Download 13.3: 24, 30, 32 24. Determine whether the function is continuous at

13.3: 24, 30, 32
24. Determine whether the function is continuous at (0, 0).
( 4 4
x +3y
(x, y) 6= (0, 0)
2
2 ,
f (x, y) = x +y
0,
(x, y) = (0, 0)
Answer. Setting x = r cos θ and y = r sin θ, we have
r4 (cos4 θ + 3 sin4 θ)
= lim r2 (cos4 θ + 3 sin4 θ) = 0
r→0
r→0
r2
lim f (x, y) = lim
(x,y)→0
because −4 ≤ (cos4 θ + 3 sin4 θ) ≤ 4 for all θ. Therefore, the function f (x, y) is
continuous at (0, 0).
30. Let f (x) = 2xy/(x2 + y 2 ). Calculate limy→0 f (0, y). Also, for any slope
−∞ < m < ∞, calculate limx→0 f (x, mx). Show that f (x, y) has a limit as
(x, y) approaches (0, 0) along any straight line path approaching the origin and
that the limit can be any number in the interval [−1, 1].
Proof. Along a straight line path to (0, 0), either x = 0 or y = mx. If x = 0,
lim
(x,y)→(0,0)
f (x, y) = lim f (0, y) = lim
y→0
y→0
0
= 0,
y2
and if y = mx,
lim
(x,y)→(0,0)
2mx2
2m
2m
= lim
=
.
x→0 x2 (1 + m2 )
x→0 1 + m2
1 + m2
f (x, y) = lim f (x, mx) = lim
x→0
2m
Now for any c ∈ [−1, 1], we want to find some m ∈ R such that c = 1+m
2 . This
2
is equivalent to finding solutions to cm + c − 2m = 0. If c √= 0, we can use
2
m = 0. If c 6= 0, the quadratic formula gives us solutions 1± c1−c , which are
real numbers precisely when c ∈ [−1, 0) ∪ (0, 1]. Therefore, f (x, y) has a limit
as (x, y) approaches (0, 0) along any straight line path approaching the origin
and that the limit can be any number in the interval [−1, 1].
32. Let f (x, y) = xy 3 /(x2 + y 6 ). Prove that along any straight line path
approaching the origin, the limit of f (x, y) is 0. However, show that there is
another path along which f (x, y) has a different limit as (x, y) approaches (0, 0).
Therefore, f (x, y) does not have a limit as (x, y) tends to (0, 0).
Proof. Along a straight line path to (0, 0), either x = 0 or y = mx. If x = 0,
lim
(x,y)→(0,0)
f (x, y) = lim f (0, y) = lim
y→0
y→0
0
= 0,
y6
and if y = mx,
lim
(x,y)→(0,0)
m3
m3 x 4
2
=
lim
x
= 0.
x→0
x→0 x2 (1 + m6 x4 )
1 + m6 x4
f (x, y) = lim f (x, mx) = lim
x→0
1
However, along the path y =
lim
(x,y)→(0,0)
√
3
x,
f (x, y) = lim f (x,
√
3
x→0
1
1
x2
= lim = .
x→0 2
x→0 2x2
2
x) = lim
Therefore, f (x, y) does not have a limit as (x, y) tends to (0, 0).
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