Since d = rt,
d
t
Average Speed
Example:
Suppose you drive 200 miles in 4 hours. What is
your average speed?
d 200 miles

t
4hours
= 50 mph

The moment you look at your speedometer,
you see your instantaneous speed.
Example
A 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.)
Example
A 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
d 16(2  t )  16(2)

t
(2  t )  2
2
16(2  t ) 2  64

t
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.
16(2  t )  64
t
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.
Algebraically:
16(2  t )  64 16(4  4t  t )  64

t
t
2
2
64  64t  16t  64

t
2
64t  16t

t
t (64  16 t )

t
2
Algebraically:
t (64  16 t )

t
 64  16 t
Now, when Δt is 0, our
average speed is
64 ft/sec

Let f be a function defined on a open interval
containing a, except possibly at a itself.
Then, there exists a
 0
such that
f ( x)  L  
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
0  x  c    f ( x)  L  
Still,
WHAT THE
CRAP??????
lim f ( x)  L
x c
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).
lim f  x   2
x 2
not 1
Properties of Limits
Limits can be added, subtracted, multiplied, multiplied by a
constant, divided, and raised to a power.
If L, M, c, and k are real numbers and
lim f ( x)  L and lim g ( x)  M then
x c
x c

f ( x)  g ( x)   L  M
1.) Sum Rule: lim
x c

f ( x)  g ( x)   L  M
2.) Difference Rule: lim
x c
3.) Product Rule:
lim f ( x)  g ( x)   L  M
x c
Properties of Limits

k  f ( x)   k  L
4.) Constant Multiple Rule: lim
x c
f ( x) L

, M0
5.) Quotient Rule: lim
x c g ( x )
M
6.) Power Rule:
If r and s are integers,s  0, then
lim f ( x) 
r/s
x c
 Lr / s

k  k and lim x  c and the properties
Use lim
xc
xc
of limits to find the following limits:
( x  4 x  3)
a.) lim
x c
3
2
 lim x 3  lim 4 x 2  lim 3
x c
x c
   
x4  x2 1
b.) lim
x c
x2  5
x c

 lim x  4 lim x  lim 3
3
x c
x c
2
x c
 c3  4c 2  3

x c


lim x 2  5
x c
x c

lim x 4  x 2  1

lim x 4  lim x 2  lim1
x c
x c
2
x c
lim x  lim 5
x c
x c
c4  c2 1

c2  5

If f is a continuous function on an open
interval containing the number a, then
lim f ( x )  f ( a )
xa
(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: lim 3x  1
x 5
 3(5) 1
4
2.) Using Properties of Limits
Ex: Find the limit: lim 3 x sin x

x 

 lim 3 x lim sin x
x 
x 
 (3 )(sin )
 (3 )(0)
0

(product rule)
3.) Factoring & Simplifying
2
x  3x  2
Ex: Find the limit: lim
x 2
x2
What happens if we
just substitute in
the limit?
( x  1)( x  2)
x2
 lim x  1
x2
 2  1
When something like
this happens, we
need to see if we can
factor & simplify!
1
HOLY
COWCULUS!!!
4.) Using the conjugate
t2  2
Ex: Find the limit: lim
t 0
t
What happens if we
just substitute in
the limit?
02  2
0
0 We must simplify
0 again.
t2  2
t2  2
 lim

t 0
t
t2  2
t 22
 lim
t 0 t ( t  2 
2)
1
 lim
t 0
t2  2

1
1

02  2
2 2
5.) Use a table or graph
Ex: Find the limit:
3 sin 2 x
lim
x 0
2x
What happens if we
just substitute in
the limit?
3 sin 2(0)
2(0)
3 sin 0
0
0
0
As 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 that
g(x) ≤ f(x) ≤ h(x) for all x in the interval
except at possibly at a itself, and
lim g ( x)  lim h( x)  L
xa
xa
then, lim f ( x)  L
xa
h(x)
f(x)
g(x)
1
Ex: Find the limit: lim x sin  
x 0
 x
2
sin oscillates between -1 and 1, so
1
Now, let’s get the problem to look like the
 1  sin    1 one given.
 x
1
2
2
 x  x sin    x 2
 x
 2  1 
2
2
lim  x  lim x sin    lim x
x 0
x 0
 x   x0



 
1
Ex: Find the limit: lim x sin  
x 0
 x
2
 2  1 
0  lim x sin    0
x 0
 x 

Therefore, by the Sandwich Theorem,
1
lim x sin    0
x 0
 x
2
0.02
0.01
-0.2
-0.1
0
-0.01
-0.02
0.1
x
0.2

In order for a limit to exist, the limit from the
left must approach the same value as the
limit from the right.
If
lim f ( x)  lim f ( x)  L
xa
then
xa
lim f ( x)  L
xa
lim f ( x)
xa
and
lim f ( x)
xa
are called one-sided limits
lim f  x 
x 1
2
does not exist because the left
and right hand limits do not
match!
1
1
At x=1:
2
3
4
lim f  x   0
left hand limit
lim f  x   1
right hand limit
x 1
x 1
f 1  1
value of the function
lim f  x   1
2
x2
because the left and right
hand limits match.
1
1
At x=2:
2
3
4
lim f  x   1
left hand limit
lim f  x   1
right hand limit
x 2
x 2
f  2  2
value of the function
lim f  x   2
2
x 3
because the left and right
hand limits match.
1
1
At x=3:
2
3
4
lim f  x   2
left hand limit
lim f  x   2
right hand limit
x 3
x 3
f  3  2
value of the function

Section 2.1 (#7, 11, 15, 19, 21, 23, 27, 3136, 37, 43, 49, 63)