CALCULUS AND ANALYTIC GEOMETRY III
SOLUTIONS TO SAMPLE TEST 2
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1. The limit does not exist, for if it did, then we would obtain the same limit when
approaching (0
,
0) along any curve. However, this is not so, since if we approach
along the line
y
=
x
3
, we have
lim
(
x,x
3
)
→
(0
,
0)
x
3
.x
3
(
x
3
)
2
+
x
6
= lim
x
→
0
x
6
2
x
6
=
1
2
,
while for the line
x
= 0, we have
lim
(0
,y
)
→
(0
,
0)
0
3
.y
0
6
+
y
2
= lim
y
→
0
0 = 0
.
Since the limit of
f
(
x, y
) does not exist at (0
,
0), no choice of the value of
f
(0
,
0)
will result in a function which is continuous at (0
,
0), since for continuity, the limit
must exist in the ﬁrst place.
2.
f
x
(
x, y
) = 4
x
3

4
y
,
f
y
(
x, y
) = 4
y
3

4
x
. These partial derivatives exist every
where, so the only critical points are the points (
x, y
) where both partials are zero.
Thus we must solve the simultaneous equations:
(1)
±
x
3

y
= 0
y
3

x
= 0
.
Substituting, we get