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