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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: