Lesson 5: Continuity

Section 1.5Continuity

V63.0121, Calculus I

February 2–3, 2009

Announcements

I Quiz 1 is this week: Tuesday (Sections 11 and 12), Thursday(Section 4), Friday (Sections 13 and 14). 15 minutes, coversSections 1.1–1.2

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I Congratulations to the Super Bowl XLIII ChampionPittsburgh Steelers!

Outline

Warmup

Grader Notes

Continuity

The Intermediate Value Theorem

Back to the Questions

Hatsumon

Here are some discussion questions to start.

I Were you ever exactly three feet tall?

I Was your height (in inches) ever equal to your weight (inpounds)?

I Is there a pair of points on opposite sides of the world at thesame temperature at the same time?

Hatsumon





Hatsumon





Outline

Warmup

Grader Notes

Continuity



But first, a word from our friendly graders

I Please turn in neat problem sets: loose-leaf paper, stapled

I Label homework with Name, Section (10 or 4), Date, ProblemSet number

I Do not turn in scratch work

Please label your graphs

Example: Graph F (x) = |2x + 1|.

incomplete

x

y

(−1/2, 0)

(0, 1)

better

Example 6= explanation

ProblemIf f is even and g is odd, what can you say about fg?

Let f (x) = x2 (even) and g(x) = x3 (odd).Then (fg)(x) = x2 · x3 = x5, and that’s odd.So the product of an even function and an oddfunction is an odd function.

Dangerous!

The trouble with proof by example

ProblemWhich odd numbers are prime?

The numbers 3, 5, and7 are odd, and all prime.So all odd numbers areprime.

Fallacious!

Use the definitions

Let f be even and g be odd. Then

(fg)(−x) = f (−x)g(−x) = f (x)(−g(x))

= −f (x)g(x) = −(fg)(x)

So fg is odd.

Better

Outline

Warmup

Grader Notes

Continuity



Recall: Direct Substitution Property

Theorem (The Direct Substitution Property)

If f is a polynomial or a rational function and a is in the domain off , then

limx→a

f (x) = f (a)

Definition of Continuity

DefinitionLet f be a function defined near a. We say that f is continuous ata if

limx→a

f (x) = f (a).

Free Theorems

Theorem

(a) Any polynomial is continuous everywhere; that is, it iscontinuous on R = (−∞,∞).

(b) Any rational function is continuous wherever it is defined; thatis, it is continuous on its domain.

Showing a function is continuous

Example

Let f (x) =√

4x + 1. Show that f is continuous at 2.

SolutionWe have

limx→a

f (x) = limx→2

√4x + 1

=√

limx→2

(4x + 1)

=√

9 = 3.

Each step comes from the limit laws.

In fact, f is continuous on (−1/4,∞). The function f is rightcontinuous at the point −1/4.


Example

Let f (x) =√


SolutionWe have

limx→a

f (x) = limx→2

√4x + 1

=√

limx→2

(4x + 1)

=√

9 = 3.




Example

Let f (x) =√


SolutionWe have

limx→a

f (x) = limx→2

√4x + 1

=√

limx→2

(4x + 1)

=√

9 = 3.


In fact, f is continuous on (−1/4,∞).

The function f is rightcontinuous at the point −1/4.


Example

Let f (x) =√


SolutionWe have

limx→a

f (x) = limx→2

√4x + 1

=√

limx→2

(4x + 1)

=√

9 = 3.



The Limit Laws give Continuity Laws

TheoremIf f and g are continuous at a and c is a constant, then thefollowing functions are also continuous at a:

I f + g

I f − g

I cf

I fg

If

g(if g(a) 6= 0)

Transcendental functions are continuous, too

TheoremThe following functions are continuous wherever they are defined:

1. sin, cos, tan, cot sec, csc

2. x 7→ ax , loga, ln

3. sin−1, tan−1, sec−1

What could go wrong?

In what ways could a function f fail to be continuous at a point a?Look again at the definition:

limx→a

f (x) = f (a)

Pitfall #1

: The limit does not exist

Example

Let

f (x) =

{x2 if 0 ≤ x ≤ 1

2x if 1 < x ≤ 2

At which points is f continuous?

SolutionAt any point a in [0, 2] besides 1, lim

x→af (x) = f (a) because f is

represented by a polynomial near a, and polynomials have thedirect substitution property. However,

limx→1−

f (x) = limx→1−

x2 = 12 = 1

limx→1+

f (x) = limx→1+

2x = 2(1) = 2

So f has no limit at 1. Therefore f is not continuous at 1.

Pitfall #1: The limit does not exist

Example

Let

f (x) =

{x2 if 0 ≤ x ≤ 1

2x if 1 < x ≤ 2


SolutionAt any point a in [0, 2] besides 1, lim

x→af (x) = f (a) because f is

represented by a polynomial near a, and polynomials have thedirect substitution property. However,

limx→1−

f (x) = limx→1−

x2 = 12 = 1

limx→1+

f (x) = limx→1+

2x = 2(1) = 2

So f has no limit at 1. Therefore f is not continuous at 1.

Graphical Illustration of Pitfall #1

x

y

−1 1 2

−1

1

2

3

4

Pitfall #2

: The function has no value

Example

Let

f (x) =x2 + 2x + 1

x + 1


SolutionBecause f is rational, it is continuous on its whole domain. Notethat −1 is not in the domain of f , so f is not continuous there.

Pitfall #2: The function has no value

Example

Let

f (x) =x2 + 2x + 1

x + 1


SolutionBecause f is rational, it is continuous on its whole domain. Notethat −1 is not in the domain of f , so f is not continuous there.


x

y

−1

1

f cannot be continuous where it has no value.

Pitfall #3

: function value 6= limit

Example

Let

f (x) =

{7 if x 6= 1

π if x = 1


Solutionf is not continuous at 1 because f (1) = π but lim

x→1f (x) = 7.

Pitfall #3: function value 6= limit

Example

Let

f (x) =

{7 if x 6= 1

π if x = 1


Solutionf is not continuous at 1 because f (1) = π but lim

x→1f (x) = 7.


x

y

π

7

1

Special types of discontinuites

removable discontinuity The limit limx→a

f (x) exists, but f is not

defined at a or its value at a is not equal to the limitat a.

jump discontinuity The limits limx→a−

f (x) and limx→a+

f (x) exist, but

are different. f (a) is one of these limits.

Graphical representations of discontinuities

x

y

π

7

1

removable

x

y

−1 1 2

−1

1

2

3

4

jump

The greatest integer function

x

y

−2

−2

−1

−1

1

1

2

2

3

3

The greatest integer function f (x) = [[x ]] has jump discontinuities.

The greatest integer function

x

y

−2

−2

−1

−1

1

1

2

2

3

3

The greatest integer function f (x) = [[x ]] has jump discontinuities.

Outline

Warmup

Grader Notes

Continuity



A Big Time Theorem

Theorem (The Intermediate Value Theorem)

Suppose that f is continuous on the closed interval [a, b] and let Nbe any number between f (a) and f (b), where f (a) 6= f (b). Thenthere exists a number c in (a, b) such that f (c) = N.

Illustrating the IVT

Suppose that f is continuous on the closed interval [a, b] and let Nbe any number between f (a) and f (b), where f (a) 6= f (b). Thenthere exists a number c in (a, b) such that f (c) = N.

x

f (x)

a b

f (a)

f (b)

N

cc1 c2 c3

Illustrating the IVTSuppose that f is continuous on the closed interval [a, b]

and let Nbe any number between f (a) and f (b), where f (a) 6= f (b). Thenthere exists a number c in (a, b) such that f (c) = N.

x

f (x)

a b

f (a)

f (b)

N

cc1 c2 c3

Illustrating the IVTSuppose that f is continuous on the closed interval [a, b]

and let Nbe any number between f (a) and f (b), where f (a) 6= f (b). Thenthere exists a number c in (a, b) such that f (c) = N.

x

f (x)

a b

f (a)

f (b)

N

cc1 c2 c3

Illustrating the IVTSuppose that f is continuous on the closed interval [a, b] and let Nbe any number between f (a) and f (b), where f (a) 6= f (b).

Thenthere exists a number c in (a, b) such that f (c) = N.

x

f (x)

a b

f (a)

f (b)

N

cc1 c2 c3

Illustrating the IVTSuppose that f is continuous on the closed interval [a, b] and let Nbe any number between f (a) and f (b), where f (a) 6= f (b). Thenthere exists a number c in (a, b) such that f (c) = N.

x

f (x)

a b

f (a)

f (b)

N

c

c1 c2 c3


x

f (x)

a b

f (a)

f (b)

N

cc1 c2 c3


x

f (x)

a b

f (a)

f (b)

N

c

c1 c2 c3

Using the IVT

Example

Prove that the square root of two exists.

Proof.Let f (x) = x2, a continuous function on [1, 2]. Note f (1) = 1 andf (2) = 4. Since 2 is between 1 and 4, there exists a point c in(1, 2) such that

f (c) = c2 = 2.

In fact, we can “narrow in” on the square root of 2 by the methodof bisections.

Using the IVT

Example


Proof.Let f (x) = x2, a continuous function on [1, 2].

Note f (1) = 1 andf (2) = 4. Since 2 is between 1 and 4, there exists a point c in(1, 2) such that

f (c) = c2 = 2.


Using the IVT

Example



f (c) = c2 = 2.


Using the IVT

Example



f (c) = c2 = 2.


Finding√

2 by bisections

x f (x) = x2

1 1

2 4

1.5 2.25

1.25 1.5625

1.375 1.8906251.4375 2.06640625

Finding√

2 by bisections

x f (x) = x2

1 1

2 4

1.5 2.25

1.25 1.5625

1.375 1.8906251.4375 2.06640625

Finding√

2 by bisections

x f (x) = x2

1 1

2 4

1.5 2.25

1.25 1.5625

1.375 1.8906251.4375 2.06640625

Finding√

2 by bisections

x f (x) = x2

1 1

2 4

1.5 2.25

1.25 1.5625

1.375 1.8906251.4375 2.06640625

Finding√

2 by bisections

x f (x) = x2

1 1

2 4

1.5 2.25

1.25 1.5625

1.375 1.890625

1.4375 2.06640625

Finding√

2 by bisections

x f (x) = x2

1 1

2 4

1.5 2.25

1.25 1.5625

1.375 1.8906251.4375 2.06640625

Using the IVT

Example

Let f (x) = x3 − x − 1. Show that there is a zero for f .

Solutionf (1) = −1 and f (2) = 5. So there is a zero between 1 and 2.

(More careful analysis yields 1.32472.)

Using the IVT

Example

Let f (x) = x3 − x − 1. Show that there is a zero for f .

Solutionf (1) = −1 and f (2) = 5. So there is a zero between 1 and 2.(More careful analysis yields 1.32472.)

Outline

Warmup

Grader Notes

Continuity




True or FalseAt one point in your life you were exactly three feet tall.

True or FalseAt one point in your life your height in inches equaled your weightin pounds.

True or FalseRight now there are two points on opposite sides of the Earth withexactly the same temperature.

Question 1: True!

Let h(t) be height, which varies continuously over time. Thenh(birth) < 3 ft and h(now) > 3 ft. So there is a point c in(birth, now) where h(c) = 3.





Question 2: True!

Let h(t) be height in inches and w(t) be weight in pounds, bothvarying continuously over time. Let f (t) = h(t)− w(t). For mostof us (call your mom), f (birth) > 0 and f (now) < 0. So there is apoint c in (birth, now) where f (c) = 0. In other words,

h(c)− w(c) = 0 ⇐⇒ h(c) = w(c).





Question 3

I Let T (θ) be the temperature at the point on the equator atlongitude θ.

I How can you express the statement that the temperature onopposite sides is the same?

I How can you ensure this is true?

Lesson 5: Continuity

Education

f continuous

f xg x

graph f x

continuous examplelet

function f

limit lim f x

x1x1 lim f x

domain of f