1551 limits and continuity

Chapter 8: Functions of Several Variables

Section 8.2Limits and Continuity

Written by Richard GillAssociate Professor of Mathematics

Tidewater Community College, Norfolk Campus, Norfolk, VA

With Assistance from a VCCS LearningWare Grant

This section will extend the properties of limits and continuity from the familiar function of one variable to the new territory of functions of two or three variables. I hate to bring up painful memories but here is the formal definition of a limit back when we were dealing with functions of one variable.

.)( then ,0if that so 0 a exists there

0each for that means )(limstatement The number. real a be Llet and c)at possibly (except

c containing intervalopen an on definedfunction a be fLet

Lxfcx

Lxfcx

In less formal language this means that, if the limit holds, then f(x) gets closer and closer to L as x gets closer and closer to c. c

cc

( )

LL

L

( ) x is the input

f(x) is the output

Just to refresh your memory, consider the following limits.

?04

4)2(22

42lim 222

xx

xGood job if you saw this as “limit does not exist” indicating a vertical asymptote at x = -2.

?00

4)2(22

42lim 222

xx

xThis limit is indeterminate. With some algebraic manipulation, the zero factors could cancel and reveal a real number as a limit. In this case, factoring leads to……

41

21lim

)2)(2(2lim

42lim

2

222

x

xxx

xx

x

xx

The limit exists as x approaches 2 even though the function does not exist. In the first case, zero in the denominator led to a vertical asymptote; in the second case the zeros cancelled out and the limit reveals a hole in the graph at (2, ¼).

x

y

42)( 2

xxxf

The concept of limits in two dimensions can now be extended to functions of two variables. The function below uses all points on the xy-plane as its domain.

3),( 22 yxyxfz

x

y

z

If the point (2,0) is the input, then 7 is the output generating the point (2,0,7).

(2,0)

(2,0,7)

If the point (-1,3) is the input, then 13 is the output generating (-1,3,13).

(-1,3)

(-1,3,13)

For the limit of this function to exist at (-1,3), values of z must get closer to 13 as points (x,y) on the xy-plane get closer and closer to (-1,3). Observe the values in the table to see if it looks like the limit will hold.

13),(lim?

)3,1(),(

yxf

yx

The concept of limits in two dimensions can now be extended to functions of two variables. The function below uses all points on the xy-plane as its domain.

3),( 22 yxyxfz

x

y

z

(2,0)

(2,0,7)

(-1,3)

(-1,3,13)

For the limit of this function to exist at (-1,3), values of z must get closer to 13 as points (x,y) on the xy-plane get closer and closer to (-1,3). Observe the values in the table to see if it looks like the limit will hold.

(x,y) (x,y,z)(0,0) (0,0,3)(-1,1) (-1,1,5)(-1,2) (-1,2,8)(-1,2.5) (-1,2.5,10.25)(-1, 2.9) (-1, 2.9, 12.41)(-.9,3) (-.9,3,12.81)(-1.1,3) (-1.1,3,13.21)

13),(lim?

)3,1(),(

yxf

yxThe table presents evidence that the limit will hold, but not proof. For proof we have to go back to epsilon and delta.

3),( 22 yxyxfz

x

y

zDefinition of a Limit

.L-y)f(x, then )()(0

ifsuch that 0 a exists there0every

for that means ),(limstatement The itself. b)(a,at possibly except b),(a,center with circle a ofinterior thethroughout

defined be variables twoof ffunction aLet

22

),(),(

byax

Lyxfbayx

13),(lim?

)3,1(),(

yxf

yx

In the context of the limit we examined, suppose that .25.

If the limit holds, we should be able to construct a circle centered at (-1,3) with as the radius and any point inside this circle will generate a z value that is closer to 13 than .25.

Center (-1,3)

(x,y)

25.13),( yxf

Definition of Continuity of a Function of Two VariablesA function of two variables is continuous at a point (a,b) in an open region R if f(a,b) is equal to the limit of f(x,y) as (x,y) approaches (a,b). In limit notation:

).,(),(lim),(),(

bafyxfbayx

The function f is continuous in the open region R if f is continuous at every point in R.

The following results are presented without proof. As was the case in functions of one variable, continuity is “user friendly”. In other words, if k is a real number and f and g are continuous functions at (a,b) then the functions below are also continuous at (a,b):

0b)g(a, if ),(),(),(/ )],()[,(),(

),(),(),( )],([),(

yxgyxfyxgfyxgyxfyxfg

yxgyxfyxgfyxfkyxkf

The conclusions in the previous slide indicate that arithmetic combinations of continuous functions are also continuous—that polynomial and rational functions are continuous on their domains.

Finally, the following theorem asserts that the composition of continuous functions are also continuous.

)).,(()),((lim and b)(a,at continuous is)),((),)((function n compositio then the

b),f(a,at continuous is g and b)(a,at continuous is f If

),(),(bafgyxfgyxfgyxfg

bayx

Example 1. Find the limit and discuss the continuity of the function.

yxx

yx 2lim

)2,1(),(

Solution

21

41

2)1(21

2lim

)2,1(),(

yxx

yx

The function will be continuous when 2x+y > 0.

Example 2. Find the limit and discuss the continuity of the function.

yxx

yx 2lim

)2,1(),(

Solution41

2)1(21

2lim

)2,1(),(

yxx

yx

The function will be continuous whenThe function will not be defined when y = -2x.

.02 yx

xy

z

Example 3. Use your calculator to fill in the values of the table below. The first table approaches (0,0) along the line y=x. The second table approaches (0,0) along the line x=0. (If different paths generate different limits, the official limit does not exist.) Use the patterns to determine the limit and discuss the continuity of the function.

)ln(21 22 yxz

)ln(

21lim 22

)0,0(),(yx

yx

(x,y) z(2,2)(1,1)(.5,.5)(.1,.1)(.01,.01)

(x,y) z(0,2)(0,1)(0,.5)(0,.1)(0,.01)

0.35

-1.04-0.35

1.964.26

-0.6900.692.304.61

Calculate these values yourself. Then click to confirm.

xy

z)ln(

21 22 yxz

)ln(21lim 22

)0,0(),(yx

yx

(x,y) z(2,2)(1,1)(.5,.5)(.1,.1)(.01,.01)

(x,y) z(0,2)(0,1)(0,.5)(0,.1)(0,.01)

0.35

-1.04-0.35

1.964.26

-0.6900.692.304.61

Solution: from the graph, it appears that z values get larger and larger as (x,y) approaches (0,0). Conceptually, we would expect values of the natural log function to approach infinity as the inputs approach 0. The numeric values in the table appear to agree. The conclusion:

This is the end of lesson 8.2. There will be three sets of exercises available for this lesson. Each lesson will have 6 exercises. Print the exercises first and work out the answers. When you submit the answers on Blackboard, each exercise worked correctly will add to your thinkwell exercise total.