S E C T I O N
15.1
Functions of Two or More Variables
(ET Section 14.1)
625
⇒
x
2
+
y
2
≥
2

xy

⇒
x
2
+
y
2
2
≥

xy

Therefore, (1) gives

c
 =
xy
x
2
+
y
2
+
1
=

xy

x
2
+
y
2
+
1
≤
x
2
+
y
2
2
x
2
+
y
2
+
1
=
1
2
x
2
+
y
2
x
2
+
y
2
+
1
<
1
2
·
1
=
1
2
That is,

c

<
1
2
. It follows that for
c
≥
1
2
the level curve is empty.
x
y
z
We now show that if

c

<
1
2
the level curve is a hyperbola.
x
y
z
Let
t
=
y
and
s
=
x

y
2
c
. Then for

c

<
1
2
the equation of the level curve (2) can be rewritten as
1
4
c
2

1
t
2

s
2
=
1
t
2
2
c
√
1

4
c
2
2

s
2
=
1
This is the equation of a hyperbola in standard position in the
ts
plane. The equation (2) for

c

<
1
2
is the equation of a
hyperbola obtained by a change of variables for a hyperbola in standard position, in the
xy
plane. For
c
=
0
.
5 the level
curve is empty:
x
y
z
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626
C H A P T E R
15
DIFFERENTIATION IN SEVERAL VARIABLES
(ET CHAPTER 14)
15.2 Limits and Continuity in Several Variables
(ET Section 14.2)
Preliminary Questions
1.
What is the difference between
D
(
P
,
r
)
and
D
*
(
P
,
r
)
?
SOLUTION
D
(
P
,
r
)
is the open disk of radius
r
and center
(
a
,
b
)
. It consists of all points distanced less than
r
from
P
,
hence
D
(
P
,
r
)
includes the point
P
.
D
*
(
P
,
r
)
consists of all points in
D
(
P
,
r
)
other than
P
itself.
2.
Suppose that
f
(
x
,
y
)
is continuous at
(
2
,
3
)
and that
f
(
2
,
y
)
=
y
3
for
y
=
3. What is the value
f
(
2
,
3
)
?
SOLUTION
f
(
x
,
y
)
is continuous at
(
2
,
3
)
, hence the following holds:
f
(
2
,
3
)
=
lim
(
x
,
y
)
→
(
2
,
3
)
f
(
x
,
y
)
Since the limit exists, we may compute it by approaching
(
2
,
3
)
along the vertical line
x
=
2. This gives
f
(
2
,
3
)
=
lim
(
x
,
y
)
→
(
2
,
3
)
f
(
x
,
y
)
=
lim
y
→
3
f
(
2
,
y
)
=
lim
y
→
3
y
3
=
3
3
=
27
We conclude that
f
(
2
,
3
)
=
27.
3.
Suppose that
Q
(
x
,
y
)
is a function such that
1
Q
(
x
,
y
)
is continuous for all
(
x
,
y
)
. Which of the following statements
are true?
(a)
Q
(
x
,
y
)
is continuous at all
(
x
,
y
)
.
(b)
Q
(
x
,
y
)
is continuous for
(
x
,
y
)
=
(
0
,
0
)
.
(c)
Q
(
x
,
y
)
=
0 for all
(
x
,
y
)
.
SOLUTION
All three statements are true. Let
f
(
x
,
y
)
=
1
Q
(
x
,
y
)
. Hence
Q
(
x
,
y
)
=
1
f
(
x
,
y
)
.
(a)
Since
f
is continuous,
Q
is continuous whenever
f
(
x
,
y
)
=
0. But by the definition of
f
it is never zero, therefore
Q
is continuous at all
(
x
,
y
)
.
(b)
Q
is continuous everywhere including at
(
0
,
0
)
.
(c)
Since
f
(
x
,
y
)
=
1
Q
(
x
,
y
)
is continuous, the denominator is never zero, that is,
Q
(
x
,
y
)
=
0 for all
(
x
,
y
)
.
Moreover, there are no points where
Q
(
x
,
y
)
=
0. (The equality
Q
(
x
,
y
)
=
(
0
,
0
)
is meaningless since the range of
Q
consists of real numbers.)
4.
Suppose that
f
(
x
,
0
)
=
3 for all
x
=
0 and
f
(
0
,
y
)
=
5 for all
y
=
0. What can you conclude about
lim
(
x
,
y
)
→
(
0
,
0
)
f
(
x
,
y
)
?
SOLUTION
We show that the limit lim
(
x
,
y
)
→
(
0
,
0
)
f
(
x
,
y
)
does not exist. Indeed, if the limit exists, it may be computed
by approaching
(
0
,
0
)
along the
x
axis or along the
y
axis. We compute these two limits:
lim
(
x
,
y
)
→
(
0
,
0
)
along
y
=
0
f
(
x
,
y
)
=
lim
x
→
0
f
(
x
,
0
)
=
lim
x
→
0
3
=
3
lim
(
x
,
y
)
→
(
0
,
0
)
along
x
=
0
f
(
x
,
y
)
=
lim
y
→
0
f
(
0
,
y
)
=
lim
y
→
0
5
=
5
Since the limits are different,
f
(
x
,
y
)
does not approach one limit as
(
x
,
y
)
→
(
0
,
0
)
, hence the limit lim
(
x
,
y
)
→
(
0
,
0
)
f
(
x
,
y
)
does not exist.
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 Spring '08
 Hubscher
 Calculus, Derivative, lim, Continuous function, Limit of a function

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