2.1Rates of Change and Limits

Suppose you drive 200 miles, and it takes you 4 hours.

Then your average speed is:mi

200 mi 4 hr 50 hr

distanceaverage speed

elapsed time

x

t

If you look at your speedometer during this trip, it might read 65 mph. This is your instantaneous speed.

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 2

0

16 2 16 2limh

h

h

The limit as h approaches zero:

2

0

4 4 416 lim

h

h h

h

2

0

4 4 416 lim

h

h h

h

0

16 lim 4h

h

0

64

Since the 16 is unchanged as h approaches zero, we can factor 16 out.

Consider:sin x

yx

What happens as x approaches zero?

Graphically:

sin /y x x

22

/ 2

WINDOW

Y=

GRAPH

sin /y x x

Looks like y=1

sin /y x x

You can scroll down to see more values.

TABLE

It appears that the limit of as x approaches zero is 1sin x

x

Limit notation: limx cf x L

“The limit of f of x as x approaches c is L.”

So:0

sinlim 1x

x

x

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

2

lim 2x

f x

not 1

Properties of Limits:

Limits can be added, subtracted, multiplied, multiplied by a constant, divided, and raised to a power.

(See your book for details.)

For a limit to exist, the function must approach the same value from both sides.

One-sided limits approach from either the left or right side only.

1 2 3 4

1

2

At x=1: 1

lim 0x

f x

1

lim 1x

f x

1 1f

left hand limit

right hand limit

value of the function

1

limxf x

does not exist because the left and right hand limits do not match!

At x=2: 2

lim 1x

f x

2

lim 1x

f x

2 2f

left hand limit

right hand limit

value of the function

2

lim 1x

f x

because the left and right hand limits match.

1 2 3 4

1

2

At x=3: 3

lim 2x

f x

3

lim 2x

f x

3 2f

left hand limit

right hand limit

value of the function

3

lim 2xf x

because the left and right hand limits match.

1 2 3 4

1

2

The Sandwich Theorem:

If for all in some interval about

and lim lim , then lim .x c x c x c

g x f x h x x c c

g x h x L f x L

Show that: 2

0

1lim sin 0xx

x

The maximum value of sine is 1, so 2 21sinx x

x

The minimum value of sine is -1, so 2 21sinx x

x

So: 2 2 21sinx x x

x

2 2 2

0 0 0

1lim lim sin limx x x

x x xx

2

0

10 lim sin 0

xx

x

2

0

1lim sin 0xx

x

By the sandwich theorem:

Y= WINDOW

p