The concept of limit is a lot harder for functions of several variables than for just one. We show the more dramatric ways that a limit can fail.

1. Section 11.2
Limits and Continuity
Math 21a
February 29, 2008
Announcements
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Midterm I Review Session 3/5, 6–7:30pm in SC Hall D
Midterm I, 3/11, 7–9pm in SC Hall D
Image: kaet44

2. Outline
Introduction and definition
Rules of limits
Complications
Showing a limit doesn’t exist
Showing a limit does exist
Continuity
Worksheet

3. Where we’re going: derivatives of multivariable functions
Recall that if f is a function that takes real numbers to real
numbers,
f (x + h) − f (x)
f (x) = lim
h→0 h

4. Where we’re going: derivatives of multivariable functions
Recall that if f is a function that takes real numbers to real
numbers,
f (x + h) − f (x)
f (x) = lim
h→0 h
We want to do the same thing in more than one variable. So we
need to take limits in more than one dimension.

5. Definition
We write
lim f (x, y ) = L
(x,y )→(a,b)
and we say that the limit of f (x, y ) as (x, y ) approaches (a, b) is
L if we can make the values of f (x, y ) as close to L as we like by
taking the point (x, y ) to be sufficiently close to (a, b).

6. easy limits
lim x =a
(x,y )→(a,b)
lim y =b
(x,y )→(a,b)
lim c =c
(x,y )→(a,b)

7. Outline
Introduction and definition
Rules of limits
Complications
Showing a limit doesn’t exist
Showing a limit does exist
Continuity
Worksheet

8. Like regular limits, limits of multivariable functions can be
added
subtracted
multiplied
composed
divided, provided the limit of the denominator is not zero.

9. Limit of a Polynomial
Example
Find lim (x 5 + 4x 3 y − 5xy 2 )
(x,y )→(5,−2)

10. Limit of a Polynomial
Example
Find lim (x 5 + 4x 3 y − 5xy 2 )
(x,y )→(5,−2)
Solution
lim (x 5 + 4x 3 y − 5xy 2 ) = (5)5 + 4(5)3 (−2) − 5(5)(−2)2
(x,y )→(5,−2)
= 3125 + 4(125)(−2) − 5(5)(4)
= 2025.

11. Limit of a Rational Expression
Example
Compute
x2
lim .
(x,y )→(1,2) x 2 + y 2

12. Limit of a Rational Expression
Example
Compute
x2
lim .
(x,y )→(1,2) x 2 + y 2
Solution
x2 (1)2
lim =
(x,y )→(1,2) x 2 + y 2 (1)2 + (2)2
1
=
5

13. Outline
Introduction and definition
Rules of limits
Complications
Showing a limit doesn’t exist
Showing a limit does exist
Continuity
Worksheet

14. The only real problem is a limit where the denominator goes to
zero.
If the numerator goes to some number and the denominator
goes to zero then the quotient cannot have a limit.

15. The only real problem is a limit where the denominator goes to
zero.
If the numerator goes to some number and the denominator
goes to zero then the quotient cannot have a limit.
If on the other hand the numerator and denominator both go
to zero we have no clue. Most “interesting” limits come from
this. e.g.,
f (x + h) − f (x)
f (x) = lim
h→0 h

16. You probably remember this statement:
Fact
For a function f (x) of one variable,
lim f (x) = L ⇐⇒ lim+ f (x) = L and lim f (x) = L
x→a x→a x→a−

17. You probably remember this statement:
Fact
For a function f (x) of one variable,
lim f (x) = L ⇐⇒ lim+ f (x) = L and lim f (x) = L
x→a x→a x→a−
For functions of two variables, “left-hand limits” and “right-hand
limits” aren’t enough.

18. Showing a limit doesn’t exist
Theorem
Suppose lim f (x, y ) = L. Then the limit of f as
(x,y )→(a,b)
(x, y ) → (a, b) is L along all paths through (a, b).

19. Showing a limit doesn’t exist
Theorem
Suppose lim f (x, y ) = L. Then the limit of f as
(x,y )→(a,b)
(x, y ) → (a, b) is L along all paths through (a, b).
There are two contrapositives to this statement:

20. Showing a limit doesn’t exist
Theorem
Suppose lim f (x, y ) = L. Then the limit of f as
(x,y )→(a,b)
(x, y ) → (a, b) is L along all paths through (a, b).
There are two contrapositives to this statement:
If there is a path through (a, b) along which the limit does not
exist, the two-dimensional limit does not exist

21. Showing a limit doesn’t exist
Theorem
Suppose lim f (x, y ) = L. Then the limit of f as
(x,y )→(a,b)
(x, y ) → (a, b) is L along all paths through (a, b).
There are two contrapositives to this statement:
If there is a path through (a, b) along which the limit does not
exist, the two-dimensional limit does not exist
If there are two paths through (a, b) along which the limits
exist but disagree, the two-dimensional limit does not exist

22. Example
x
Show lim does not exist.
(x,y )→(0,0) x 2 + y2

23. Example
x
Show lim does not exist.
(x,y )→(0,0) x 2 + y2
Solution
Follow a path towards (0, 0) along the line y = 0.
x 1
lim f (x, 0) = lim = lim
x→0 x→0 x2 +0 2 x→0 x
x
which does not exist. So lim does not exist.
(x,y )→(0,0) x 2 + y2

24. Example
x
Show lim does not exist.
(x,y )→(0,0) x 2 + y2
Solution
Follow a path towards (0, 0) along the line y = 0.
x 1
lim f (x, 0) = lim = lim
x→0 x→0 x2 +0 2 x→0 x
x
which does not exist. So lim does not exist.
(x,y )→(0,0) x 2 + y2

25. We can see the problems in a graph.
2 2
x 1 1
2 + y2
= c ⇐⇒ x− + y2 =
x c 2c

26. We can see the problems in a graph.
2 2
x 1 1
2 + y2
= c ⇐⇒ x− + y2 =
x c 2c
1
0.5
0
-0.5
-1
-1 -0.5 0 0.5 1

27. We can see the problems in a graph.
2 2
x 1 1
2 + y2
= c ⇐⇒ x− + y2 =
x c 2c
1
0.5
0 10
5 1
0
-5 0.5
-0.5 -10
-1 0
-0.5
0 -0.5
-1 0.5
-1 -0.5 0 0.5 1
1 -1

28. Example
x2
Show lim does not exist.
(x,y )→(0,0) x 2 + y 2

29. Example
x2
Show lim does not exist.
(x,y )→(0,0) x 2 + y 2
Solution
Follow a path towards (0, 0) along the line y = 0:

30. Example
x2
Show lim does not exist.
(x,y )→(0,0) x 2 + y 2
Solution
Follow a path towards (0, 0) along the line y = 0:
x2
lim f (x, 0) = lim = lim 1 = 1
x→0 x→0 x 2 + 02 x→0

31. Example
x2
Show lim does not exist.
(x,y )→(0,0) x 2 + y 2
Solution
Follow a path towards (0, 0) along the line y = 0:
x2
lim f (x, 0) = lim = lim 1 = 1
x→0 x→0 x 2 + 02 x→0
Now follow a path towards (0, 0) along the line x = 0:

32. Example
x2
Show lim does not exist.
(x,y )→(0,0) x 2 + y 2
Solution
Follow a path towards (0, 0) along the line y = 0:
x2
lim f (x, 0) = lim = lim 1 = 1
x→0 x→0 x 2 + 02 x→0
Now follow a path towards (0, 0) along the line x = 0:
02
lim f (0, y ) = lim = lim 0 = 0
y →0 x→0 02 + y 2 x→0
So the limit as (x, y ) → (0, 0) cannot exist.

33. Again, we can see the problems in a graph.
x2 1−c
= c ⇐⇒ y = ± x
x2 + y2 c

34. Again, we can see the problems in a graph.
x2 1−c
= c ⇐⇒ y = ± x
x2 + y2 c
1
0.5
0
-0.5
-1
-1 -0.5 0 0.5 1

35. Again, we can see the problems in a graph.
x2 1−c
= c ⇐⇒ y = ± x
x2 + y2 c
1
0.5
0 1
0.75 1
0.5
0.25 0.5
-0.5 0
-1 0
-0.5
0 -0.5
-1 0.5
-1 -0.5 0 0.5 1 1 -1

36. Showing a limit does exist
This is often harder. No single method always works.
Example
Show
x3
lim = 0.
(x,y )→(0,0) x 2 + y 2

37. Showing a limit does exist
This is often harder. No single method always works.
Example
Show
x3
lim = 0.
(x,y )→(0,0) x 2 + y 2
Solution
From the last problem we know that
x2
0≤ ≤1
x2 + y2
for all x and y , not both 0.

38. Showing a limit does exist
This is often harder. No single method always works.
Example
Show
x3
lim = 0.
(x,y )→(0,0) x 2 + y 2
Solution
From the last problem we know that
x2
0≤ ≤1
x2 + y2
for all x and y , not both 0. So if x > 0,
x3
0≤ ≤ x.
x2 + y2

39. Showing a limit does exist
This is often harder. No single method always works.
Example
Show
x3
lim = 0.
(x,y )→(0,0) x 2 + y 2
Solution
From the last problem we know that
x2
0≤ ≤1
x2 + y2
for all x and y , not both 0. So if x > 0,
x3
0≤ ≤ x.
x2 + y2
As x → 0+ , the fraction in the middle must go to 0!

40. Another way
Switch to polar coordinates!
x3 (r cos θ)3
lim = lim
(x,y )→(0,0) x 2 + y 2 (r ,θ)→(0,0) (r cos θ)2 + (r sin θ)2
r 3 cos3 θ
= lim
(r ,θ)→(0,0) r 2 cos2 θ + r 2 sin2 θ
= lim r cos3 θ = 0 · 1 = 0.
(r ,θ)→(0,0)

41. Outline
Introduction and definition
Rules of limits
Complications
Showing a limit doesn’t exist
Showing a limit does exist
Continuity
Worksheet

42. Continuity
Definition
A function f of two variables is called continuous at (a, b) if
lim f (x, y ) = f (a, b).
(x,y )→(a,b)
We say f is continuous on D if f is continuous at every point
(a, b) in D.

43. Outline
Introduction and definition
Rules of limits
Complications
Showing a limit doesn’t exist
Showing a limit does exist
Continuity
Worksheet

44. Worksheet
Image: Erick Cifuentes