Operations on the continuity of a function of two variables

 Operations on continuity of a function of two variables:

Theorem 1

The sum of continuous functions is continuous

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

Theorem 2

The product of continuous functions is continuous

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

Theorem 3

The composition of continuous functions is continuous

Let g be a function of two variables from

Let's suppose g continuous at  (x₀.y₀) and let z₀ = g(x₀.y₀). Let f be a function that relates elements from R to R such that z₀ is in the domain of f. Then 

Example

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

Solution

a) Continuity of f(x.y) = 4x³y²

We have f(x) = 4x³ is continuous for every real number; f(y) = y² is continuous for every real number.. Therefore f(x,y) = f(x).f(y) as product of two continuous functions is continuous for every real number.

b) Continuity of g(x, y) =  cos(4x³y²) 

We have f(x, y) =4x³y² which is continuous for every element of  R²; h(x) = cosx is continuous for every element of R . Let Let f(x,y) = z. We have h(z) = cosz ; h[f(x,y)] = cos[f(x, y)] ; foh[(x,y)] = cos (4x³y²) is continuous for every element of R² as a composition of 2 functions. We can see that foh(x,y) is nothing more than the function f(x,y), Therefore  f(x.y) = 4x³y² is continuous for every element of R².

Practice

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

Continuity of a function of two variables

 Goal: State the conditions for continuity of a function of two variables

Definition

The definition of the continuity of a function of two variables is similar to that of a function of one variable.. 

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

Example 1

Show that the function:

                                                                

Solution
In order for f(a,b) to exist the denominator must be different of zero. We have x + y + 1 = 5 - 3 + 1 =3. Since the denominator is different of zero, let's calculate f(a, b):

Let's calculate limf(x, y) when (x,y) approaches (5, -3):

We have :

All three condiitions are verified. Therefore the function is continuous at the point (5, -3).

Practice

Show that the function:

Limit of a function of two variables at a boundary point

 Objective: Determine the limit of a function at a boundary point

Before to define the limit of a function of two variables at a boundary point, let's define interior and boundary point.

Interior point.

Let's consider a subset S of R². A point P is an interior point of S if any δ disk centered at P is located completely inside of S. Example: the point (-1, 1) in the figure below is an interior point

Boundary point

A point P is a boundary point if any δ disk centered at P is not completely located inside of S. It means that  some points of the disk are located inside of S and others are located outside of S. Example: the point (2,3) is a boundary point.

Definition of the limit of a function f at a boundary point

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

if for any ϵ > 0. there exists δ>0 such that for any point (x,y) located inside the domain of and within a suitable small distance δ of (a,b), the value of f(x,y) is no more than ϵ away from L. Using symbols we write: for any number δ >0, there exists ϵ>0 such that

Example

Solution

The domain of the function:

 

 is a circle of radius 5 centered at the origin along its interior as shown in the figure below

Let's calculate the limit of the function by applying the limit rules which apply to the boundary of the domain of the function as well the interior points of the domain.

Here is the graph of the function                                                                                                              :

                                                                                                                       
Practice                                                                                                                                                

Evaluate the following limit:                                                                                                                

 

                                                                                                                                               

Limit of a function of two variables

 Goal: define the limit of a function of two variables

The limit of a function of two variables is based on the limit of a function of one variable. Let's recall the definition:

Let f(x) be defined for all x≠a in an open interval containing a . Let L be a real number. Then 

if  for every ϵ> 0 there exists ẟ> 0 such that if 0< ❙x - a❙ < δ for all x in the domain of f, then

The idea of an open interval in a function of two variables is similar to the open interval in a single variable. Let's define an open interval in a function of two variables.

Definition 
An open disk δ centered at a point (a, b) ε R² and of radius  δ is defined by:

In a function of one variable, when x is is very close to a, we have ❘x-a❘ < δ meaning that the distance of all x to a is very small. In a disk, the distance of all circles are smaller than the square of the radius δ

If we take the square of both sides and taking into account that the first side is strictly positive, we have:

Definition

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

if for every ϵ>0, there exists a small enough δ>0 such that for all points (x, y) in a delta disk around (a, b) except for (a, b) itself, the value of f(x,y) is no more away than ε from L. Using symbols, we rite:

for every ϵ > 0, there exists ẟ > 0 such that

 Theorem

Limit rules

Let f(x,y) and g(x,y) be defined for all (x, y) ≠ (a, b) in a neighborhood around (a, b), and assume the neighborhood is completely inside the domain of f. Assume L and M are real numbers such that

,

Then each of the following statements is true:

Constant rule

Identity rules

Sum rule

Difference rule

Constant multiple rule

Product rule

Quotient rule

Power rule

for any positive integer n                                                                                                                               

Root rule:

for all L if n is odd and positive and for all L⪈ 0 if n is even and positive for L≽ 0 for all (x,y) ≠ (a, b)

in neighborhood of (a, b)

Example 

Find each of the following limits: 

Solution

Let's first use the sum and difference rules to separate the terms:

                                                                                                                     

Let's use the constant multiple rule on the second, third, fourth and fifth limit:

Let's use the power rule on the second and third limit then the product rule on the second limit

Let's use the identity rule on the first six limits and the the constant rule on the last limit:

b. In this example we need to make sure that the limit of the denominator is different of zero. Upon calculating the limit of the denominator we find:

It is different of zero. Now let's calculate the limit of the numerator:

Finally applying the quotient rule, we find:

Practice

Evaluate the following limit:

Level curve. Making a contour map (continued)

 In this post we are going to solve another example of level curve and contour map.

Example

Solution

Let's substitute c by 0 in the function:

Let's square both sides and multiply by -1                                                                                                       

                                                                                                           

Let's put the x terms together, the y terms together and 8 to both sides:                                                               

  

Let's group the pairs of terms containing the same variable in parentheses and put 4 in factor                   

from the first  pair                                                                                                                                            

Let's complete the square of the expressions in parentheses and add the correct terms to the                       

 hand right side                                                                                                                                                                    

 

Let's factor the left hand side and  simplify the right hand:                                                                        

Let's divide both sides by 16:                                                                                                                        

This equation equation describes an ellipse centered at (-1,2). Here is its graph:                                         

                           

                                                     

                                                                              

We can continue to find level curves for c=1, c= 2, c= 3, c= 4.                                                            

For example for c= 1, the equation of the ellipse is:                                                                                 

    (x - 1)²/4 + (y + 2)²/4 = 11/4. We have a different equation but still an ellipse centered at (1, -2)               

For c = 4, the level curve is the point (-1, 2)                                                                                          

The domain is the set of (x,y) such as the expression under the radical is greater than or equal to 0 meaning that 8+8x-4y-4x²-y≥ 0. Solving this inequality leads to (x - 1)²/4 + (y + 2)²/4 ≤ 4.

 All pair(x,y) that satisfy this inequality are found inside the ellipse described by the expression 

 (x - 1)²/4 + (y + 2)²/4 . 

The domain is the set of pairs (x,y) that are located inside the ellipse. The range is the the set of values   for c. These values are 0, 1, 3, 3. 4. They constitute the range of the function.                                                                                                                                                                      

Practice 

Find and graph the level curve of the following function: g(x,y) =  x² + y² - 6x +2y corresponding 

to c = 15