2.1

Rates of Change and Limits

A rock falls from a high cliff.

The position of the rock is given by:216y t

After 2 seconds:216 2 64y

average speed: av

64 ft ft32

2 sec secV

What is the instantaneous speed at 2 seconds?

instantaneous

yV

t

for some very small change in t

2 216 2 16 2h

h

where h = some very small change in t

We can use the TI-89 to evaluate this expression for smaller and smaller values of h.

instantaneous

yV

t

2 2

16 2 16 2h

h

hy

t

1 80

0.1 65.6

.01 64.16

.001 64.016

.0001 64.0016

.00001 64.0002

16 2 ^ 2 64 1,.1,.01,.001,.0001,.00001h h h

We can see that the velocity approaches 64 ft/sec as h becomes very small.

We say that the velocity has a limiting value of 64 as h approaches zero.

(Note that h never actually becomes zero.)

2

0

16 2 64limh

h

h

The limit as h approaches zero:

2

0

16 4 4 64limh

h h

h

2

0

64 64 16 64limh

h h

h

0lim 64 16h

h

0

64

1 2

( )The of

with respe

Definition: avera

ct to over the

ge rat

interv

e of change

al [ , ] is

y f x

x x x

2 1 1 1

2 1

( ) ( ) ( ) ( )f x f x f x h f xy

x x x h

Secant Line

y = f(x)

1 1( , ( ))P x f x

1x 2x

x

y

2 2( , ( ))Q x f x

x

y

2 1

1 2

where

the length of the

interval [ , ]

h x x

x x

which is the slope of

the secant line .PQ

3

Find the averge rate of change of the

function over the int

Ex

er

ample

val ( ) 1

:

2, .[ 3]f x x Solution:

3 32 1

2 1

( ) ( ) ((3) 1) (( 2) 1)

3 ( 2)

f x f xy

x x x

28 8 36

5 5

Example: The function gives the height (in feet) of a ball thrown straight up as a function of time, t (in seconds).

2( ) 16 100 6s t t t

(a) Find the average rate of change of the height of the ball between 1 and t seconds.

st

s t st

t

( ) ( )

,1

11

s t t t( ) 16 100 62

s( ) ( ) ( )1 16 1 100 1 6 902

Solution:

(b) Use the result found in part (a) to find the average rate of change of the height of the ball between 1 and 2 seconds.

(a)

2( ) (1) 16 100 6 90

1 1

s s t s t t

t t t

16 100 841

2t tt

4 4 25 21

1

2t t

t

4 4 25 21

1

2t t

t

4 4 21 1

1( )( )t t

t

4 4 21( )t

If t = 2, the average rate of change between 1 second and 2 seconds is: -4(4(2) - 21) = 52 ft/second.

The average rate of change between 1 second and t seconds is: -4(4t - 21)

(b)

Example: How does the the function f(x)= x2-x+2 behave near x=2.

x<2 f(x) x>2 f(x) 1.0 2.000000 3.0 8.0000001.5 2.750000 2.5 5.7500001.8 3.440000 2.2 4.6400001.9 3.710000 2.1 4.3100001.95 3.852500 2.05 4.1525001.99 3.970100 2.01 4.0301001.995 3.985025 2.001 4.003001

Before we give a definition of limit, let us look at the following examples.

Solution:

• We see that when x is close to 2 (x>2 or x<2), f(x) is close to 4. Then we can say that the limit of the function f(x) = x2-x+2 as x approaches 2 is equal to 4

•And we write this as the notation:

2

2lim( 2) 4x

x x

Value of x f(x)=(x2–1)/(x – 1) 0.99

1.99 1.01

2.01 0.999 1.999

1.001 2.0010.999999 1.9999991.000001 2.000001

y

-1 1 2 x

1

2

3

0

1

1)(

2

x

xxf

2

: How does the function

behave near ?

1(E )

1

xam

1

2

plex

f xx

x

Solution:

•The function f(x) is defined for all x except x=1

• For any x=1 we simplify the formula of the function by

2 ( 1)( for 1

1)( ) 1

1 1

1 x xf x x

x

x

xx

• So the graph of the function f is the line y=x+1 with the removable point (1,2)

•From the table we say that f(x) approaches the limit 2 as x approaches 1

•And we write this as the notation2

1

1lim 2

1x

x

x

Definition: Informal Definition of limit

lim ( )x a

f x L

Let f(x) be defined on an open interval about a ,

except possibly at a itself. If we can make the

values of f(x) arbitrarily close to L by taking x to

be sufficiently close to a but not equal to a. And

we say “the limit of f(x) equals L” as x approaches a.

And we write

The limit of a function refers to the value that the function approaches, not the actual value (if any).

2

lim 2x

f x

But not 1

3 By graphing, find:

77, if 3

( ) 6 7,

Exa lim ( ) if mple:

if 3

x

x xf x

x

f x

2

4

6

8

2 4 6

(3, 7)

y

x

4)(lim3

xfx

Solution:

From graph we get

Try This

Find:

f(0)is undefined; and

2( )

1 1

xf x

x

0lim ( )x

f x

0lim ( ) 2x

f x

Solution:

Try This

, 0( ) 1 1

1, 0

xx

f x xx

2

1

0lim ( )x

f x

Solution:f(0)is undefined; and

0lim ( ) 2x

f x

Find:

Try This

Find the limit of f(x) as x approaches 3 where f is defined by:

2, 3( )

3, 3

xf x

x

3lim ( ) 2x

f x

Solution:

2

2 3 Examp

2 Find lim ( ) if ( )

2 3 2

if it ex

le:

ist.

x

x xf x f x

x x

(2, 7)

(2, -1)

2lim ( )x

f x

does not exist, there is a jump at x =2.

Solution:

Example: Consider

3 1, 1

( ) 14, 1

xx

f x xx

3

1

1lim ?

1x

x

x

Solution:

The limit is 3

Example: Discuss the behavior of the following functions as x 0

0, 0a) ( )

1 0

xU x

x

The unit step function, it values jump at x=0.For negative values of x close to 0, U(x)=0.For positive values of x close to 0, U(x)=1.

0 lim ( ) Therefore does not exist.

xU x

jump infinite

oscillating

1/ , 0b) g( )

0, 0

x xx

x

1x

0, 0c) ( )

sin 0

xf x

x

The values of g grow too large in the absolute valuesAs x 0 and do not stay close to any real number

0 lim ( ) Therefore does not exist.

xg x

0 lim ( ) Therefore does not exist.

xf x

The function’s values oscillate between +1 and -1 in every interval containing 0.

oscillating

1( ) sinf x

x

The function’s values oscillate between +1 and -1 in every interval containing 0.

Example: Discuss the behavior of the following functions as x 0

0 lim ( ) Therefore does not exist.

xf x

Solution: