Transcript Limits and Continuity - Tidewater Community College

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
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.
Let f be a functiondefined on an openintervalcontainingc
(exceptpossibly at c) and let L be a real number.
T hestatementlim f ( x)  L means thatfor each   0
x c
thereexistsa   0 so thatif
L
L 
x is
the
input
)
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.
L
(
0  x  c   , then f ( x)  L   .
f(x) is the
output
(
c 
c
)
c 
Just to refresh your memory, consider the following limits.
x2
22
4


?
x 2 x 2  4
(2) 2  4
0
lim
lim
x 2
x2
22
0


?
x 2  4 (2) 2  4 0
x2
x2

lim

x2 x 2  4
x  2 ( x  2)( x  2)
1
1
lim

x2 x  2
4
lim
Good job if you saw this as “limit does not exist”
indicating a vertical asymptote at x = -2.
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……
y
f ( x) 
x2
x2  4
x
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).
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.
z
z  f ( x, y)  x 2  y 2  3
If the point (2,0) is the input, then 7 is
the output generating the point (2,0,7).
If the point (-1,3) is the input, then
13 is the output generating (-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.
(-1,3,13)
(2,0,7)
(-1,3)
(2,0)
?
lim
( x , y )  ( 1, 3)
f ( x, y )  13
x
y
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.
z
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.
?
lim
The table
presents
evidence
that the
limit will
hold, but
not proof.
For proof
we have
to go back
to epsilon
and delta.
( x , y )  ( 1, 3)
z  f ( x, y)  x 2  y 2  3
(-1,3,13)
f ( x, y )  13
(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)
(2,0,7)
(-1,3)
(2,0)
x
y
Definition of a Limit
z  f ( x, y)  x 2  y 2  3
z
Let a functionf of two variablesbe defined
throughouttheinteriorof a circle with center
(a, b), exceptpossibly at (a, b) itself.T he
statement lim
( x , y )  ( a ,b )
f ( x, y )  L means thatfor
every  0 thereexistsa   0 such thatif
f ( x, y) 13  .25
0  ( x  a)  ( y  b)   then f(x,y) - L   .
2
2
In the context of the limit we examined,
?
lim
( x , y )  ( 1, 3)
f ( x, y )  13
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.
(x,y)
Center
(-1,3)
x
y
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 f ( x, y )  f (a, b).
( x , y )  ( a ,b )
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):
kf ( x, y )  k[ f ( x, y )]
fg ( x, y )  f ( x, y )[g ( x, y )]
f  g ( x , y )  f ( x, y )  g ( x, y )
f ( x, y )
f / g ( x, y ) 
if g(a, b)  0
g ( x, y )
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.
If f is continuousat (a, b) and g is continuousat f(a,b),
then thecomposition function( g  f )(x, y )  g ( f ( x, y ))
is continuousat (a, b) and
lim
( x , y )  ( a ,b )
g ( f ( x, y ))  g ( f (a, b)).
Example 1. Find the limit and discuss the continuity of
the function.
x
lim
( x , y )(1, 2 )
2x  y
Solution
lim
( x , y )(1, 2 )
x
1
1
1



2x  y
2(1)  2
4 2
The function will be continuous when 2x+y > 0.
Example 2. Find the limit and discuss the continuity of
x
the function.
lim
( x , y ) (1, 2 ) 2 x  y
Solution
x
1
1
lim


( x , y )(1, 2 ) 2 x  y
2(1)  2 4
The function will be continuous when 2 x  y  0.
The function will not be defined when y = -2x.
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.
z
1
ln( x 2  y 2 )
( x , y )( 0 , 0 ) 2
lim
(x,y)
z
(x,y)
(2,2)
-1.04
(0,2)
-0.69
(1,1)
-0.35
(0,1)
0
(.5,.5)
0.35
(0,.5)
0.69
(.1,.1)
1.96
(0,.1)
2.30
(.01,.01) 4.26
z
1
ln( x 2  y 2 )
2
z
(0,.01) 4.61
y
x
Calculate these values yourself. Then click to confirm.
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:
z
1
lim
ln( x 2  y 2 )  
( x , y ) ( 0 , 0 ) 2
(x,y)
z
(x,y)
(2,2)
-1.04
(0,2)
-0.69
(1,1)
-0.35
(0,1)
0
(.5,.5)
0.35
(0,.5)
0.69
(.1,.1)
1.96
(0,.1)
2.30
(.01,.01) 4.26
z
(0,.01) 4.61
z
1
ln( x 2  y 2 )
2
y
x
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.