13 Functions of Several Variables

13.2 Limits and Continuity of Multivariable Functions

We continue with the pattern we have established in this text: after defining a new kind of function, we apply calculus ideas to it. The previous section defined functions of two and three variables; this section investigates what it means for these functions to be “continuous.”

We begin with a series of definitions. We are used to “open intervals” such as (1,3), which represents the set of all x such that 1<x<3, and “closed intervals” such as [1,3], which represents the set of all x such that 1x3. We need analogous definitions for open and closed sets in the x-y plane.

margin: P1P2xy (a)P1P2xy (b)P1P2xy (c) Figure 13.2.1: Illustrating open and closed sets in the x-y plane. Λ
Definition 13.2.1 Open Disk, Boundary and Interior Points, Open and Closed Sets, Bounded Sets

An open disk B in 2 centered at (x0,y0) with radius r is the set of all points (x,y) such that (x-x0)2+(y-y0)2<r.

Let S be a set of points in 2. A point P in 2 is a boundary point of S if all open disks centered at P contain both points in S and points not in S.

A point P in S is an interior point of S if there is an open disk centered at P that contains only points in S.

A set S is open if every point in S is an interior point.

A set S is closed if it contains all of its boundary points.

A set S is bounded if there is an M>0 such that the open disk, centered at the origin with radius M, contains S. A set that is not bounded is unbounded.

Figure 13.2.1 shows several sets in the x-y plane. In each set, point P1 lies on the boundary of the set as all open disks centered there contain both points in, and not in, the set. In contrast, point P2 is an interior point for there is an open disk centered there that lies entirely within the set.

The set depicted in Figure 13.2.1(a) is a closed set as it contains all of its boundary points. The set in (b) is open, for all of its points are interior points (or, equivalently, it does not contain any of its boundary points). The set in (c) is neither open nor closed as it contains some of its boundary points.

Example 13.2.1 Determining open/closed, bounded/unbounded

Determine if the domain of the function f(x,y)=1-x29-y24 is open, closed, or neither, and if it is bounded.

SolutionThis domain of this function was found in Example 13.1.2 to be D={(x,y)|x29+y241}, the region bounded by the ellipse x29+y24=1. Since the region includes the boundary (indicated by the use of “”), the set contains all of its boundary points and hence is closed. The region is bounded as a disk of radius 4, centered at the origin, contains D.

Example 13.2.2 Determining open/closed, bounded/unbounded

Determine if the domain of f(x,y)=1x-y is open, closed, or neither.

SolutionAs we cannot divide by 0, we find the domain to be D={(x,y)|x-y0}. In other words, the domain is the set of all points (x,y) not on the line y=x.

margin: xy Figure 13.2.2: Sketching the domain of the function in Example 13.2.2. Λ

The domain is sketched in Figure 13.2.2. Note how we can draw an open disk around any point in the domain that lies entirely inside the domain, and also note how the only boundary points of the domain are the points on the line y=x. We conclude the domain is an open set. The set is unbounded.

Limits

Recall a pseudo-definition of the limit of a function of one variable: “limxcf(x)=L” means that if x is “really close” to c, then f(x) is “really close” to L. A similar pseudo-definition holds for functions of two variables. We’ll say that

“​lim(x,y)(x0,y0)f(x,y)=L

means “if the point (x,y) is really close to the point (x0,y0), then f(x,y) is really close to L.” The formal definition is given below.

Definition 13.2.2 Limit of a Function of Two Variables

Let S be an open set containing (x0,y0), and let f be a function of two variables defined on S, except possibly at (x0,y0). The limit of f(x,y) as (x,y) approaches (x0,y0) is L, denoted

lim(x,y)(x0,y0)f(x,y)=L,

means that given any ϵ>0, there exists δ>0 such that for all (x,y)(x0,y0), if (x,y) is in the open disk centered at (x0,y0) with radius δ, then |f(x,y)-L|<ϵ.

margin:
(fullscreen)
Figure 13.2.3: Illustrating the definition of a limit. The open disk in the x-y plane has radius δ. Let (x,y) be any point in this disk; f(x,y) is within ϵ of L. Λ

The concept behind Definition 13.2.2 is sketched in Figure 13.2.3. Given ϵ>0, find δ>0 such that if (x,y) is any point in the open disk centered at (x0,y0) in the x-y plane with radius δ, then f(x,y) should be within ϵ of L.

Computing limits using this definition is rather cumbersome. The following theorem allows us to evaluate limits much more easily.

Theorem 13.2.1 Basic Limit Properties of Functions of Two Variables

Let b, x0, y0, L and K be real numbers, let n be a positive integer, and let f and g be functions with the following limits:

lim(x,y)(x0,y0)f(x,y)=L and lim(x,y)(x0,y0)g(x,y)=K.

The following limits hold.
1.  Constants: lim(x,y)(x0,y0)b=b 2.  Identity: lim(x,y)(x0,y0)x=x0;  lim(x,y)(x0,y0)y=y0 3.  Sums/Differences: lim(x,y)(x0,y0)(f(x,y)±g(x,y))=L±K 4.  Scalar Multiples: lim(x,y)(x0,y0)bf(x,y)=bL 5.  Products: lim(x,y)(x0,y0)f(x,y)g(x,y)=LK 6.  Quotients: lim(x,y)(x0,y0)f(x,y)/g(x,y)=L/K, (K0) 7.  Powers: lim(x,y)(x0,y0)f(x,y)n=Ln 8.  Roots: lim(x,y)(x0,y0)f(x,y)n=Ln   (when n is odd or L0)

This theorem can be proved by the same arguments as the analogous results for functions of one variable in Theorem 1.3.1. Combined with Theorems 1.3.3 and 1.3.4 of Section 1.3, this allows us to evaluate many limits.

Example 13.2.3 Evaluating a limit

Evaluate the following limits:

1.lim(x,y)(1,π)(yx+cos(xy))    2.lim(x,y)(0,0)3xyx2+y2

Solution

  1. (a)

    The aforementioned theorems allow us to simply evaluate y/x+cos(xy) when x=1 and y=π. If an indeterminate form is returned, we must do more work to evaluate the limit; otherwise, the result is the limit. Therefore

    lim(x,y)(1,π)yx+cos(xy) =π1+cosπ
    =π-1.
  2. (b)

    We attempt to evaluate the limit by substituting 0 in for x and y, but the result is the indeterminate form “0/0.” To evaluate this limit, we must “do more work,” but we have not yet learned what “kind” of work to do. Therefore we cannot yet evaluate this limit.

When dealing with functions of a single variable we also considered one-sided limits and stated

limxcf(x)=Lif and only if bothlimxc+f(x)=Landlimxc-f(x)=L.

That is, the limit is L if and only if f(x) approaches L when x approaches c from either direction, the left or the right.

In the plane, there are infinite directions from which (x,y) might approach (x0,y0). In fact, we do not have to restrict ourselves to approaching (x0,y0) from a particular direction, but rather we can approach that point along any possible path. It is possible to arrive at different limiting values by approaching (x0,y0) along different paths. If this happens, we say that lim(x,y)(x0,y0)f(x,y) does not exist (this is analogous to the left and right hand limits of single variable functions not being equal).

Our theorems tell us that we can evaluate most limits quite simply, without worrying about paths. When indeterminate forms arise, the limit may or may not exist. If it does exist, it can be difficult to prove this as we need to show the same limiting value is obtained regardless of the path chosen. The case where the limit does not exist is often easier to deal with, for we can often pick two paths along which the limit is different.

Example 13.2.4 Showing limits do not exist


(a) Show lim(x,y)(0,0)3xyx2+y2 does not exist by finding the limits along the lines y=mx. (b) Show lim(x,y)(0,0)sin(xy)x+y does not exist by finding the limit along the path y=-sinx.

Solution

  1. (a)

    Evaluating lim(x,y)(0,0)3xyx2+y2 along the lines y=mx means replace all y’s with mx and evaluating the resulting limit:

    lim(x,mx)(0,0)3x(mx)x2+(mx)2 =limx03mx2x2(m2+1)
    =limx03mm2+1
    =3mm2+1.

    While the limit exists for each choice of m, we get a different limit for each choice of m. That is, along different lines we get differing limiting values, meaning the limit does not exist.

  2. (b)

    Let f(x,y)=sin(xy)x+y. We are to show that lim(x,y)(0,0)f(x,y) does not exist by finding the limit along the path y=-sinx. First, however, consider the limits found along the lines y=mx as done above.

    lim(x,mx)(0,0)sin(x(mx))x+mx =limx0sin(mx2)x(m+1)
    =limx0sin(mx2)x1m+1.

    By applying L’Hôpital’s Rule, we can show this limit is 0 except when m=-1, that is, along the line y=-x. This line is not in the domain of f, so we have found the following fact: along every line y=mx in the domain of f, lim(x,y)(0,0)f(x,y)=0. Now consider the limit along the path y=-sinx:

    lim(x,-sinx)(0,0)sin(-xsinx)x-sinx =limx0sin(-xsinx)x-sinx

    Now apply L’Hôpital’s Rule twice:

    =limx0cos(-xsinx)(-sinx-xcosx)1-cosx(=0/0)
    =limx0-sin(-xsinx)(-sinx-xcosx)2+cos(-xsinx)(-2cosx+xsinx)sinx
    =-2/0the limit does not exist.

    Step back and consider what we have just discovered. Along any line y=mx in the domain of the f(x,y), the limit is 0. However, along the path y=-sinx, which lies in the domain of f(x,y) for all x0, the limit does not exist. Since the limit is not the same along every path to (0,0), we say lim(x,y)(0,0)sin(xy)x+y does not exist.

Example 13.2.5 Finding a limit

Let f(x,y)=5x2y2x2+y2. Find lim(x,y)(0,0)f(x,y).

SolutionIt is relatively easy to show that along any line y=mx, the limit is 0. This is not enough to prove that the limit exists, as demonstrated in the previous example, but it tells us that if the limit does exist then it must be 0.

To prove the limit is 0, we apply Definition 13.2.2. Let ϵ>0 be given. We want to find δ>0 such that if (x-0)2+(y-0)2<δ, then |f(x,y)-0|<ϵ.

Set δ<ϵ/5. Note that |5y2x2+y2|<5 for all (x,y)(0,0), and that if x2+y2<δ, then x2<δ2.

Let (x-0)2+(y-0)2=x2+y2<δ. Consider |f(x,y)-0|:

|f(x,y)-0| =|5x2y2x2+y2-0|
=|x25y2x2+y2|
<δ25
<ϵ55
=ϵ.

Thus if (x-0)2+(y-0)2<δ then |f(x,y)-0|<ϵ, which is what we wanted to show. Thus lim(x,y)(0,0)5x2y2x2+y2=0.

We also have a multivariable version of the squeeze theorem.

Theorem 13.2.2 Squeeze Theorem

Let S be an open set containing (x0,y0). Suppose f(x,y), g(x,y), and h(x,y) are defined on S except possibly at (x0,y0) and both

lim(x,y)(x0,y0)g(x,y)=Landlim(x,y)(x0,y0)h(x,y)=L.

If g(x,y)f(x,y)h(x,y) for all (x,y)S except possibly at (x0,y0), then

lim(x,y)(x0,y0)f(x,y)=L.

This theorem provides other proofs of the previous example.

Example 13.2.6 Finding a limit using the Squeeze Theorem

We have

05x2y2x2+y25x2y2+5y4x2+y2=5y2(x2+y2)x2+y2=5y2.

Since 00 and 5y20 as (x,y)(0,0) we have

lim(x,y)(0,0)5x2y2x2+y2=0

by the Squeeze Theorem.

If we set x=rcosθ and y=rsinθ we have (x,y)0 as r0. Now

5x2y2x2+y2=5r4cos2θsin2θr2(cos2θ+sin2θ)=5r2cos2θsin2θ.

Thus

05x2y2x2+y2=5r2cos2θsin2θ5r2.

Since 5r20 as r0 we have again

lim(x,y)(0,0)5x2y2x2+y2=0

by the Squeeze Theorem.

Continuity

Definition 1.6.1 defines what it means for a function of one variable to be continuous. In brief, it meant that the function always equaled its limit. We define continuity for functions of two variables in a similar way as we did for functions of one variable.

Definition 13.2.3 Continuous

Let a function f(x,y) be defined on an open disk B containing the point (x0,y0).

  1. (a)

    f is continuous at (x0,y0) if lim(x,y)(x0,y0)f(x,y)=f(x0,y0).

  2. (b)

    f is continuous on an open set S if f is continuous at each points in S. (We say that f is continuous everywhere if f is continuous on 2.)

It follows that if f is a continuous function of one variable, then f(x) (or f(y)) is a continuous function of the variables (x,y).

Example 13.2.7 Continuity of a function of two variables

Let f(x,y)={cosysinxxx0cosyx=0. Is f continuous at (0,0)? Is f continuous everywhere?

SolutionTo determine if f is continuous at (0,0), we need to compare lim(x,y)(0,0)f(x,y) to f(0,0).

Applying the definition of f, we see that f(0,0)=cos0=1.

We now consider the limit lim(x,y)(0,0)f(x,y). Substituting 0 for x and y in (cosysinx)/x returns the indeterminate form “0/0”, so we need to do more work to evaluate this limit.

Consider two related limits: lim(x,y)(0,0)cosy and lim(x,y)(0,0)sinxx. The first limit does not contain x, and since cosy is continuous,

lim(x,y)(0,0)cosy=limy0cosy=cos0=1.

The second limit does not contain y. By Theorem 1.3.6 we can say

lim(x,y)(0,0)sinxx=limx0sinxx=1.

Finally, Theorem 13.2.1 of this section states that we can combine these two limits as follows:

lim(x,y)(0,0)cosysinxx =lim(x,y)(0,0)(cosy)(sinxx)
=(lim(x,y)(0,0)cosy)(lim(x,y)(0,0)sinxx)
=(1)(1)
=1.

We have found that lim(x,y)(0,0)cosysinxx=f(0,0), so f is continuous at (0,0).

margin:
(fullscreen)
Figure 13.2.4: A graph of f(x,y) in Example 13.2.7. Λ

A similar analysis shows that f is continuous at all points in 2. As long as x0, we can evaluate the limit directly; when x=0, a similar analysis shows that the limit is cosy. Thus we can say that f is continuous everywhere. A graph of f is given in Figure 13.2.4. Notice how it has no breaks, jumps, etc.

The following theorems are very similar to Theorems 1.6.1 and 1.6.2, giving us ways to combine continuous functions to create other continuous functions.

Theorem 13.2.3 Properties of Continuous Functions

Let f and g be continuous on an open set S, let c be a real number, and let n be a positive integer. The following functions are continuous on S.
1.  Sums/Differences: f±g 2.  Constant Multiples: cf 3.  Products: fg 4.  Quotients: f/g   (as longs as g0 on B) 5.  Powers: fn 6.  Roots: fn   (if f0 on B or n is odd)

Theorem 13.2.4 Continuity of Compositions

Let f be continuous on S, where the range of f on S is J, and let g be a single variable function that is continuous on J. Then

(gf)(x,y)=g(f(x,y)),

is continuous on S.

Example 13.2.8 Establishing continuity of a function

Let f(x,y)=sin(x2cosy). Show f is continuous everywhere.

SolutionWe will apply Theorems 1.6.1, 13.2.3, and 13.2.4. We let f1(x,y)=x2. Since y is not actually used in the function, and polynomials are continuous (by Theorem 1.6.1), we conclude f1 is continuous everywhere. A similar statement can be made about f2(x,y)=cosy. Part 3 of Theorem 13.2.3 states that f3=f1f2 is continuous everywhere, and Theorem 13.2.4 states the composition of sine with f3 is continuous: that is, sin(f3)=sin(x2cosy) is continuous everywhere.

Functions of Three Variables

The definitions and theorems given in this section can be extended in a natural way to definitions and theorems about functions of three (or more) variables. We cover the key concepts here; some terms from Definitions 13.2.1 and 13.2.3 are not redefined but their analogous meanings should be clear to the reader.

Definition 13.2.4 Open Balls, Limit, Continuous


(a) An open ball in 3 centered at (x0,y0,z0) with radius r is the set of all points (x,y,z) such that (x-x0)2+(y-y0)2+(z-z0)2=r. (b) Let D be an open set in 3 containing (x0,y0,z0), and let f(x,y,z) be a function of three variables defined on D, except possibly at (x0,y0,z0). The limit of f(x,y,z) as (x,y,z) approaches (x0,y0,z0) is L, denoted lim(x,y,z)(x0,y0,z0)f(x,y,z)=L, means that given any ϵ>0, there is a δ>0 such that for all (x,y,z)(x0,y0,z0), if (x,y,z) is in the open ball centered at (x0,y0,z0) with radius δ, then |f(x,y,z)-L|<ϵ. (c) Let f(x,y,z) be defined on an open ball B containing (x0,y0,z0). Then f is continuous at (x0,y0,z0) if lim(x,y,z)(x0,y0,z0)f(x,y,z)=f(x0,y0,z0).

These definitions can also be extended naturally to apply to functions of four or more variables. Theorems 13.2.3 and 13.2.4 also applies to function of three or more variables, allowing us to say that the function

f(x,y,z)=ex2+yy2+z2+3sin(xyz)+5

is continuous everywhere.

When considering single variable functions, we studied limits, then continuity, then the derivative. In our current study of multivariable functions, we have studied limits and continuity. In the next section we study derivation, which takes on a slight twist as we are in a multivariable context.

Exercises 13.2

 

Terms and Concepts

  1. 1.

    Describe in your own words the difference between boundary and interior points of a set.

  2. 2.
    Use your own words to describe (informally) what lim(x,y)(1,2)f(x,y)=17 means.
  3. 3.

    Give an example of a closed, bounded set.

  4. 4.

    Give an example of a closed, unbounded set.

  5. 5.

    Give an example of a open, bounded set.

  6. 6.

    Give an example of a open, unbounded set.

Problems

In Exercises 7–10., a set S is given.

  1. (a)

    Give one boundary point and one interior point, when possible, of S.

  2. (b)

    State whether S is open, closed, or neither.

  3. (c)

    State whether S is bounded or unbounded.

  1. 7.

    S={(x,y)|(x-1)24+(y-3)291}

  2. 8.

    S={(x,y)|yx2}

  3. 9.

    S={(x,y)|x2+y2=1}

  4. 10.

    S={(x,y)|y>sinx}

In Exercises 11–18.:

  1. (a)

    Find the domain D of the given function.

  2. (b)

    State whether D is an open or closed set.

  3. (c)

    State whether D is bounded or unbounded.

  1. 11.

    f(x,y)=9-x2-y2

  2. 12.

    f(x,y)=y-x2

  3. 13.

    f(x,y)=1y-x2

  4. 14.

    f(x,y)=x2-y2x2+y2

  5. 15.

    f(x,y)=x2+3xy+y2

  6. 16.

    f(x,y)=x2+y+1x-2y+3

  7. 17.

    f(x,y)=11-x2-y2

  8. 18.

    f(x,y)=ln(xy)

In Exercises 19–24., a limit is given. Evaluate the limit along the paths given, then state why these results show the given limit does not exist.

  1. 19.
    lim(x,y)(0,0)x2-y2x2+y2 (a) Along the path y=0. (b) Along the path x=0.
  2. 20.
    lim(x,y)(0,0)x+yx-y (a) Along the path y=mx.
  3. 21.
    lim(x,y)(0,0)xy-y2y2+x (a) Along the path y=mx. (b) Along the path x=0.
  4. 22.
    lim(x,y)(0,0)sin(x2)y (a) Along the path y=mx. (b) Along the path y=x2.
  5. 23.
    lim(x,y)(1,2)x+y-3x2-1 (a) Along the path y=2. (b) Along the path y=x+1.
  6. 24.
    lim(x,y)(π,π/2)sinxcosy (a) Along the path x=π. (b) Along the path y=x-π/2.
  1. 25.
    Use the Squeeze Theorem to show lim(x,y)(0,0)sin(xy)cos(1x2+y2)=0.
  2. 26.
    Let f(x,y)=x2yx4+y2. (a) Show that f(x,y)0 along every line through the origin. (b) Despite this, show that lim(x,y)(0,0)f(x,y) does not exist.
  3. 27.
    Show that lim(x,y)(0,0)x4-9y4x2+3y2=0.
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