2012 Pearson Education, Inc.

Chapter 2

2.3

Section 2.3

Continuity

Limits and Continuity

Slide 2.3- 2 2012 Pearson Education, Inc.

Quick Review

2

31

1 1

1

2

2 2

2

3 2 11. Find lim

4

2. Let int . Find each limit.

a lim b lim

c lim d 1

4 5, 23. Let

4 , 2

Find each limit.

a lim b lim

c lim d 2

x

x x

x

x x

x

x x

x

f x x

f x f x

f x f

x x xf x

x x

f x f x

f x f

Slide 2.3- 3 2012 Pearson Education, Inc.

Quick Review

2 2

In Exercises 4 – 6, find the remaining functions in the list of functions:

, , , .

2 1 14. , 1

5

5. , sin , domain of [0, )

f g f g g f

xf x g x

x x

f x x g f x x g

Slide 2.3- 4 2012 Pearson Education, Inc.

Quick Review

2

3

16. 1, , 0

7. Use factoring to solve 2 9 5 0

8. Use graphing to solve 2 1 0

g x x g f x xx

x x

x x

Slide 2.3- 5 2012 Pearson Education, Inc.

Quick Review

2

5 , 3In Exercises 9 and 10, let

6 8, 3

9. Solve the equation 4

10. Find a value of for which the equation

has no solution.

x xf x

x x x

f x

c f x c

Slide 2.3- 6 2012 Pearson Education, Inc.

Quick Review Solutions

2

31

1 1

1

2

2 2

2

2

2 1

no limi

3 2 11. Find lim

4

2. Let int . Find each limit.

a lim b lim

c lim d 1

4 5, 23. Let

4 , 2

Find each limit.

a lim b lim

l

t

1 2

i

1

c m

x

x x

x

x x

x

x x

x

f x x

f x f x

f x f

x x xf x

x x

f x f x

dno limit 22 f x f

Slide 2.3- 7 2012 Pearson Education, Inc.

Quick Review Solutions

2 2

2

In Exercises 4 – 6, find the remaining functions in the list of functions:

, , , .

2 1 14. , 1

5

5. , sin , domain of [0, )

2 3 4, 0 , 5

6 1 2 1

sin , 0 sin , 0

f g f g g f

xf

x xf g x x

x g xx x

f x

g f x xx x

g x x x f g x x x

x g f x x g

Slide 2.3- 8 2012 Pearson Education, Inc.

Quick Review Solutions

2

2

3

16. 1, , 0

7. Use factoring to solve 2 9 5 0

8. Use graphing to solve 2 1 0

11, 0 , 1

1

1, 5

2

0.453

xf x x f g x

g x x g f x xx

x x

x x

xxx

x

x

Slide 2.3- 9 2012 Pearson Education, Inc.

Quick Review Solutions

2

5 , 3In Exercises 9 and 10, let

6 8, 3

9. Solve the equation 4

10. Find a value of for which the equation

has no solution

1

Any in [1,2. )

x xf x

x x x

f x

c f

x

c

x c

Slide 2.3- 10 2012 Pearson Education, Inc.

What you’ll learn about

Continuity at a Point Continuous Functions Algebraic Combinations Composites Intermediate Value Theorem for Continuous Functions

…and whyContinuous functions are used to describe how a body moves through space and how the speed of a chemical reaction changes with time.

Slide 2.3- 11 2012 Pearson Education, Inc.

Continuity at a Point

Any function whose graph can be sketched in

one continuous motion without lifting the pencil is an

example of a continuous function.

y f x

Slide 2.3- 12 2012 Pearson Education, Inc.

Example Continuity at a Point

Find the points at which the given function is continuous and the points at

which it is discontinuous.o

Points at which is continuousfAt 0 x

At 6 x

At 0 < < 6 but not 2 3 c cPoints at which is discontinuousfAt 2xAt 0, 2 3, 6 c c c

0

lim 0

x

f x f

6

lim 6

x

f x f

lim

x c

f x f c

2

lim does not existxf x

these points are not in the domain of f

Slide 2.3- 13 2012 Pearson Education, Inc.

Continuity at a Point

Interior Point: A function is continuous at an interior point of its

domain if lim

Endpoint: A function is continuous at a left

endpoint or is continuous

x c

y f x c

f x f c

y f x

a

at a right endpoint of its domain if

lim or lim respectively.x a x b

b

f x f a f x f b

Slide 2.3- 14 2012 Pearson Education, Inc.

Continuity at a Point

If a function   is   , we say that 

is    and   is a point of discontinuity of   . 

Note that need not be in the domain of   .

f f

c f

c f

not continuous at a point

discontinuous at

c

c

Slide 2.3- 15 2012 Pearson Education, Inc.

Continuity at a Point

The typical discontinuity types are:

a) Removable (2.21b and 2.21c)

b) Jump (2.21d)

c) Infinite(2.21e)

d) Oscillating (2.21f)

Slide 2.3- 16 2012 Pearson Education, Inc.

Continuity at a Point

Slide 2.3- 17 2012 Pearson Education, Inc.

Example Continuity at a Point

There is an infinite discontinuity at 1.x

2

3Find and identify the points of discontinuity of

1y

x

[5,5] by [5,10]

Slide 2.3- 18 2012 Pearson Education, Inc.

Continuous Functions

A function is if and only if

it is continuous at every point of the interval. A

is one that is continuous at every

point of its domain.  A continuous funct

continuous on an interval

continuous function

ion need not be

continuous on every interval.

Slide 2.3- 19 2012 Pearson Education, Inc.

Continuous Functions

2

2

2y

x

[5,5] by [5,10]

The given function is a continuous function because it is

continuous at every point of its domain. It does have a

point of discontinuity at 2 because it is not defined there.x

Slide 2.3- 20 2012 Pearson Education, Inc.

If the functions and are continuous at , then the

following combinations are continuous at .

1. Su ms:

2. Differences:

3. Products:

4. Constant multiples: , for any number

5. Quotients: , pr

f g x c

x c

f g

f g

f g

k f k

f

g

ovided 0g c

Properties of Continuous Functions

Slide 2.3- 21 2012 Pearson Education, Inc.

Composite of Continuous Functions

If is continuous at and is continuous at , then the

composite is continuous at .

f c g f c

g f c

Slide 2.3- 22 2012 Pearson Education, Inc.

Intermediate Value Theorem for Continuous Functions

0 0

A function that is continuous on a closed interval [ , ]

takes on every value between and . In other words,

if is between and , then for some in [ , ].

y f x a b

f a f b

y f a f b y f c c a b

Slide 2.3- 23 2012 Pearson Education, Inc.

The Intermediate Value Theorem for Continuous Functions is the reason why the graph of a function continuous on an interval cannot have any breaks. The graph will be connected, a single, unbroken curve. It will not have jumps or separate branches.

Intermediate Value Theorem for Continuous Functions