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Show that

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

does not exist.

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

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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 2 ( [link] ).

  1. A point P 0 is called an interior point    of S if there is a δ disk centered around P 0 contained completely in S .
  2. A point P 0 is called a boundary point    of S if every δ disk centered around P 0 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.
In the set S shown, ( −1 , 1 ) is an interior point and ( 2 , 3 ) is a boundary point.

Definition

Let S be a subset of 2 ( [link] ).

  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

Let S be a subset of 2 ( [link] ).

  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 ( [link] ). Using symbols, we can write: For any ε > 0 , there exists a number δ > 0 such that

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

Limit of a function at a boundary point

Prove lim ( x , y ) ( 4 , 3 ) 25 x 2 y 2 = 0 .

The domain of the function f ( x , y ) = 25 x 2 y 2 is { ( x , y ) 2 | x 2 + y 2 25 } , which is a circle of radius 5 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.
Domain of the function f ( x , y ) = 25 x 2 y 2 .

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 ) 25 x 2 y 2 = lim ( x , y ) ( 4 , 3 ) ( 25 x 2 y 2 ) = lim ( x , y ) ( 4 , 3 ) 25 lim ( x , y ) ( 4 , 3 ) x 2 lim ( x , y ) ( 4 , 3 ) y 2 = 25 4 2 3 2 = 0.

See the following graph.

The upper hemisphere in xyz space with radius 5 and center the origin.
Graph of the function f ( x , y ) = 25 x 2 y 2 .
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Evaluate the following limit:

lim ( x , y ) ( 5 , −2 ) 29 x 2 y 2 .

lim ( x , y ) ( 5 , −2 ) 29 x 2 y 2

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Practice Key Terms 8

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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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