4.2 Limits and Continuity

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 x ≠ a x ≠ a in an open interval containing a . a . Let L L be a real number. Then

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

Definition

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

as shown in the following graph.

The idea of a δ δ disk appears in the definition of the limit of a function of two variables. If δ δ is small, then all the points ( x , y ) ( x , y ) in the δ δ disk are close to ( a , b ) . ( a , b ) . This is completely analogous to x x being close to a a 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 δ δ disk.

Definition

Let f f be a function of two variables, x x 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

if for each ε > 0 ε > 0 there exists a small enough δ > 0 δ > 0 such that for all points ( x , y ) ( x , y ) in a δ δ 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 ε ε away from L L (Figure 4.15). Using symbols, we write the following: For any ε > 0 , ε > 0 , there exists a number δ > 0 δ > 0 such that

| f ( x , y ) − L | < ε whenever 0 < ( x − a ) 2 + ( y − b ) 2 < δ . | f ( x , y ) − L | < ε whenever 0 < ( x − a ) 2 + ( y − b ) 2 < δ .

Figure 4.15 The limit of a function involving two variables requires that f ( x , y ) f ( x , y ) be within ε ε of L L whenever ( x , y ) ( x , y ) is within δ δ 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 L L and M M are real numbers such that lim ( x , y ) → ( a , b ) f ( x , y ) = L lim ( 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 c c be a constant. Then each of the following statements holds:

Constant Law:

Identity Laws:

Sum Law:

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

Constant Multiple Law:

Product Law:

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

Quotient Law:

lim ( x , y ) → ( a , b ) f ( x , y ) g ( x , y ) = L M for M ≠ 0 lim ( x , y ) → ( a , b ) f ( x , y ) g ( x , y ) = L M for M ≠ 0

Power Law:

for any positive integer n . n .

Root Law:

for all L L if n n is odd and positive, and for L ≥ 0 L ≥ 0 if n n is even and positive provided that f ( x , y ) ≥ 0 f ( x , y ) ≥ 0 for all ( x , y ) ≠ ( a , b ) ( x , y ) ≠ ( a , b ) in neighborhood of ( a , b ) ( a , b ) .

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 ) ( x 2 − 2 x y + 3 y 2 − 4 x + 3 y − 6 ) lim ( x , y ) → ( 2 , −1 ) ( x 2 − 2 x y + 3 y 2 − 4 x + 3 y − 6 )
2. lim ( x , y ) → ( 2 , −1 ) 2 x + 3 y 4 x − 3 y lim ( x , y ) → ( 2 , −1 ) 2 x + 3 y 4 x − 3 y

Solution

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

lim ( x , y ) → ( 2 , −1 ) ( x 2 − 2 x y + 3 y 2 − 4 x + 3 y − 6 ) = ( lim ( x , y ) → ( 2 , −1 ) x 2 ) − ( lim ( x , y ) → ( 2 , −1 ) 2 x y ) + ( lim ( x , y ) → ( 2 , −1 ) 3 y 2 ) − ( lim ( x , y ) → ( 2 , −1 ) 4 x ) + ( lim ( x , y ) → ( 2 , −1 ) 3 y ) − ( lim ( x , y ) → ( 2 , −1 ) 6 ) . lim ( x , y ) → ( 2 , −1 ) ( x 2 − 2 x y + 3 y 2 − 4 x + 3 y − 6 ) = ( lim ( x , y ) → ( 2 , −1 ) x 2 ) − ( lim ( x , y ) → ( 2 , −1 ) 2 x y ) + ( lim ( x , y ) → ( 2 , −1 ) 3 y 2 ) − ( lim ( x , y ) → ( 2 , −1 ) 4 x ) + ( lim ( x , y ) → ( 2 , −1 ) 3 y ) − ( lim ( x , y ) → ( 2 , −1 ) 6 ) .

= ( lim ( x , y ) → ( 2 , −1 ) x 2 ) − 2 ( lim ( x , y ) → ( 2 , −1 ) x y ) + 3 ( lim ( x , y ) → ( 2 , −1 ) y 2 ) − 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 ) x 2 ) − 2 ( lim ( x , y ) → ( 2 , −1 ) x y ) + 3 ( lim ( x , y ) → ( 2 , −1 ) y 2 ) − 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 ) x ) 2 − 2 ( lim ( x , y ) → ( 2 , −1 ) x ) ( lim ( x , y ) → ( 2 , −1 ) y ) + 3 ( lim ( x , y ) → ( 2 , −1 ) y ) 2 − 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 ) x ) 2 − 2 ( lim ( x , y ) → ( 2 , −1 ) x ) ( lim ( x , y ) → ( 2 , −1 ) y ) + 3 ( lim ( x , y ) → ( 2 , −1 ) y ) 2 − 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 ) ( 4 x − 3 y ) = lim ( x , y ) → ( 2 , −1 ) 4 x − lim ( x , y ) → ( 2 , −1 ) 3 y = 4 ( lim ( x , y ) → ( 2 , −1 ) x ) − 3 ( lim ( x , y ) → ( 2 , −1 ) y ) = 4 ( 2 ) − 3 ( −1 ) = 11. lim ( x , y ) → ( 2 , −1 ) ( 4 x − 3 y ) = lim ( x , y ) → ( 2 , −1 ) 4 x − lim ( x , y ) → ( 2 , −1 ) 3 y = 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 ) ( 2 x + 3 y ) = lim ( x , y ) → ( 2 , −1 ) 2 x + lim ( x , y ) → ( 2 , −1 ) 3 y = 2 ( lim ( x , y ) → ( 2 , −1 ) x ) + 3 ( lim ( x , y ) → ( 2 , −1 ) y ) = 2 ( 2 ) + 3 ( −1 ) = 1. lim ( x , y ) → ( 2 , −1 ) ( 2 x + 3 y ) = lim ( x , y ) → ( 2 , −1 ) 2 x + lim ( x , y ) → ( 2 , −1 ) 3 y = 2 ( lim ( x , y ) → ( 2 , −1 ) x ) + 3 ( lim ( x , y ) → ( 2 , −1 ) y ) = 2 ( 2 ) + 3 ( −1 ) = 1.

lim ( x , y ) → ( 2 , −1 ) 2 x + 3 y 4 x − 3 y = lim ( x , y ) → ( 2 , −1 ) ( 2 x + 3 y ) lim ( x , y ) → ( 2 , −1 ) ( 4 x − 3 y ) = 1 11 . lim ( x , y ) → ( 2 , −1 ) 2 x + 3 y 4 x − 3 y = lim ( x , y ) → ( 2 , −1 ) ( 2 x + 3 y ) lim ( x , y ) → ( 2 , −1 ) ( 4 x − 3 y ) = 1 11 .

Checkpoint 4.6

Evaluate the following limit:

lim ( x , y ) → ( 5 , −2 ) x 2 − y y 2 + x − 1 3 . lim ( x , y ) → ( 5 , −2 ) x 2 − y y 2 + x − 1 3 .

Since we are taking the limit of a function of two variables, the point ( a , b ) ( a , b ) is in ℝ 2 , ℝ 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 ) 2 x y 3 x 2 + y 2 lim ( x , y ) → ( 0 , 0 ) 2 x y 3 x 2 + y 2
2. lim ( x , y ) → ( 0 , 0 ) 4 x y 2 x 2 + 3 y 4 lim ( x , y ) → ( 0 , 0 ) 4 x y 2 x 2 + 3 y 4

Solution

1. The domain of the function f ( x , y ) = 2 x y 3 x 2 + y 2 f ( x , y ) = 2 x y 3 x 2 + y 2 consists of all points in the x y -plane x y -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 = 0 y = 0 in the x y -plane. x y -plane. Substituting y = 0 y = 0 into f ( x , y ) f ( x , y ) gives
   

for any value of x . x . Therefore the value of f f remains constant for any point on the x -axis, x -axis, and as y y approaches zero, the function remains fixed at zero.
Next, consider the line y = x . y = x . Substituting y = x y = x into f ( x , y ) f ( x , y ) gives

f ( x , x ) = 2 x ( x ) 3 x 2 + x 2 = 2 x 2 4 x 2 = 1 2 . f ( x , x ) = 2 x ( x ) 3 x 2 + x 2 = 2 x 2 4 x 2 = 1 2 .

This is true for any point on the line y = x . y = x . If we let x x approach zero while staying on this line, the value of the function remains fixed at 1 2 , 1 2 , regardless of how small x x is.
Choose a value for ε ε that is less than 1 / 2 1 / 2 —say, 1 / 4 . 1 / 4 . Then, no matter how small a δ δ disk we draw around ( 0 , 0 ) , ( 0 , 0 ) , the values of f ( x , y ) f ( x , y ) for points inside that δ δ disk will include both 0 0 and 1 2 . 1 2 . Therefore, the definition of limit at a point is never satisfied and the limit fails to exist.

Figure 4.16 Graph of the function f ( x , y ) = ( 2 x y ) / ( 3 x 2 + y 2 ) . f ( x , y ) = ( 2 x y ) / ( 3 x 2 + y 2 ) . 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 1 2 . 1 2 .

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 x -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 = k x . y = k x . Then the limit becomes

lim ( x , y ) → ( 0 , 0 ) 4 x y 2 x 2 + 3 y 4 = lim ( x , y ) → ( 0 , 0 ) 4 x ( k x ) 2 x 2 + 3 ( k x ) 4 = lim ( x , y ) → ( 0 , 0 ) 4 k 2 x 3 x 2 + 3 k 4 x 4 = lim ( x , y ) → ( 0 , 0 ) 4 k 2 x 1 + 3 k 4 x 2 = lim ( x , y ) → ( 0 , 0 ) ( 4 k 2 x ) lim ( x , y ) → ( 0 , 0 ) ( 1 + 3 k 4 x 2 ) = 0 lim ( x , y ) → ( 0 , 0 ) 4 x y 2 x 2 + 3 y 4 = lim ( x , y ) → ( 0 , 0 ) 4 x ( k x ) 2 x 2 + 3 ( k x ) 4 = lim ( x , y ) → ( 0 , 0 ) 4 k 2 x 3 x 2 + 3 k 4 x 4 = lim ( x , y ) → ( 0 , 0 ) 4 k 2 x 1 + 3 k 4 x 2 = lim ( x , y ) → ( 0 , 0 ) ( 4 k 2 x ) lim ( x , y ) → ( 0 , 0 ) ( 1 + 3 k 4 x 2 ) = 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 = y 2 . x = y 2 . Substituting y 2 y 2 in place of x x in f ( x , y ) f ( x , y ) gives

lim ( x , y ) → ( 0 , 0 ) 4 x y 2 x 2 + 3 y 4 = lim ( x , y ) → ( 0 , 0 ) 4 ( y 2 ) y 2 ( y 2 ) 2 + 3 y 4 = lim ( x , y ) → ( 0 , 0 ) 4 y 4 y 4 + 3 y 4 = lim ( x , y ) → ( 0 , 0 ) 1 = 1. lim ( x , y ) → ( 0 , 0 ) 4 x y 2 x 2 + 3 y 4 = lim ( x , y ) → ( 0 , 0 ) 4 ( y 2 ) y 2 ( y 2 ) 2 + 3 y 4 = lim ( x , y ) → ( 0 , 0 ) 4 y 4 y 4 + 3 y 4 = lim ( x , y ) → ( 0 , 0 ) 1 = 1.