    3. 6.2 Line Integrals
    4. 6.3 Conservative Vector Fields
    5. 6.4 Green’s Theorem
    6. 6.5 Divergence and Curl
    7. 6.6 Surface Integrals
    8. 6.7 Stokes’ Theorem
    9. 6.8 The Divergence Theorem
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Chapter Review Exercises
  8. 7 Second-Order Differential Equations
    1. Introduction
    2. 7.1 Second-Order Linear Equations
    3. 7.2 Nonhomogeneous Linear Equations
    4. 7.3 Applications
    5. 7.4 Series Solutions of Differential Equations
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  9. A | Table of Integrals
  10. B | Table of Derivatives
  11. C | Review of Pre-Calculus
  12. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
  13. Index

Learning Objectives

  • 4.2.1. Calculate the limit of a function of two variables.
  • 4.2.2. Learn how a function of two variables can approach different values at a boundary point, depending on the path of approach.
  • 4.2.3. State the conditions for continuity of a function of two variables.
  • 4.2.4. Verify the continuity of a function of two variables at a point.
  • 4.2.5. 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 f(x)f(x) be defined for all xaxa in an open interval containing a.a. Let LL be a real number. Then

limxaf(x)=Llimxaf(x)=L

if for every ε>0,ε>0, there exists a δ>0,δ>0, such that if 0<|xa|<δ0<|xa|<δ for all xx in the domain of f,f, then

|f(x)L|<ε.|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

Consider a point (a,b)2.(a,b)2. A δδ disk centered at point (a,b)(a,b) is defined to be an open disk of radius δδ centered at point (a,b)(a,b)—that is,

{(x,y)2|(xa)2+(yb)2<δ2}{(x,y)2|(xa)2+(yb)2<δ2}

as shown in the following graph.

On the xy plane, the point (2, 1) is shown, which is the center of a circle of radius δ.
Figure 4.14 A δδ disk centered around the point (2,1).(2,1).

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

aδ<x<a+δ.aδ<x<a+δ.

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

Definition

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

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

if for each ε>0ε>0 there exists a small enough δ>0δ>0 such that for all points (x,y)(x,y) in a δδ disk around (a,b),(a,b), except possibly for (a,b)(a,b) itself, the value of f(x,y)f(x,y) is no more than εε away from LL (Figure 4.15). Using symbols, we write the following: For any ε>0,ε>0, there exists a number δ>0δ>0 such that

|f(x,y)L|<εwhenever0<(xa)2+(yb)2<δ.|f(x,y)L|<εwhenever0<(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 4.15 The limit of a function involving two variables requires that f(x,y)f(x,y) be within εε of LL whenever (x,y)(x,y) is within δδ of (a,b).(a,b). The smaller the value of ε,ε, the smaller the value of δ.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 Limit Laws.

Theorem 4.1

Limit laws for functions of two variables

Let f(x,y)f(x,y) and g(x,y)g(x,y) be defined for all (x,y)(a,b)(x,y)(a,b) in a neighborhood around (a,b),(a,b), and assume the neighborhood is contained completely inside the domain of f.f. Assume that LL and MM are real numbers such that lim(x,y)(a,b)f(x,y)=Llim(x,y)(a,b)f(x,y)=L and lim(x,y)(a,b)g(x,y)=M,lim(x,y)(a,b)g(x,y)=M, and let cc be a constant. Then each of the following statements holds:

Constant Law:

lim(x,y)(a,b)c=clim(x,y)(a,b)c=c
4.2

Identity Laws:

lim(x,y)(a,b)x=alim(x,y)(a,b)x=a
4.3
lim(x,y)(a,b)y=blim(x,y)(a,b)y=b
4.4

Sum Law:

lim(x,y)(a,b)(f(x,y)+g(x,y))=L+Mlim(x,y)(a,b)(f(x,y)+g(x,y))=L+M
4.5

Difference Law:

lim(x,y)(a,b)(f(x,y)g(x,y))=LMlim(x,y)(a,b)(f(x,y)g(x,y))=LM
4.6

Constant Multiple Law:

lim(x,y)(a,b)(cf(x,y))=cLlim(x,y)(a,b)(cf(x,y))=cL
4.7

Product Law:

lim(x,y)(a,b)(f(x,y)g(x,y))=LMlim(x,y)(a,b)(f(x,y)g(x,y))=LM
4.8

Quotient Law:

lim(x,y)(a,b)f(x,y)g(x,y)=LMforM0lim(x,y)(a,b)f(x,y)g(x,y)=LMforM0
4.9

Power Law:

lim(x,y)(a,b)(f(x,y))n=Lnlim(x,y)(a,b)(f(x,y))n=Ln
4.10

for any positive integer n.n.

Root Law:

lim(x,y)(a,b)f(x,y)n=Lnlim(x,y)(a,b)f(x,y)n=Ln
4.11

for all LL if nn is odd and positive, and for L0L0 if nn 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 4.8

Finding the Limit of a Function of Two Variables

Find each of the following limits:

  1. lim(x,y)(2,−1)(x22xy+3y24x+3y6)lim(x,y)(2,−1)(x22xy+3y24x+3y6)
  2. lim(x,y)(2,−1)2x+3y4x3ylim(x,y)(2,−1)2x+3y4x3y

Solution

  1. 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).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.=(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.=(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.lim(x,y)(2,−1)(x22xy+3y24x+3y6)=(2)22(2)(−1)+3(−1)24(2)+3(−1)6=−6.
  2. 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.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.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.lim(x,y)(2,−1)2x+3y4x3y=lim(x,y)(2,−1)(2x+3y)lim(x,y)(2,−1)(4x3y)=111.
Checkpoint 4.6

Evaluate the following limit:

lim(x,y)(5,−2)x2yy2+x13.lim(x,y)(5,−2)x2yy2+x13.

Since we are taking the limit of a function of two variables, the point (a,b)(a,b) is in 2,2, 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).(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 4.9

Limits That Fail to Exist

Show that neither of the following limits exist:

  1. lim(x,y)(0,0)2xy3x2+y2lim(x,y)(0,0)2xy3x2+y2
  2. lim(x,y)(0,0)4xy2x2+3y4lim(x,y)(0,0)4xy2x2+3y4

Solution

  1. The domain of the function f(x,y)=2xy3x2+y2f(x,y)=2xy3x2+y2 consists of all points in the xy-planexy-plane except for the point (0,0)(0,0) (Figure 4.16). To show that the limit does not exist as (x,y)(x,y) approaches (0,0),(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).(0,0). First, consider the line y=0y=0 in the xy-plane.xy-plane. Substituting y=0y=0 into f(x,y)f(x,y) gives
    f(x,0)=2x(0)3x2+02=0f(x,0)=2x(0)3x2+02=0

    for any value of x.x. Therefore the value of ff remains constant for any point on the x-axis,x-axis, and as yy approaches zero, the function remains fixed at zero.
    Next, consider the line y=x.y=x. Substituting y=xy=x into f(x,y)f(x,y) gives
    f(x,x)=2x(x)3x2+x2=2x24x2=12.f(x,x)=2x(x)3x2+x2=2x24x2=12.

    This is true for any point on the line y=x.y=x. If we let xx approach zero while staying on this line, the value of the function remains fixed at 12,12, regardless of how small xx is.
    Choose a value for εε that is less than 1/21/2—say, 1/4.1/4. Then, no matter how small a δδ disk we draw around (0,0),(0,0), the values of f(x,y)f(x,y) for points inside that δδ disk will include both 00 and 12.12. Therefore, the definition of limit at a point is never satisfied and the limit fails to exist.
    In xyz space, the function f(x, y) = 2xy/(3x2 + y2) is shown, which is a slightly twisted plane, with values of 0 along the line y = 0 and values of ½ along the line y = x.
    Figure 4.16 Graph of the function f(x,y)=(2xy)/(3x2+y2).f(x,y)=(2xy)/(3x2+y2). Along the line y=0,y=0, the function is equal to zero; along the line y=x,y=x, the function is equal to 12.12.

    In a similar fashion to a., we can approach the origin along any straight line passing through the origin. If we try the x-axisx-axis (i.e., y=0),y=0), then the function remains fixed at zero. The same is true for the y-axis.y-axis. Suppose we approach the origin along a straight line of slope k.k. The equation of this line is y=kx.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)=0lim(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.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.x=y2. Substituting y2y2 in place of xx in f(x,y)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.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 a., it is impossible to find a δδ disk around the origin that satisfies the definition of the limit for any value of ε<1.ε<1. Therefore, lim(x,y)(0,0)4xy2x2+3y4lim(x,y)(0,0)4xy2x2+3y4 does not exist.
Checkpoint 4.7

Show that

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

does not 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

Let S be a subset of 22 (Figure 4.17).

  1. A point P0P0 is called an interior point of SS if there is a δδ disk centered around P0P0 contained completely in S.S.
  2. A point P0P0 is called a boundary point of SS if every δδ disk centered around P0P0 contains points both inside and outside S.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 4.17 In the set SS shown, (−1,1)(−1,1) is an interior point and (2,3)(2,3) is a boundary point.

Definition

Let S be a subset of 22 (Figure 4.17).

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

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

Definition

Let S be a subset of 22 (Figure 4.17).

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

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

Definition

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

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

if for any ε>0,ε>0, there exists a number δ>0δ>0 such that for any point (x,y)(x,y) inside the domain of ff and within a suitably small distance positive δδ of (a,b),(a,b), the value of f(x,y)f(x,y) is no more than εε away from LL (Figure 4.15). Using symbols, we can write: For any ε>0,ε>0, there exists a number δ>0δ>0 such that

|f(x,y)L|<εwhenever0<(xa)2+(yb)2<δ.|f(x,y)L|<εwhenever0<(xa)2+(yb)2<δ.

Example 4.10

Limit of a Function at a Boundary Point

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

Solution

The domain of the function f(x,y)=25x2y2f(x,y)=25x2y2 is {(x,y)2|x2+y225},{(x,y)2|x2+y225}, which is a circle of radius 55 centered at the origin, along with its interior as shown in the following graph.

A circle with radius 5 centered at the origin with its interior filled in.
Figure 4.18 Domain of the function f(x,y)=25x2y2.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.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 4.19 Graph of the function f(x,y)=25x2y2.f(x,y)=25x2y2.
Checkpoint 4.8

Evaluate the following limit:

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

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

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

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

Definition

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

  1. f(a,b)f(a,b) exists.
  2. lim(x,y)(a,b)f(x,y)lim(x,y)(a,b)f(x,y) exists.
  3. lim(x,y)(a,b)f(x,y)=f(a,b).lim(x,y)(a,b)f(x,y)=f(a,b).

Example 4.11

Demonstrating Continuity for a Function of Two Variables

Show that the function f(x,y)=3x+2yx+y+1f(x,y)=3x+2yx+y+1 is continuous at point (5,−3).(5,−3).

Solution

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

  1. f(a,b)f(a,b) exists. This is true because the domain of the function ff consists of those ordered pairs for which the denominator is nonzero (i.e., x+y+10).x+y+10). Point (5,−3)(5,−3) satisfies this condition. Furthermore,
    f(a,b)=f(5,−3)=3(5)+2(−3)5+(−3)+1=1562+1=3.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)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.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).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.
Checkpoint 4.9

Show that the function f(x,y)=262x2y2f(x,y)=262x2y2 is continuous at point (2,−3).(2,−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)(x0,y0) in its domain if for every ε>0ε>0 there exists a δ>0δ>0 such that, whenever (xx0)2+(yy0)2<δ(xx0)2+(yy0)2<δ it is true, |f(x,y)f(a,b)|<ε.|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 4.2

The Sum of Continuous Functions Is Continuous

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

Theorem 4.3

The Product of Continuous Functions Is Continuous

If g(x)g(x) is continuous at x0x0 and h(y)h(y) is continuous at y0,y0, then f(x,y)=g(x)h(y)f(x,y)=g(x)h(y) is continuous at (x0,y0).(x0,y0).

Theorem 4.4

The Composition of Continuous Functions Is Continuous

Let gg be a function of two variables from a domain D2D2 to a range R.R. Suppose gg is continuous at some point (x0,y0)D(x0,y0)D and define z0=g(x0,y0).z0=g(x0,y0). Let ff be a function that maps to such that z0z0 is in the domain of f.f. Last, assume ff is continuous at z0.z0. Then fgfg is continuous at (x0,y0)(x0,y0) as shown in the following figure.

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 4.20 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 4.12

More Examples of Continuity of a Function of Two Variables

Show that the functions f(x,y)=4x3y2f(x,y)=4x3y2 and g(x,y)=cos(4x3y2)g(x,y)=cos(4x3y2) are continuous everywhere.

Solution

The polynomials g(x)=4x3g(x)=4x3 and h(y)=y2h(y)=y2 are continuous at every real number, and therefore by the product of continuous functions theorem, f(x,y)=4x3y2f(x,y)=4x3y2 is continuous at every point (x,y)(x,y) in the xy-plane.xy-plane. Since f(x,y)=4x3y2f(x,y)=4x3y2 is continuous at every point (x,y)(x,y) in the xy-planexy-plane and g(x)=cosxg(x)=cosx is continuous at every real number x,x, the continuity of the composition of functions tells us that g(x,y)=cos(4x3y2)g(x,y)=cos(4x3y2) is continuous at every point (x,y)(x,y) in the xy-plane.xy-plane.

Checkpoint 4.10

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

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)f(x,y,z) that gives the temperature at a physical location (x,y,z)(x,y,z) in three dimensions. Or perhaps a function g(x,y,z,t)g(x,y,z,t) can indicate air pressure at a location (x,y,z)(x,y,z) at time t.t. How can we take a limit at a point in 3?3? 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 δδ 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

Let (x0,y0,z0)(x0,y0,z0) be a point in 3.3. Then, a δδ ball in three dimensions consists of all points in 33 lying at a distance of less than δδ from (x0,y0,z0)(x0,y0,z0)—that is,

{(x,y,z)3|(xx0)2+(yy0)2+(zz0)2<δ}.{(x,y,z)3|(xx0)2+(yy0)2+(zz0)2<δ}.

To define a δδ 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)P=(w0,x0,y0,z0) in 4,4, a δδ ball around PP can be described by

{(w,x,y,z)4|(ww0)2+(xx0)2+(yy0)2+(zz0)2<δ}.{(w,x,y,z)4|(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),(x0,y0,z0), it suffices to show that for any point in a δδ ball centered at (x0,y0,z0),(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 4.13

Finding the Limit of a Function of Three Variables

Find lim(x,y,z)(4,1,−3)x2y3z2x+5yz.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.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, 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.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.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.
Checkpoint 4.11

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

Section 4.2 Exercises

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

60.

lim(x,y)(1,2)xlim(x,y)(1,2)x

61.

lim(x,y)(1,2)5x2yx2+y2lim(x,y)(1,2)5x2yx2+y2

62.

Show that the limit lim(x,y)(0,0)5x2yx2+y2lim(x,y)(0,0)5x2yx2+y2 exists and is the same along the paths: y-axisy-axis and x-axis,x-axis, and along y=x.y=x.

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

63.

lim(x,y)(0,0)4x2+10y2+44x210y2+6lim(x,y)(0,0)4x2+10y2+44x210y2+6

64.

lim(x,y)(11,13)1xylim(x,y)(11,13)1xy

65.

lim(x,y)(0,1)y2sinxxlim(x,y)(0,1)y2sinxx

66.

lim(x,y)(0,0)sin(x8+y7xy+10)lim(x,y)(0,0)sin(x8+y7xy+10)

67.

lim(x,y)(π/4,1)ytanxy+1lim(x,y)(π/4,1)ytanxy+1

68.

lim(x,y)(0,π/4)secx+23xtanylim(x,y)(0,π/4)secx+23xtany

69.

lim(x,y)(2,5)(1x5y)lim(x,y)(2,5)(1x5y)

70.

lim(x,y)(4,4)xlnylim(x,y)(4,4)xlny

71.

lim(x,y)(4,4)ex2y2lim(x,y)(4,4)ex2y2

72.

lim(x,y)(0,0)9x2y2lim(x,y)(0,0)9x2y2

73.

lim(x,y)(1,2)(x2y3x3y2+3x+2y)lim(x,y)(1,2)(x2y3x3y2+3x+2y)

74.

lim(x,y)(π,π)xsin(x+y4)lim(x,y)(π,π)xsin(x+y4)

75.

lim(x,y)(0,0)xy+1x2+y2+1lim(x,y)(0,0)xy+1x2+y2+1

76.

lim(x,y)(0,0)x2+y2x2+y2+11lim(x,y)(0,0)x2+y2x2+y2+11

77.

lim(x,y)(0,0)ln(x2+y2)lim(x,y)(0,0)ln(x2+y2)

For the following exercises, complete the statement.

78.

A point (x0,y0)(x0,y0) in a plane region RR is an interior point of RR if _________________.

79.

A point (x0,y0)(x0,y0) in a plane region RR is called a boundary point of RR if ___________.

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

80.

lim(x,y)(2,1)xy1xy1lim(x,y)(2,1)xy1xy1

81.

lim(x,y)(0,0)x44y4x2+2y2lim(x,y)(0,0)x44y4x2+2y2

82.

lim(x,y)(0,0)x3y3xylim(x,y)(0,0)x3y3xy

83.

lim(x,y)(0,0)x2xyxylim(x,y)(0,0)x2xyxy

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

84.

lim(x,y,z)(1,2,3)xz2y2zxyz1lim(x,y,z)(1,2,3)xz2y2zxyz1

85.

lim(x,y,z)(0,0,0)x2y2z2x2+y2z2lim(x,y,z)(0,0,0)x2y2z2x2+y2z2

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.

86.

lim(x,y)(0,0)xy+y3x2+y2lim(x,y)(0,0)xy+y3x2+y2

  1. Along the x-axisx-axis (y=0)(y=0)
  2. Along the y-axisy-axis (x=0)(x=0)
  3. Along the path y=2xy=2x
87.

Evaluate lim(x,y)(0,0)xy+y3x2+y2lim(x,y)(0,0)xy+y3x2+y2 using the results of previous problem.

88.

lim(x,y)(0,0)x2yx4+y2lim(x,y)(0,0)x2yx4+y2

  1. Along the x-axis (y=0)(y=0)
  2. Along the y-axis (x=0)(x=0)
  3. Along the path y=x2y=x2
89.

Evaluate lim(x,y)(0,0)x2yx4+y2lim(x,y)(0,0)x2yx4+y2 using the results of previous problem.

Discuss the continuity of the following functions. Find the largest region in the xy-planexy-plane in which the following functions are continuous.

90.

f(x,y)=sin(xy)f(x,y)=sin(xy)

91.

f(x,y)=ln(x+y)f(x,y)=ln(x+y)

92.

f(x,y)=e3xyf(x,y)=e3xy

93.

f(x,y)=1xyf(x,y)=1xy

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

94.

f(x,y)=x2yx2+y2f(x,y)=x2yx2+y2

95.

f(x,y)={x2yx2+y2if(x,y)(0,0)0if(x,y)=(0,0)}f(x,y)={x2yx2+y2if(x,y)(0,0)0if(x,y)=(0,0)}

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

96.

f(x,y)=sin(x2+y2)x2+y2f(x,y)=sin(x2+y2)x2+y2

97.

Determine whether g(x,y)=x2y2x2+y2g(x,y)=x2y2x2+y2 is continuous at (0,0).(0,0).

98.

Create a plot using graphing software to determine where the limit does not exist. Determine the region of the coordinate plane in which f(x,y)=1x2yf(x,y)=1x2y is continuous.

99.

Determine the region of the xy-planexy-plane in which the composite function g(x,y)=arctan(xy2x+y)g(x,y)=arctan(xy2x+y) is continuous. Use technology to support your conclusion.

100.

Determine the region of the xy-planexy-plane in which f(x,y)=ln(x2+y21)f(x,y)=ln(x2+y21) is continuous. Use technology to support your conclusion. (Hint: Choose the range of values for xandyxandy carefully!)

101.

At what points in space is g(x,y,z)=x2+y22z2g(x,y,z)=x2+y22z2 continuous?

102.

At what points in space is g(x,y,z)=1x2+z21g(x,y,z)=1x2+z21 continuous?

103.

Show that lim(x,y)(0,0)1x2+y2lim(x,y)(0,0)1x2+y2 does not exist at (0,0)(0,0) by plotting the graph of the function.

104.

[T] Evaluate lim(x,y)(0,0)xy2x2+y4lim(x,y)(0,0)xy2x2+y4 by plotting the function using a CAS. Determine analytically the limit along the path x=y2.x=y2.

105.

[T]

  1. Use a CAS to draw a contour map of z=9x2y2.z=9x2y2.
  2. What is the name of the geometric shape of the level curves?
  3. Give the general equation of the level curves.
  4. What is the maximum value of z?z?
  5. What is the domain of the function?
  6. What is the range of the function?
106.

True or False: If we evaluate lim(x,y)(0,0)f(x)lim(x,y)(0,0)f(x) along several paths and each time the limit is 1,1, we can conclude that lim(x,y)(0,0)f(x)=1.lim(x,y)(0,0)f(x)=1.

107.

Use polar coordinates to find lim(x,y)(0,0)sinx2+y2x2+y2.lim(x,y)(0,0)sinx2+y2x2+y2. You can also find the limit using L’Hôpital’s rule.

108.

Use polar coordinates to find lim(x,y)(0,0)cos(x2+y2).lim(x,y)(0,0)cos(x2+y2).

109.

Discuss the continuity of f(g(x,y))f(g(x,y)) where f(t)=1/tf(t)=1/t and g(x,y)=2x5y.g(x,y)=2x5y.

110.

Given f(x,y)=x24y,f(x,y)=x24y, find limh0f(x+h,y)f(x,y)h.limh0f(x+h,y)f(x,y)h.

111.

Given f(x,y)=x24y,f(x,y)=x24y, find limh0f(1+h,y)f(1,y)h.limh0f(1+h,y)f(1,y)h.

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