# Rates of Change and Limits Contributed by: TOPICS DISCUSSED:
1. rates of change of limits
2. the sandwich theorem
1. 2.1
Rates of Change
and Limits
Grand Teton National Park, Wyoming
Photo by Vickie Kelly, 2007 Greg Kelly, Hanford High School, Richland, Washington
2. Suppose you drive 200 miles, and it takes you 4 hours.
mi
Then your average speed is: 200 mi  4 hr  50
hr
distance x
average speed  
elapsed time t
If you look at your speedometer during this trip, it might

3. A rock falls from a high cliff.
2
The position of the rock is given by: y 16t
After 2 seconds: y 16 22 64
64 ft ft
average speed: Vav  32
2 sec sec
What is the instantaneous speed at 2 seconds?

4. 2 2
y 16  2  h   16  2 
Vinstantaneous  
t h
for some very small where h = some very
change in t small change in t
We can use the TI-89 to evaluate this expression for
smaller and smaller values of h.

5. 2 2
y 16  2  h   16  2 
Vinstantaneous  
t h
 16   2  h  ^ 2  64  h h  1,.1,.01,.001,.0001,.00001
y
We can see that the velocity
approaches 64 ft/sec as h becomes
h t
very small.
1 80
We say that the velocity has a limiting
0.1 65.6
value of 64 as h approaches zero.
.01 64.16
(Note that h never actually becomes .001 64.016
zero.) 64.0016
.0001
.00001 64.0002

6. 2
The limit as h 16  2  h   16 22
approaches zero:
lim
h 0 h
16 lim
 4  4 h  h 2
 4
Since the 16 is
unchanged as h
h 0 h
approaches zero,
we can factor 16 4  4h  h 2  4
out. 16 lim
h 0 h
0
16 lim  4  h  64
h 0

7. sin x
Consider: y
x
What happens as x approaches zero?
Y= y sin  x  / x
 2
2
 /2
WINDOW
GRAPH

8. y sin  x  / x
Looks like y=1

9. y sin  x  / x
Numerically:
TblSet
TABLE
You can scroll
down to see
more values.

10. y sin  x  / x
sin x
It appears that the limit of as x approaches zero is 1
x
TABLE
You can scroll
down to see
more values.

11. Limit notation: lim f  x  L
x c
“The limit of f of x as x approaches c is L.”
sin x
So: lim 1
x 0 x

12. The limit of a function refers to the value that the
function approaches, not the actual value (if any).
lim f  x  2
x 2
not 1

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

14. lim f  x 
2
x 1 does not exist
because the left and
1
right hand limits do not
match!
1 2 3 4
At x=1: lim f  x  0 left hand limit
x 1
lim f  x  1 right hand limit
x 1
f  1 1 value of the function

15. 2 lim f  x  1
x 2
1
because the left and
right hand limits match.
1 2 3 4
At x=2: lim f  x  1 left hand limit
x 2
lim f  x  1 right hand limit
x 2
f  2  2 value of the function

16. 2 lim f  x  2
x 3
1
because the left and
right hand limits match.
1 2 3 4
At x=3: lim f  x  2 left hand limit
x 3
lim f  x  2 right hand limit
x 3
f  3 2 value of the function

17. The Sandwich Theorem:
If g  x   f  x  h  x  for all x c in some interval about c
and lim g  x  lim h  x  L, then lim f  x  L.
x c x c x c
2 1
Show that: lim x sin   0
x 0
 x
1
The maximum value of sine is 1, so x sin    x 2
2
 x
1
The minimum value of sine is -1, so x sin    x 2
2
 x
1
So:  x  x sin    x 2
2 2
 x 
18. 1
lim  x lim x sin   lim x 2
2 2
x 0 x 0
 x  x 0
2 1
0 lim x sin   0
x 0
 x
2 1
By the sandwich theorem: lim x sin   0
x 0
 x
Y= WINDOW

19.