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Functions of Several Variables
Limits and Continuity
Written by Richard Gill
Associate 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 ,0
if 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
( )
L
L
L
( )
x is the input
f(x) is the output
Just to refresh your memory, consider the following limits.
?0
4
4)2(
22
4
2lim
222
x
xx
Good job if you saw this as “limit does not exist” indicating a vertical asymptote at x = -2.
?0
0
4)2(
22
4
2lim
222
x
xx
This 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……
4
1
2
1lim
)2)(2(
2lim
4
2lim
2
222
x
xx
x
x
x
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
4
2)(
2
x
xxf
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
z
Definition 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 ),(
),(),(/ )],()[,(),(
),(),(),( )],([),(
yxg
yxfyxgfyxgyxfyxfg
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
),(),(bafgyxfg
yxfgyxfg
bayx
Example 1. Find the limit and discuss the continuity of the function.
yx
xyx 2lim
)2,1(),(
Solution
2
1
4
1
2)1(2
1
2lim
)2,1(),(
yx
xyx
The function will be continuous when 2x+y > 0.
Example 2. Find the limit and discuss the continuity of the function.
yx
xyx 2lim
)2,1(),(
Solution
4
1
2)1(2
1
2lim
)2,1(),(
yx
xyx
The function will be continuous when
The function will not be defined when y = -2x.
.02 yx
x
y
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(2
1 22 yxz
)ln(2
1lim 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
1.96
4.26
-0.69
0
0.69
2.30
4.61
Calculate these values yourself. Then click to confirm.
x
y
z)ln(
2
1 22 yxz
)ln(
2
1lim 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
1.96
4.26
-0.69
0
0.69
2.30
4.61
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:
For comments on this presentation you may email the author Professor Richard Gill [email protected] or the publisher of the VML, Dr. Julia Arnold at [email protected].
