Lesson05 Continuity Slides+Notes

Section 2.4Continuity

Math 1a

October 3, 2007

Questions

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

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

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

Questions

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

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

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

Questions

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

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

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

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)

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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 its whole domain, which is[−1

4 ,∞).

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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 its whole domain, which is[−1

4 ,∞).

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 its whole domain, which is[−1

4 ,∞).

Math 1a - October 03, 2007.GWBWednesday, Oct 3, 2007

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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:

1. f + g

2. f − g

3. cf

4. fg

5.f

g(if g(a) 6= 0)

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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

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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.

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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.

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Pitfall #2

: The function has no value

Example

Let

f (x) =x2 + 2x + 1

x + 1

At which points is f continuous?

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.

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Pitfall #2: The function has no value

Example

Let

f (x) =x2 + 2x + 1

x + 1

At which points is f continuous?

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 #3

: function value 6= limit

Example

Let

f (x) =

{46 if x 6= 1

π if x = 1

At which points is f continuous?

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

x→1f (x) = 46.

Pitfall #3: function value 6= limit

Example

Let

f (x) =

{46 if x 6= 1

π if x = 1

At which points is f continuous?

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

x→1f (x) = 46.

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.

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

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.

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

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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

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

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

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

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

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.