ASSIGNMENT 3
·
SOLUTIONS
MAT 473
·
SPRING 2012
Exercise 4.2.
Suppose
f
:
R
2
→
R
is defned by
f
(
x, y
)=
xy
2
x
2
+
y
2
iF (
x, y
)
°
=(0
,
0)
0i
F
(
x, y
)=(0
,
0)
.
Show that all partial derivatives oF
f
exist everywhere, and
f
is continuous everywhere, but
f
°
(0
,
0) still does not exist.
Proof.
Continuity at (
x, y
)
°
=(0
,
0) is clear, since
f
is defned as a rational Function at such
points. ±or continuity at (0
,
0), note that For (
x, y
)
°
=(0
,
0),

f
(
x, y
)

=

x

y
2
x
2
+
y
2
≤
x

,
so
f
(
x, y
)
→
0as(
x, y
)
→
0.
±or the partial derivatives, again there is no issue at (
x, y
)
°
=(
0
,
0). ±or the partial
derivatives at (0
,
0), compute directly:
D
1
f
(0
,
0) = lim
t
→
0
f
(0 +
t,
0)
−
f
(0
,
0)
t
=l
im
t
→
0
0
t
=0
and
D
2
f
(0
,
0) = lim
t
→
0
f
(0
,t
)
−
f
(0
,
0)
t
=l
im
t
→
0
0
t
=0
.
Thus, iF
f
were di²erentiable at (0
,
0), we’d have
[
f
°
(0
,
0)] =
°
00
±
,
which would imply that
f
°
(0
,
0) = 0, which in turn would imply that
lim
(
h,k
)
→
(0
,
0)
f
(0 +
h,
0+
k
)
−
f
(0
,
0)
−
0(
h, k
)
±
(
h, k
)
±
=0
.
However, this is not the case, since For example
lim
(
h,h
)
→
(0
,
0)
f
(0 +
h,
0+
h
)
−
f
(0
,
0)
−
0(
h, h
)
±
(
h, h
)
±
=l
im
h