Average Speed
t
d
Average Speed
Example:Suppose you drive 200 miles in 4 hours. What is your average speed?
hours
miles
t
d
4
200
Since d = rt,
= 50 mph
The moment you look at your speedometer, you see your instantaneous speed.
Instantaneous Speed
ExampleA rock breaks loose from the top of a tall cliff. What is the speed of the rock at 2 seconds?
We can calculate the average speed of the rock from 2 seconds to a time slightly later than 2 seconds (t = 2 + Δt, where Δt is a slight change in time.)
Instantaneous SpeedExampleA rock breaks loose from the top of a tall cliff. What is the speed of the rock at 2 seconds?
Free fall equation: y = 16t2
2)2(
)2(16)2(16 22
t
t
t
d
t
t
64)2(16 2
We cannot use this formula to calculate the speed at the exact instant t = 2 because that would require letting Δt = 0, and that would give 0/0. However, we can look at what is happening when Δt is close to 0.
Instantaneous Speed
t
t
64)2(16 2
Length of Δt (seconds)
Average Speed (ft/sec)
1 80
0.1 65.6
0.01 64.16
0.001 64.016
0.0001 64.0016
0.00001 64.00016
What is happening?As Δt gets smaller, the rock’s average speed gets closer to 64 ft/sec.
Instantaneous SpeedAlgebraically:
t
tt
t
t
64)44(1664)2(16 22
t
tt
64166464 2
t
tt
21664
t
tt
)1664(
Instantaneous SpeedAlgebraically:
t
tt
)1664(
t 1664
Now, when Δt is 0, our average speed is64 ft/sec
Let f be a function defined on a open interval containing a, except possibly at a itself. Then, there exists a
such that
Limits
0
Lxf )(WHAT THE CRAP??????
The function f has a limit L as x approaches c if any positive number (ε), there is a positive number σ such that
Limits
Lxfcx )(0Still, WHAT THE CRAP?????? Lxf
cx
)(lim
We read, “The limit as x approaches c of a function is L.”
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
If L, M, c, and k are real numbers and and then
Limits can be added, subtracted, multiplied, multiplied by a constant, divided, and raised to a power.
Lxfcx
)(lim Mxgcx
)(lim
MLxgxfcx
)()(lim
MLxgxfcx
)()(lim
MLxgxfcx
)()(lim
1.) Sum Rule:
2.) Difference Rule:
3.) Product Rule:
Properties of Limits
Lkxfkcx
)(lim
0M ,)(
)(lim M
L
xg
xfcx
srsr
cxLxf //)(lim
then0, s integers, are s andr If
4.) Constant Multiple Rule:
5.) Quotient Rule:
6.) Power Rule:
Use and and the properties of limits to find the following limits:
Examplekk
cx
lim cx
cx
lim
a.) )34(lim 23
xxcx
3lim4limlim 23
cxcxcxxx
34 23 cc
b.) 5
1lim
2
24
x
xxcx
5lim
1lim2
24
x
xx
cx
cx
5limlim
1limlimlim2
24
cxcx
cxcxcx
x
xx
5
12
24
c
cc
3limlim4lim 23
cxcxcx
cxxx
If f is a continuous function on an open interval containing the number a, then
Evaluating Limits
)()(lim afxfax
(In other words, you can many times substitute the number x is approaching into the function to find the limit.)
Techniques for Evaluating Limits: 1.) Substituting Directly
Ex: Find the limit: 13lim5
xx
1)5(3
4
Limiting Techniques:2.) Using Properties of Limits
Ex: Find the limit: xxx
sin3lim
xxxx
sinlim3lim
(product rule)
))(sin3(
)0)(3(
0
2
)2)(1(
x
xx
Limiting Techniques:3.) Factoring & Simplifying
Ex: Find the limit:2
23lim
2
2
x
xxx
What happens if we just substitute in the limit?
When something like this happens, we need to see if we can factor & simplify!
HOLY COWCULUS!
!!
1lim2
x
x
12
1
Limiting Techniques4.) Using the conjugate
Ex: Find the limit:t
tt
22lim
0
What happens if we just substitute in the limit?
0
220
0
0 We must simplify again.
22
2222lim
0
t
t
t
tt
)22(
22lim
0
tt
tt
22
1lim
0
tt
220
1
22
1
4
2
Limiting Techniques5.) Use a table or graph
Ex: Find the limit:x
xx 2
2sin3lim
0
What happens if we just substitute in the limit?
)0(2
)0(2sin3
0
0sin3
0
0As x approaches 0, you can see that the graph of f(x) approaches 3. Therefore the limit is 3.(You can also see this in your table.)
If f, g, and h are functions defined on some open interval containing a such thatg(x) ≤ f(x) ≤ h(x) for all x in the interval except at possibly at a itself, and
6. Sandwich (Squeeze) Theorem
Lxhxgaxax
)(lim)(lim
then, Lxfax
)(lim h(x)
g(x)
f(x)
Sandwich (Squeeze) Theorem
Ex: Find the limit:
xx
x
1sinlim 2
0
sin oscillates between -1 and 1, so
11
sin1
x
Now, let’s get the problem to look like the one given.
222 1sin x
xxx
2
0
2
0
2
0lim
1sinlimlim x
xxx
xxx
Sandwich (Squeeze) Theorem
Ex: Find the limit:
xx
x
1sinlim 2
0
01
sinlim0 2
0
xx
x
01
sinlim 2
0
xx
x
Therefore, by the Sandwich Theorem,
In order for a limit to exist, the limit from the left must approach the same value as the limit from the right.
Existence of a Limit
Lxfxfaxax
)(lim)(limIf
then Lxfax
)(lim
)(lim xfax
)(lim xfax
and
are called one-sided limits
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