Suppose that the values of f (x, y) lie arbitrarily close to a fixed real number L for all points (x, y) sufficiently close to a point (x ₀, y₀). If this is true, then we can say that f approaches the limit L as (x, y) ® (x ₀, y₀). This explanation is quite similar to the description given in calculus I for a limit. Recall that when we were investigating the limit of a function in one variable, we could only approach x ₀ from the left or the right. (See figure 1) Now, for functions of more than one variable, we can approach it in many directions, called paths. (See figure 2) Sometimes, we will have to evaluate the limit along a specific path, but we will discuss this later in this set of notes.

Now I will state the formal definition of a limit.

figure 2

DEFINITION:

We say that a function f (x, y) approaches the limit L as (x, y) approaches (x ₀, y₀), and write

if, for every number e > 0, there exists a corresponding d > 0 such that for all (x, y) in the domain of f

Notice that the formal definition is similar to the one discussed in calculus I, and the properties stated for the limits of one- variable functions carry over to limits of functions of two or more variables.

EXAMPLE 1:

Evaluate the following limit.

SOLUTION:

Since the point causes no problem in the function, just evaluate the function at that point.

EXAMPLE 2:

Evaluate the following limit.

SOLUTION:

EXAMPLE 3:

Evaluate the following limit.

SOLUTION:

How do we handle a limit that states that we cannot use a certain path because the function is undefined there? Sometime you can rewrite the fraction so that the problem path does not cause a problem. Here are some examples of how to do this.

EXAMPLE 4:

Evaluate the following limit.

SOLUTION:

Since the factor, x - y, causes the problem along the path y = x, I will try to simplify the rational expression by factoring.

EXAMPLE 5:

Evaluate the following limit.

SOLUTION:

The problem that these two paths cause is the fact that if I put the point (2, -4) into the function, I come up with the indeterminate form 0/0. I am going to use factor by grouping to simplify the denominator.

EXAMPLE 6:

Evaluate the following limit.

SOLUTION:

Both the denominator and the numerator will be 0 when the point (2, 2) is plugged in. Therefore, I will rationalize the denominator in the hopes that this simplifies the rational expression.

 EXAMPLE 7:

Evaluate the following limit.

SOLUTION:

EXAMPLE 8:

Evaluate the following limit

SOLUTION:

 DEFINITION:

A function f (x, y) is continuous at the point (x ₀, y₀) if

1.

f is defined at (x ₀, y₀)

2.

3.

A function is continuous if it is continuous at every point of its domain.

 Notice that the definition of continuity is not much different than the definition we used in calculus I.

The question that we should be asking ourselves is how do we determine where a function is continuous? If the function has a denominator, then you will need to determine what points make it zero. If the function has an even root in it, then you will need to determine what points make the radicand negative. So let us look at some examples.

 EXAMPLE 9:

At what points (x, y) in the plane is the function f (x, y) = ln (x ² + y ²) is continuous?

SOLUTION:

Remember that the argument of the natural log has to be greater than zero, so this function is continuous for all (x, y) except for the point (0, 0).

EXAMPLE 10:

At what points (x, y) in the plane is the function

is continuous?

SOLUTION:

Since the denominator will never equal zero, this function is continuous for all (x, y) in the xy-plane.

EXAMPLE 11:

At what points (x, y) in the plane is the function

is continuous?

SOLUTION:

I will have to determine what values will make the denominator zero, so I will set the denominator equal to zero and solve for x.

x ² - 3x + 2 = 0 ® (x - 2)(x - 1) = 0 ® x = 2 or x = 1.

This function will be continuous for all (x, y) in the plane except when x = 1 and x = 2.

EXAMPLE 12:

At what points (x, y, z) in space if the function

is continuous?

SOLUTION:

Since we are dealing with an even root, we will have to determine where the radicand is greater than or equal to zero. Then any values that make the radicand less than zero will be points in which the function will not be continuous at.

x ² + y ² - 1 ³ 0 ® x ² + y ² ³ 1

This function is continuous for all (x, y, z) except for points that lie in the interior of the cylinder x ² + y ² = 1.

EXAMPLE 13:

At what points (x, y, z) in space is the function

is continuous?

SOLUTION:

When y = 0 and z = 0, the denominator will be zero, so this function will be continuous for all (x, y, z) except for the points of the form (x, 0, 0).

Remember that when are evaluating a limit of f (x, y) or f (x, y, z), we are approaching the limit from many different ways (or paths). Therefore, if we need to show that a limit does not exist, we will have to show that we will get different limits on different paths. The paths that I use to show this are generic in form. They are y = mx, y = mx ², y = mx ³, etc., and the one I use, depends on what the function looks like.

EXAMPLE 14:

SOLUTION:

Since the function has a x ² in it, the path that I will use is y = mx ².

As m varies, so will the limit. Therefore, the limit does not exist.

EXAMPLE 15:

SOLUTION:

Since there is a x ² and a y ² in the radical, and I would like to be able to factor out a x ² out of the radical, I will us the path y = mx.

As m varies, so will the limit. Therefore, the limit does not exist.