Ch 3.1 Limits Solve limits algebraically using direct substitution and by factoring/cancelling Solve...

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Ch 3.1 Limits Solve limits algebraically using direct substitution and by factoring/cancelling Solve limits graphically – including the wall method Solve limits using Derive/Maple

Transcript of Ch 3.1 Limits Solve limits algebraically using direct substitution and by factoring/cancelling Solve...

Ch 3.1 Limits

• Solve limits algebraically using direct substitution and by factoring/cancelling

• Solve limits graphically – including the wall method

• Solve limits using Derive/Maple

Slide 1-2

THE LIMIT (L) OF A FUNCTION IS THE VALUE THE FUNCTION (y) APPROACHES AS THE VALUE OF (x) APPROACHES A GIVEN VALUE.

ax Lxf )(lim

Slide 1-3

Direct Substituion

•Easiest way to solve a limit •Can’t use if it gives an undefined answer

Slide 1-4

Table method and direct substitution method.

3lim ?x 2x

because as x gets closer and closerto 2, x cubed gets closer and closerto 8. (using table on next slide)

Slide 1-5

Can use a table to find a limit.

X 1.8 1.9 1.99 1.99999

2 2.00001

2.001 2.1 2.2

y 5.832 6.859 7.8805

7.9998 8.0001 8.012 9.261 10.648

This is just for explanation purposes. We won’t use this method.

Slide 1-6

Graphically y =3x

Slide 1-7

In the previous example it was fairly evident that the limit was 8, because when we replaced x with 2 (using direct substitution) the function had a value of 8. This is not always as evident. Find the limit below.

2

1

2 3lim

1x

x x

x

Slide 1-8

Rewrite before substituting

Factor and cancel common factors – then do direct substitution.

The answer is 4.

Slide 1-9

means x approaches a from the right

means x approaches a from the left

ax

ax

Slide 1-10

When finding the limit of a function it is important to let x approach a from both the right and left. If the same value of L is approached by the function then the limit exist and

Lxf )(limax

Slide 1-11

THEOREM:As x approaches a, the limit of f (x) is L, if the limit from the left exists and the limit from the right exists and both limits are L. That is, If

1)

and 2)

Then

limx a

f (x) L,

limx a

f (x) L,

limx a

f (x) L,

Slide 1-12

Graphs can be used to determine the limit of a function. Find the following limits.

1

1lim

2

x

x1x

Slide 1-13

Limit Graphically example 1 Find:

(DNE)

1.1 Limits: A Numerical and Graphical Approach

limx1

H (x)

Slide 1-14

1lim ( ) 'x

H x doesn t exist

1lim ( ) 4x

H x

1lim ( ) 2x

H x

(DNE)

Therefore,

Example 1 - Answer

Slide 1-15

The “Wall” Method:

As an alternative approach to Example 1, we can draw a “wall” at x = 1, as shown in blue on the following graphs. We then follow the curve from left to rightwith pencil until we hit the wall and mark the locationwith an × , assuming it can be determined. Then we follow the curve from right to left until we hit the walland mark that location with an ×. If the locations arethe same, we have a limit. Otherwise, the limit does not exist.

Slide 1-16

Thus for Example 1: now try this one

Find3

lim ( ) ?x

H x

limx1

H (x)

1.1 Limits: A Numerical and Graphical Approach

Slide 1-17

a) Limit GraphicallyObserve on the graph

that: 1)

and 2)

Therefore,

limx 3

f (x) 4

limx 3

f (x) 4

limx 3

f (x) 4.

1.1 Limits: A Numerical and Graphical Approach

Slide 1-18

We can also use Derive to evaluate limits.

Slide 1-19

Use Derive to find the indicated limit.

12204 25

0lim

xxxx

3

22

2 4

1limx

x

x

3

92

3lim

x

x

x

Slide 1-20

Limits at infinitySometimes we will be concerned with the value of a function as the value of x increases without bound. These cases are referred to as limits at infinity and are denoted:

( ) and ( )lim limx x

f x f x

Slide 1-21

For polynomial functions the limit will be + or – infinity. Use the leading term to identify the end behavior. Use it to determine whether the answer is + or – infinity.

Slide 1-22

For rational functions, the limit at infinity is the same as the horizontal asymptote (y = L) of the function. Recall the method of finding the horizontal asymptote depends on the degrees of the numerator and denominator

Slide 1-23Copyright © 2008 Pearson Education, Inc. Publishing as

Pearson Addison-Wesley

Slide 1.1- 23

Limit GraphicallyObserve on the graph

that,again, you can only approach ∞ from the left.

Therefore,

Answer : 3

lim ( ) ?x

f x

Slide 1-24

Find each limit

100 50) limx

x

xb

3 2

2

2 3 1)

2 4limx

x xc

x x

2

2

5 17

2) lim

x

x

xa

Slide 1-25

Answers

a) 5b) 100c) No limit exists, but graph to see if Answer to c is

or

Slide 1-26

Infinite limits

When considering f(x) may increase or decrease without bound (becomes infinite) as x approaches a. In these cases the limit is infinite and

)(lim xfax

)(lim xfax

Slide 1-27

When this occurs the line x = a is a vertical asymptote.

Polynomial functions do not have vertical asymptotes, but rational functions may have vertical asymptotes.

Slide 1-28

b) Limit GraphicallyObserve on the graph

that: 1)

and 2)

Therefore, does not exist.

limx 2

f (x)

limx 2

f (x)

limx 2

f (x)

Slide 1-29

We will use graphs to find the answers to these. Graph each on the computer to determine the answer.

0

3) limx x

a 1

3

1) lim

xx

b

1

3

1) lim

xx

c

Slide 1-30

Answers

a) DNEb)c)

Slide 1-31

The cost (in dollars) for manufacturing a particular videotape is

xxC 615000)(

)(xC

lim ( )x

C x

where x is the number of tapesproduced. The average cost pertape, denoted by is foundby dividing C(x) by x. Find and interpret

Slide 1-32

Answers

As the number of tapes increases, the average cost per tape approaches $6.