Instructor: Longfei Li
Math 243 Lecture Notes
14.2 Limits and Continuity
Limits
We use the notation
lim
f (x, y) = L
(x,y)→(a,b)
to indicate that the value of L approach the number L as the point (x, y) approaches the point (a, b)
along any path that stays in the domain of f
Definition: Let f be a function of two variables whose domain D includes points arbitrarily close to
(a, b). Then we say that the limit of f (x, y) as (x, y) approaches (a, b) is L and we write
lim
f (x, y) = L
(x,y)→(a,b)
if forpevery number > 0 there is a corresponding number δ > 0 such that if (x, y) ∈ D and
0 < (x − a)2 + (y − b)2 < δ then |f (x, y) − L| ≤ .
Remark: Other notations lim f (x, y) = L and f (x, y) → L as (x, y) → (a, b).
x→0
y→0
Recall: for functions of a single variable, if
lim f (x) 6= lim f (x),
x→a−
x→a+
then lim f (x) does not exist (DNE).
x→a
For functions of two variables, the definition only refers to the distance from (x, y) to (a, b), does NOT
refer to the direction of approach.
If f (x, y) → L as (x, y) → (a, b) along any path C, then
lim
f (x, y) exists.
(x,y)→(a,b)
If f (x, y) → L1 as (x, y) → (a, b) along a path C1 and f (x, y) → L2 as (x, y) → (a, b) along a path C2 ,
where L1 6= L1 , then
lim f (x, y) does NOT exist.
(x,y)→(a,b)
Squeeze theorem also holds from functions of two variables.
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Example: Find
3x2 y
+ y2
lim
(x,y)→(0,0) x2
Solution: Since
3x2 y 2
≤ 3|y|
x + y2
We know that
lim
3|y| = 0
(x,y)→(0,0)
So by squeeze theorem
3x2 y
=0
(x,y)→(0,0) x2 + y 2
lim
Remark: This example shows how we can show limit exists and find the limiting value.
Example: Show that
x2 − y 2
(x,y)→(0,0) x2 + y 2
lim
does not exist.
Solution:
First, let (x, y) approaches (0, 0) along the x−axis, then y = 0 gives f (x, 0) =
x2
x2
= 1 for all x 6= 0. So
f (x, y) → 1 as (x, y) → (0, 0) along the x−axis
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Second, let (x, y) → (0, 0) along y−axis. Then x = 0, we have f (0, y) = −y
= −1 for all y 6= 0. So
y2
f (x, y) → −1 in this path.
f has two different limits along two different lines, thus the given limit does not exit.
Remark: This example shows how we normally do to show limit DNE
Example: If f (x, y) =
x2
xy
, does
+ y2
lim
f (x, y) exis?
(x,y)→(0,0)
Solution: If (x, y) → (0, 0) along any non-vertical line y = kx, then we have
f (x, y) = f (x, kx) =
So
f (x, y) →
k
kx2
=
2
2
2
x +k x
1 + k2
k
as (x, y) → (0, 0) along any line y = kx
1 + k2
The limit is different as (x, y) → (0, 0) along different lines y = kx. So the limit does NOT exist.
Sometimes, approaching the given point (a, b) along straight lines doesn’t work, we need to approach
(a, b) along some other curves.
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Example:
Show that
x2 y
DNE
+ y2
lim
(x,y)→(0,0) x4
Solution:
(x, y) → (0, 0) along lines y = kx, we have
f (x, y) =
kx3
→ 0, as (x, kx) → (0, 0)
x4 + k 2 x2
f has the same limiting value along every non-vertical lines through the origin. But this does not show
the given limit is 0, for if we let (x, y) → (0, 0) along y = x2 , we will get
f (x, x2 ) =
x4
1
= → 0 as (x, x2 ) → (0, 0)
4
4
x +x
2
So we got a different limiting value when approaching along the curve y = x2 . Therefore, the given
limit DNE.
Continuity
Definition: A function f is continuous at (a, b) if
lim
f (x, y) = f (a, b).
(x,y)→(0,0)
We say f is continuous on D if f is continuous at every point (a, b) ∈ D.
Polynomial Function: is a sum of terms of the form cxm y n , where c is constant and m, n are
nonnegative integers.
f (x, y) = x3 + 3x2 y 4 + 4y 2 − 6xy + 5
Polynomial functions are continuous on R2 .
Rational Function: is a ratio of polynomials.
x2 y 2 + 3x
x2 + y 2
g(x, y) =
Rational functions are continuous on its domain.
Example: evaluate
lim
(x2 y 3 − x3 y 2 + 3x + 2y)
(x,y)→(1,2)
Solution:
Polynomial is continuous on R2 , so the limit is
lim
(x2 y 3 − x3 y 2 + 3x + 2y) = 12 · 23 − 13 · 22 + 3 · 1 + 2 · 2 = 11
(x,y)→(1,2)
Example: Where is the function continuous:
(
f (x, y) =
3x2 y
,
x2 +y 2
if (x, y) 6= (0, 0)
0,
if (x, y) = (0, 0)
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Solution:
We know f (x, y) is continuous when (x, y) 6= (0, 0) since it’s a rational function there.
In addition, we knew from the first example
3x2 y
= 0 = f (0, 0)
(x,y)→(0,0) x2 + y 2
lim
So the function f (x, y) is continuous on R2
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