Mathematics (from Greek μάθημα máthēma, “knowledge, study, learning”) is the study of topics such as quantity (numbers), structure, space, and change. There is a range of views among mathematicians and philosophers as to the exact scope and definition of mathematics
1. Chapter 8: Functions of
Several Variables
Section 8.2
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
2. 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
ifthatso0aexiststhere
0eachforthatmeans)(limstatementThe
number.realabeLletandc)atpossibly(except
ccontainingintervalopenanondefinedfunctionabefLet
εδ
δ
ε
<−<−<
>
>=
→
Lxfcx
Lxf
cx
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
δ+cδ−c
( )
L
ε+L
ε−L
()
x is
the
input
f(x) is the
output
3. Just to refresh your memory, consider the following limits.
?
0
4
4)2(
22
4
2
lim 222
→
−
=
−−
−−
=
−
−
−→ x
x
x
Good job if you saw this as “limit does not exist”
indicating a vertical asymptote at x = -2.
?
0
0
4)2(
22
4
2
lim 222
→=
−
−
=
−
−
→ x
x
x
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
1
lim
)2)(2(
2
lim
4
2
lim
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
x
xf
4. 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
5. 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
yx
The table
presents
evidence
that the
limit will
hold, but
not proof.
For proof
we have
to go back
to epsilon
and delta.
6. 3),( 22
++== yxyxfz
x
y
z
Definition of a Limit
.L-y)f(x,then)()(0
ifsuch that0aexiststhere0every
forthatmeans),(limstatement
Theitself.b)(a,atpossiblyexceptb),(a,
centerwithcircleaofinteriorthethroughout
definedbevariablestwoofffunctionaLet
22
),(),(
εδ
δε
<<−+−<
>>
=
→
byax
Lyxf
bayx
13),(lim
?
)3,1(),(
=
−→
yxf
yx
In the context of the limit we examined,
suppose that .25.=ε
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.
δ
Center
(-1,3)
(x,y)
25.13),( <−yxf
7. Definition of Continuity of a Function of Two Variables
A 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
),(),(
bafyxf
bayx
=
→
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
yxf
yxgfyxgyxfyxfg
yxgyxfyxgfyxfkyxkf
8. 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.
)).,(()),((limandb)(a,atcontinuousis
)),((),)((functionncompositiothen the
b),f(a,atcontinuousisgandb)(a,atcontinuousisfIf
),(),(
bafgyxfg
yxfgyxfg
bayx
=
=
→
9. Example 1. Find the limit and discuss the continuity of
the function.
yx
x
yx +→ 2
lim
)2,1(),(
Solution
2
1
4
1
2)1(2
1
2
lim
)2,1(),(
==
+
=
+→ yx
x
yx
The function will be continuous when 2x+y > 0.
Example 2. Find the limit and discuss the continuity of
the function.
yx
x
yx +→ 2
lim
)2,1(),(
Solution
4
1
2)1(2
1
2
lim
)2,1(),(
=
+
=
+→ yx
x
yx
The function will be continuous when
The function will not be defined when y = -2x.
.02 ≠+ yx
10. 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
1
lim 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.
11. x
y
z
)ln(
2
1 22
yxz +
−
=∞=+
−
→
)ln(
2
1
lim 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:
12. 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.