636
C H A P T E R
15
DIFFERENTIATION IN SEVERAL VARIABLES
(ET CHAPTER 14)

2
y
x
z
0
0
0
2
1
2
1
2

1

1

2

2
The graph supports our result in part (a).
For
g
(
x
,
y
)
=
x y
x
2
+
y
2
:
(a)
g
(
x
,
y
)
=
x y
x
2
+
y
2
is deﬁned for
(
x
,
y
)
y=
(
0
,
0
)
. In Exercise 37 we showed that
lim
(
x
,
y
)
→
(
0
,
0
)
x
a
y
b
x
2
+
y
2
does not exist
if
a
+
b
≤
2. In particular the limit does not exist if
a
=
b
=
1, that is,
lim
(
x
,
y
)
→
(
0
,
0
)
g
(
x
,
y
)
does not exist. It follows that
the domain of
g
(
x
,
y
)
cannot be extended to all of
R
2
to result in a continuous function.
(b)
The graph of
g
(
x
,
y
)
conﬁrms our observation:
y
x
z
0.5
0.25
0
0
0
1
2

1

2
1
2

1

2

0.25

0.5
43.
The function
f
(
x
,
y
)
=
x
2
y
x
4
+
y
2
provides an interesting example where the limit as
(
x
,
y
)
→
(
0
,
0
)
does
not exist, even though the limit along every line
y
=
mx
exists and is zero (Figure 9).
(a)
Show that the limit along any line
y
=
exists and is equal to 0.
(b)
Calculate
f
(
x
,
y
)
at the points
(
10

1
,
10

2
)
,
(
10

5
,
10

10
)
,
(
10

20
,
10

40
)
. Do not use a calculator.
(c)
Show that
lim
(
x
,
y
)
→
(
0
,
0
)
f
(
x
,
y
)
does not exist.
Hint:
Compute the limit along the parabola
y
=
x
2
.
SOLUTION
(a)
Substituting
y
=
in
f
(
x
,
y
)
=
x
2
y
x
4
+
y
2
, we get
f
(
x
,
)
=
x
2
·
x
4
+
(
)
2
=
3
x
2
(
x
2
+
m
2
)
=
x
2
+
m
2
We compute the limit as
x
→
0 by substitution:
lim
x
→
0
f
(
x
,
)
=
lim
x
→
0
x
2
+
m
2
=
m
·
0
0
2
+
m
2
=
0
(b)
We compute
f
(
x
,
y
)
at the given points:
f
(
10

1
,
10

2
)
=
10

2
·
10

2
10

4
+
10

4
=
10

4
2
·
10

4
=
1
2
f
(
10

5
,
10

10
)
=
10

10
·
10

10
10

20
+
10

20
=
10

20
2
·
10

20
=
1
2
f
(
10

20
,
10

40
)
=
10

40
·
10

40
10

80
+
10

80
=
10

80
2
·
10

80
=
1
2