Download Sertöz Theorem Example 1: Example 2: Example 3:

Sertöz Theorem
Theorem: [Sertöz Theorem :
d be positive real numbers.
Version 1
] Let a and b be nonnegative real numbers, and c and
a b
• If
+ > 1 , then
c d
|x|a |y|b
=0.
lim
(x,y)→(0,0) |x|c + |y|d
a b
• If
+ ≤ 1 , then
c d
|x|a |y|b
lim
does not exist.
(x,y)→(0,0) |x|c + |y|d
We will prove this version below. Here is a slightly dierent, but equivalent version:
Theorem: [Sertöz Theorem :
positive even integers.
Version 2
] Let a and b be nonnegative integers, and c and d be
• If
a b
+ > 1 , then
c d
xa y b
=0.
(x,y)→(0,0) xc + y d
• If
a b
+ ≤ 1 , then
c d
xa y b
does not exist.
(x,y)→(0,0) xc + y d
lim
lim
◦::::◦
Example 1:
lim
(x,y)→(0,0)
xy/(x2 + y 2 ) does not exist by the
Sertöz Theorem
as a = 1, b = 1,
c = 2, d = 2, and a/c + b/d = 1 ≤ 1.
Example 2:
lim
(x,y)→(0,0)
xy 2 /(x2 + y 2 ) = 0 by the Sertöz Theorem as a = 1, b = 2, c = 2, d = 2,
and a/c + b/d = 3/2 > 1.
Example 3:
lim
(x,y)→(0,0)
xy 2 /(x2 + y 4 ) does not exist by the
Sertöz Theorem
as a = 1, b = 2,
c = 2, d = 4, and a/c + b/d = 1 ≤ 1.
Example 4: Let f (x, y) = xy 2 /(x2 − xy + y 2 ) . Neither version of the Sertöz Theorem can
be used for lim f (x, y) . Instead we observe that x2 − xy + y 2 = (x − y/2)2 + 3y 2 /4 ≥
(x,y)→(0,0)
3y /4 ≥ 0, and hence
2
2
xy 2
≤ |x|y = 4 |x|
0≤ 2
x − xy + y 2 3y 2 /4
3
for all (x, y) ̸= (0, 0). (Note that although we need y ̸= 0 for the middle step, the
nal inequality is true for all (x, y) ̸= (0, 0).) Now the Sandwich/Squeeze Theorem gives
lim xy 2 /(x2 − xy + y 2 ) = 0 .
(x,y)→(0,0)
Example 5: Let f (x, y) = x2 y 2 /(x3 + y 3 ) . Again neither version of the Sertöz Theorem can
be used for lim f (x, y) . Consider the curve dened by the equation y = −x + x3 . Then
(x,y)→(0,0)
lim
(x,y)→(0,0)
y=−x+x3
along the curve
(x2 − 1)2
x2 (−x + x3 )2
=
lim
x→0 x(x4 − 3x2 + 3)
x→0 x3 + (−x + x3 )3
f (x, y) = lim f (x, −x + x3 ) = lim
x→0
and this one-variable limit does not exist as the numerator goes to 1, but the denominator
goes to 0 as x goes to 0. Therefore the two-variable limit lim x2 y 2 /(x3 + y 3 ) does not
(x,y)→(0,0)
exist by the 1-Path Test .
Remark 1: Here is another way to show that this limit does not exist:
lim
x2 · 02
= lim 0 = 0
x→0 x3 + 03
x→0
f (x, y) = lim f (x, 0) = lim
x→0
(x,y)→(0,0)
x-axis
along the
lim
(x,y)→(0,0)
y=−x+x2
x2 (−x + x2 )2
(x − 1)2
1
=
lim
=
3
2
3
2
x→0 x + (−x + x )
x→0 x − 3x + 3
3
f (x, y) = lim f (x, −x+x2 ) = lim
x→0
along the curve
As 1/3 ̸= 0, the two-variable limit
Remark 2: Note that
lim
(x,y)→(0,0)
lim
(x,y)→(0,0)
x2 y 2 /(x3 + y 3 ) does not exist by the 2-Path
x2 y 2 /(|x|3 + |y|3 ) = 0 by the rst version of the
Test
.
Sertöz
as a = 2, b = 2, c = 3, d = 3, and a/c + b/d = 4/3 > 1. But this is the limit
of a dierent function from the one considered in Example 5.
Theorem
◦::::◦
Now we prove the theorem:
Proof of the Sertöz Theorem : Version 1 : Let f (x, y) = |x|a |y|b /(|x|c + |y|d ) where a and b
are nonnegative real numbers, and c and d are positive real numbers.
Case 1 : Assume that a/c + b/d < 1. Then:
lim
along
f (x, y) = lim f (x, |x|
c/d
(x,y)→(0,0)
c/d
the curve y=|x|
x→0
(
|x|a · |x|c/d )b
1
lim |x|(a/c+b/d−1)c = ∞
) = lim
)d = x→0
x→0
c
c/d
2
|x| + (|x|
as (a/c + b/d − 1)c < 0. As the limit as we approach (0, 0) along the curve y = |x|c/d does not
exist, the 2-variable limit at (0, 0) does not exist either by the 1-Path Test .
Case 2 : Assume that a/c + b/d = 1. Then:
lim
along
( c/d b
a
|x|
·
|x| )
1
1
= lim |x|(a/c+b/d−1)c =
f (x, y) = lim f (x, |x|c/d ) = lim
)
d
x→0
x→0
2 x→0
2
|x|c + (|x|c/d
c/d
(x,y)→(0,0)
the curve y=|x|
as (a/c + b/d − 1)c = 0. On the other hand,
lim
(x,y)→(0,0)
x-axis
along the
|x|a · 0b
= lim 0 = 0
x→0 |x|c + 0d
x→0
f (x, y) = lim f (x, 0) = lim
x→0
if b > 0, and as 1/2 ̸= 0, the 2-variable limit at (0, 0) does not exist by the 2-Path
b = 0, then the limit along the y -axis is 1/2 ̸= 0, and the same conclusion follows.
. If
Test
: Assume that a/c + b/d > 1. In this case we will show that the limit is 0 using the
Sandwich/Squeeze Theorem . First assume that b > d. Then we have the following sandwich
Case 3
|x|a |y|b
|y|d
a
b−d
=
|x|
|y|
≤ |x|a |y|b−d
|x|c + |y|d
|x|c + |y|d
and the result follows. Now assume that b ≤ d. This time we have the following sandwich
)
(
c/d b
|y|/|x|
|x|a |y|b
(a/c+b/d−1)c
= |x|(a/c+b/d−1)c
0≤ c
)d ≤ |x|
(
c/d
|x| + |y|d
1 + |y|/|x|
0≤
and again the result follows. (Note that although the middle step will require x ̸= 0, the nal
inequality is still true when x = 0.) Here we used the fact that 0 ≤ tb /(1 + td ) ≤ 1 for t ≥ 0
and 0 ≤ b ≤ d. This can be seen as follows:
• If 0 ≤ t ≤ 1 then 0 ≤ tb ≤ 1 ≤ 1 + td and hence 0 ≤ tb /(1 + td ) ≤ 1.
• If t > 1, then 1 < tb ≤ td < 1 + td and hence 0 ≤ tb /(1 + td ) ≤ 1.
This nishes the proof.
Last revision: March 14, 2016