Lesson 11: Limits and Continuity

Section 11.2

Limits and Continuity

Math 21a

February 29, 2008

Announcements

I Problem Sessions: Mon, 8:30; Thur, 7:30; SC 103b

I Office hours Tuesday, Wednesday 2–4pm SC 323

I Midterm I Review Session 3/5, 6–7:30pm in SC Hall D

I Midterm I, 3/11, 7–9pm in SC Hall D

Image: kaet44

Outline

Introduction and definition

Rules of limits

ComplicationsShowing a limit doesn’t existShowing a limit does exist

Continuity

Worksheet

Where we’re going: derivatives of multivariable functions

Recall that if f is a function that takes real numbers to realnumbers,

f ′(x) = limh→0

f (x + h)− f (x)

h

We want to do the same thing in more than one variable. So weneed to take limits in more than one dimension.

Where we’re going: derivatives of multivariable functions

Recall that if f is a function that takes real numbers to realnumbers,

f ′(x) = limh→0

f (x + h)− f (x)

h

We want to do the same thing in more than one variable. So weneed to take limits in more than one dimension.

DefinitionWe write

lim(x ,y)→(a,b)

f (x , y) = L

and we say that the limit of f (x , y) as (x , y) approaches (a, b) isL if we can make the values of f (x , y) as close to L as we like bytaking the point (x , y) to be sufficiently close to (a, b).

easy limits

I lim(x ,y)→(a,b)

x = a

I lim(x ,y)→(a,b)

y = b

I lim(x ,y)→(a,b)

c = c

Outline

Introduction and definition

Rules of limits

ComplicationsShowing a limit doesn’t existShowing a limit does exist

Continuity

Worksheet

Like regular limits, limits of multivariable functions can be

I added

I subtracted

I multiplied

I composed

I divided, provided the limit of the denominator is not zero.

Limit of a Polynomial

Example

Find lim(x ,y)→(5,−2)

(x5 + 4x3y − 5xy2)

Solution

lim(x ,y)→(5,−2)

(x5 + 4x3y − 5xy2) = (5)5 + 4(5)3(−2)− 5(5)(−2)2

= 3125 + 4(125)(−2)− 5(5)(4)

= 2025.

Limit of a Polynomial

Example

Find lim(x ,y)→(5,−2)

(x5 + 4x3y − 5xy2)

Solution

lim(x ,y)→(5,−2)

(x5 + 4x3y − 5xy2) = (5)5 + 4(5)3(−2)− 5(5)(−2)2

= 3125 + 4(125)(−2)− 5(5)(4)

= 2025.

Limit of a Rational Expression

Example

Compute

lim(x ,y)→(1,2)

x2

x2 + y2.

Solution

lim(x ,y)→(1,2)

x2

x2 + y2=

(1)2

(1)2 + (2)2

=1

5

Limit of a Rational Expression

Example

Compute

lim(x ,y)→(1,2)

x2

x2 + y2.

Solution

lim(x ,y)→(1,2)

x2

x2 + y2=

(1)2

(1)2 + (2)2

=1

5

Outline

Introduction and definition

Rules of limits

ComplicationsShowing a limit doesn’t existShowing a limit does exist

Continuity

Worksheet

The only real problem is a limit where the denominator goes tozero.

I If the numerator goes to some number and the denominatorgoes to zero then the quotient cannot have a limit.

I If on the other hand the numerator and denominator both goto zero we have no clue. Most “interesting” limits come fromthis. e.g.,

f ′(x) = limh→0

f (x + h)− f (x)

h

The only real problem is a limit where the denominator goes tozero.

I If the numerator goes to some number and the denominatorgoes to zero then the quotient cannot have a limit.

I If on the other hand the numerator and denominator both goto zero we have no clue. Most “interesting” limits come fromthis. e.g.,

f ′(x) = limh→0

f (x + h)− f (x)

h

You probably remember this statement:

FactFor a function f (x) of one variable,

limx→a

f (x) = L ⇐⇒ limx→a+

f (x) = L and limx→a−

f (x) = L

For functions of two variables, “left-hand limits” and “right-handlimits” aren’t enough.

You probably remember this statement:

FactFor a function f (x) of one variable,

limx→a

f (x) = L ⇐⇒ limx→a+

f (x) = L and limx→a−

f (x) = L

For functions of two variables, “left-hand limits” and “right-handlimits” aren’t enough.

Showing a limit doesn’t exist

TheoremSuppose lim

(x ,y)→(a,b)f (x , y) = L. Then the limit of f as

(x , y)→ (a, b) is L along all paths through (a, b).

There are two contrapositives to this statement:

I If there is a path through (a, b) along which the limit does notexist, the two-dimensional limit does not exist

I If there are two paths through (a, b) along which the limitsexist but disagree, the two-dimensional limit does not exist

Showing a limit doesn’t exist

TheoremSuppose lim

(x ,y)→(a,b)f (x , y) = L. Then the limit of f as

(x , y)→ (a, b) is L along all paths through (a, b).

There are two contrapositives to this statement:

I If there is a path through (a, b) along which the limit does notexist, the two-dimensional limit does not exist

I If there are two paths through (a, b) along which the limitsexist but disagree, the two-dimensional limit does not exist

Showing a limit doesn’t exist

TheoremSuppose lim

(x ,y)→(a,b)f (x , y) = L. Then the limit of f as

(x , y)→ (a, b) is L along all paths through (a, b).

There are two contrapositives to this statement:

I If there is a path through (a, b) along which the limit does notexist, the two-dimensional limit does not exist

I If there are two paths through (a, b) along which the limitsexist but disagree, the two-dimensional limit does not exist

Showing a limit doesn’t exist

TheoremSuppose lim

(x ,y)→(a,b)f (x , y) = L. Then the limit of f as

(x , y)→ (a, b) is L along all paths through (a, b).

There are two contrapositives to this statement:

I If there is a path through (a, b) along which the limit does notexist, the two-dimensional limit does not exist

I If there are two paths through (a, b) along which the limitsexist but disagree, the two-dimensional limit does not exist

Example

Show lim(x ,y)→(0,0)

x

x2 + y2does not exist.

SolutionFollow a path towards (0, 0) along the line y = 0.

limx→0

f (x , 0) = limx→0

x

x2 + 02= lim

x→0

1

x

which does not exist. So lim(x ,y)→(0,0)

x

x2 + y2does not exist.

Example

Show lim(x ,y)→(0,0)

x

x2 + y2does not exist.

SolutionFollow a path towards (0, 0) along the line y = 0.

limx→0

f (x , 0) = limx→0

x

x2 + 02= lim

x→0

1

x

which does not exist. So lim(x ,y)→(0,0)

x

x2 + y2does not exist.

Example

Show lim(x ,y)→(0,0)

x

x2 + y2does not exist.

SolutionFollow a path towards (0, 0) along the line y = 0.

limx→0

f (x , 0) = limx→0

x

x2 + 02= lim

x→0

1

x

which does not exist. So lim(x ,y)→(0,0)

x

x2 + y2does not exist.

We can see the problems in a graph.

x

x2 + y2= c ⇐⇒

(x − 1

c

)2

+ y2 =

(1

2c

)2

-1 -0.5 0 0.5 1-1

-0.5

0

0.5

1

-1

-0.5

0

0.5

1 -1

-0.5

0

0.5

1

-10

-5

0

5

10

-1

-0.5

0

0.5

We can see the problems in a graph.

x

x2 + y2= c ⇐⇒

(x − 1

c

)2

+ y2 =

(1

2c

)2

-1 -0.5 0 0.5 1-1

-0.5

0

0.5

1

-1

-0.5

0

0.5

1 -1

-0.5

0

0.5

1

-10

-5

0

5

10

-1

-0.5

0

0.5

We can see the problems in a graph.

x

x2 + y2= c ⇐⇒

(x − 1

c

)2

+ y2 =

(1

2c

)2

-1 -0.5 0 0.5 1-1

-0.5

0

0.5

1

-1

-0.5

0

0.5

1 -1

-0.5

0

0.5

1

-10

-5

0

5

10

-1

-0.5

0

0.5

Example

Show lim(x ,y)→(0,0)

x2

x2 + y2does not exist.

SolutionFollow a path towards (0, 0) along the line y = 0:

limx→0

f (x , 0) = limx→0

x2

x2 + 02= lim

x→01 = 1

Now follow a path towards (0, 0) along the line x = 0:

limy→0

f (0, y) = limx→0

02

02 + y2= lim

x→00 = 0

So the limit as (x , y)→ (0, 0) cannot exist.

Example

Show lim(x ,y)→(0,0)

x2

x2 + y2does not exist.

SolutionFollow a path towards (0, 0) along the line y = 0:

limx→0

f (x , 0) = limx→0

x2

x2 + 02= lim

x→01 = 1

Now follow a path towards (0, 0) along the line x = 0:

limy→0

f (0, y) = limx→0

02

02 + y2= lim

x→00 = 0

So the limit as (x , y)→ (0, 0) cannot exist.

Example

Show lim(x ,y)→(0,0)

x2

x2 + y2does not exist.

SolutionFollow a path towards (0, 0) along the line y = 0:

limx→0

f (x , 0) = limx→0

x2

x2 + 02= lim

x→01 = 1

Now follow a path towards (0, 0) along the line x = 0:

limy→0

f (0, y) = limx→0

02

02 + y2= lim

x→00 = 0

So the limit as (x , y)→ (0, 0) cannot exist.

Example

Show lim(x ,y)→(0,0)

x2

x2 + y2does not exist.

SolutionFollow a path towards (0, 0) along the line y = 0:

limx→0

f (x , 0) = limx→0

x2

x2 + 02= lim

x→01 = 1

Now follow a path towards (0, 0) along the line x = 0:

limy→0

f (0, y) = limx→0

02

02 + y2= lim

x→00 = 0

So the limit as (x , y)→ (0, 0) cannot exist.

Example

Show lim(x ,y)→(0,0)

x2

x2 + y2does not exist.

SolutionFollow a path towards (0, 0) along the line y = 0:

limx→0

f (x , 0) = limx→0

x2

x2 + 02= lim

x→01 = 1

Now follow a path towards (0, 0) along the line x = 0:

limy→0

f (0, y) = limx→0

02

02 + y2= lim

x→00 = 0

So the limit as (x , y)→ (0, 0) cannot exist.

Again, we can see the problems in a graph.

x2

x2 + y2= c ⇐⇒ y = ±

√1− c

cx

-1 -0.5 0 0.5 1-1

-0.5

0

0.5

1

-1

-0.5

0

0.5

1 -1

-0.5

0

0.5

1

0

0.25

0.5

0.75

1

-1

-0.5

0

0.5

Again, we can see the problems in a graph.

x2

x2 + y2= c ⇐⇒ y = ±

√1− c

cx

-1 -0.5 0 0.5 1-1

-0.5

0

0.5

1

-1

-0.5

0

0.5

1 -1

-0.5

0

0.5

1

0

0.25

0.5

0.75

1

-1

-0.5

0

0.5

Again, we can see the problems in a graph.

x2

x2 + y2= c ⇐⇒ y = ±

√1− c

cx

-1 -0.5 0 0.5 1-1

-0.5

0

0.5

1

-1

-0.5

0

0.5

1 -1

-0.5

0

0.5

1

0

0.25

0.5

0.75

1

-1

-0.5

0

0.5

Showing a limit does existThis is often harder. No single method always works.

Example

Show

lim(x ,y)→(0,0)

x3

x2 + y2= 0.

SolutionFrom the last problem we know that

0 ≤ x2

x2 + y2≤ 1

for all x and y, not both 0. So if x > 0,

0 ≤ x3

x2 + y2≤ x .

As x → 0+, the fraction in the middle must go to 0!

Showing a limit does existThis is often harder. No single method always works.

Example

Show

lim(x ,y)→(0,0)

x3

x2 + y2= 0.

SolutionFrom the last problem we know that

0 ≤ x2

x2 + y2≤ 1

for all x and y, not both 0.

So if x > 0,

0 ≤ x3

x2 + y2≤ x .

As x → 0+, the fraction in the middle must go to 0!

Showing a limit does existThis is often harder. No single method always works.

Example

Show

lim(x ,y)→(0,0)

x3

x2 + y2= 0.

SolutionFrom the last problem we know that

0 ≤ x2

x2 + y2≤ 1

for all x and y, not both 0. So if x > 0,

0 ≤ x3

x2 + y2≤ x .

As x → 0+, the fraction in the middle must go to 0!

Showing a limit does existThis is often harder. No single method always works.

Example

Show

lim(x ,y)→(0,0)

x3

x2 + y2= 0.

SolutionFrom the last problem we know that

0 ≤ x2

x2 + y2≤ 1

for all x and y, not both 0. So if x > 0,

0 ≤ x3

x2 + y2≤ x .

As x → 0+, the fraction in the middle must go to 0!

Another way

Switch to polar coordinates!

lim(x ,y)→(0,0)

x3

x2 + y2= lim

(r ,θ)→(0,0)

(r cos θ)3

(r cos θ)2 + (r sin θ)2

= lim(r ,θ)→(0,0)

r3 cos3 θ

r2 cos2 θ + r2 sin2 θ

= lim(r ,θ)→(0,0)

r cos3 θ = 0 · 1 = 0.

Outline

Introduction and definition

Rules of limits

ComplicationsShowing a limit doesn’t existShowing a limit does exist

Continuity

Worksheet

Continuity

DefinitionA function f of two variables is called continuous at (a, b) if

lim(x ,y)→(a,b)

f (x , y) = f (a, b).

We say f is continuous on D if f is continuous at every point(a, b) in D.

Outline

Introduction and definition

Rules of limits

ComplicationsShowing a limit doesn’t existShowing a limit does exist

Continuity

Worksheet

Worksheet

Image: Erick Cifuentes