Download 1-1:Introduction To Limits

1.2:Rates of Change & Limits
Learning Goals:
•Calculate average &
instantaneous speed
•Define, calculate & apply
properties of limits
•Use Sandwich Theorem
©2009 Mark Pickering
Important Ideas
•Limits are what make
calculus different from
algebra and trigonometry
•Limits are fundamental to
the study of calculus
•Limits are related to rate of
change
•Rate of change is important
in engineering & technology
Theorem 1
Limits have the following
properties:
if lim f ( x)  L & lim g ( x)  M
x c
then:
x c
f ( x)  g ( x)   L  M

1. lim
x c
Theorem 1
Limits have the following
properties:
if lim f ( x)  L & lim g ( x)  M
x c
then:
2.
x c
lim  f ( x)  g ( x)   L  M
x c
Theorem 1
Limits have the following
properties:
if lim f ( x)  L & lim g ( x)  M
x c
then:
3.
x c
lim  f ( x) g ( x)   L M
x c
Theorem 1
Limits have the following
properties:
if lim f ( x)  L & k a constant
x c
then:
k f ( x)   k L

4. lim
x c
Theorem 1
Limits have the following
properties:
if lim f ( x)  L & lim g ( x)  M
x c
then:
x c
f ( x) L
 ,M  0
5. lim
x c g ( x )
M
Theorem 1
Limits have the following
properties:
if lim f ( x)  L & r & s are
x c
integers, then:
r
s
r
s
6. lim  f ( x)   L
x c
Theorem 1
Limits have the following
properties:
if f ( x)  k where k is a
constant, then:
f ( x)  lim k  k
7. lim
x c
x c
(not in your text as Th. 1)
Theorem 2
For polynomial and rational
functions:
a. lim f ( x)  f (c)
x c
f ( x ) f (c )
lim

,
g
(
c
)

0
b. xc g ( x) g (c)
Limits may be found by
substitution
Example
Solve using limit properties
and substitution:
lim  2 x  3 x  2 
2
x 3
Try This
Solve using limit properties
and substitution:
2
x x4
lim
x 2
x3
6
Example
Sometimes limits do not
exist. Consider:
3
x 3
lim
x 2 x  2
If substitution gives a
constant divided by 0, the
limit does not exist (DNE)
Example
Trig functions may have
limits.
lim(sin x)
x

2
Try This
lim(cos x)
x

2
lim(cos
x
)

cos

x
2

2
0
Example
Find the limit if it exists:
x 1
lim
x  1 x  1
3
Try
substitution
Example
Find the limit if it exists:
Substitution
3
x  1 doesn’t
lim
x  1 x  1 work…does
this mean the
limit doesn’t
exist?
Important Idea
3
2
x  1 ( x  1)( x  x  1)

x 1
x 1
and
x  x 1
2
are the same except at
x=-1
Important Idea
The functions have the
same limit as x-1
Procedure
1.Try substitution
2. Factor and cancel if
substitution doesn’t work
3.Try substitution again
The factor & cancellation
technique
Try This
Find the limit if it exists:
x  x6
lim
x 3
x3
2
5
Try This
Find the limit if it exists:
2 x
lim 2
x2 x  4
1

4
Try This
Find the limit if it exists:
x  x6
lim
x  3
x3
Confirm by graphing
The limit doesn’t exist
2
Definition
When substitution results
in a 0/0 fraction, the result
is called an indeterminate
form.
Important Idea
The limit of an
indeterminate form exists,
but to find it you must use
a technique, such as factor
and cancel.
Try This
Find the limit if it exists:
2x  x  3
lim
x  1
x 1
2
-5
Try This
x 1
Graph Y1 
and
x 1
2
Y2  x  x  1 on the same
axes. What is the difference
between these graphs?
3
Analysis
Why is
there a
“hole” in
the graph
at x=1?
x 1
f ( x) 
x 1
3
Consider
x 1
f ( x) 
x 1
Example
3
for (,1)
and
f ( x)  4
for x=1
(1, )
x 1
lim
=?
x 1 x  1
3
Try This
f ( x) if
Find: lim
x 1
f ( x)  x  2, x  1
2
f ( x)  1, x  1
lim f ( x)  3
x 1
Important Idea
The existence or nonexistence of f(x) as x
approaches c has no bearing
on the existence of the limit
of f(x) as x approaches c.
Important Idea
What matters is…what value
does f(x) get very, very
close to as x gets very,very
close to c. This value is the
limit.
Try This
f ( x)
Find: lim
x0
x
f ( x) 
x 1 1
2
f(0)is undefined; 2 is the
limit
Find: lim f ( x)
x0
f ( x) 
x
x 1 1
,x  0
f ( x)  1, x  0
Try This
2
1
f(0) is defined; 2 is the limit
Try This
Find the limit of f(x) as x
approaches 3 where f is
defined by:
2 , x  3
f ( x)  
3 , x  3
lim f ( x)  2
x 3
Try This
Graph and find the limit (if it
exists):
3
lim
x 3 x  3
DNE
Theorem 3: One-sided & Two
Sided limits
if lim f ( x)  L (limit from right)

x c
and
lim f ( x)  L (limit from left)

x c
f ( x)  L (overall limit)
then xlim
c
Theorem 3: One-sided & Two
Sided limits (Converse)
if lim f ( x)  L (limit from right)

x c
and
lim f ( x)  M (limit from left)

x c
then lim f ( x)   (DNE)
x c
Example
x 1
, x 1
Consider f ( x) 
x 1
3
What happens at x=1?
Let x get close to 1 from the
left:
x .75 .9 .99 .999
f(x)
Try This
3
x 1
, x 1
Consider f ( x) 
x 1
Let x get close to 1 from the
right:
x
f(x)
1.25 1.1
1.01 1.001
Try This
What number does f(x)
approach as x approaches
1 from the left and from
the right?
x 1
lim
3
x 1 x  1
3
Try This
Find the limit if it exists:
lim
x0
DNE
x
x
Example
Find the limit if it exists:
1
lim sin  
x 0
x
Example
1
1.Graph sin   using a
 x
friendly window:
2. Zoom at x=0
3. Wassup at x=0?
Important Idea
If f(x) bounces from one
value to another (oscillates)
as x approaches c, the limit
of f(x) does not exist at c:
Theorem 4: Sandwich
(Squeeze) Theorem
Let f(x) be between g(x) &
h(x) in an interval containing
c. If lim g ( x)  lim h( x)  L
x c
x c
then: lim f ( x)  L
x c
f(x) is “squeezed” to L
Example
Find the limit if it exists:
lim
 0
sin 

Where  is in radians and
  
in the interval   , 

2 2
Example
Find the limit if it exists:
lim
 0
sin 

Substitution gives the
indeterminate form…
Example
Find the limit if it exists:
lim
 0
sin 

Factor and cancel doesn’t
work…
Example
Find the limit if it exists:
lim
 0
sin 

Maybe…the squeeze
theorem…
Example
g()=1
f ( ) 
sin 

h()=cos
Example
lim1  1
 0
&
lim cos   1
 0
therefore…
lim
 0
sin 

1
Two Special Trig Limits
lim
 0
lim
 0
sin 

1
1  cos

0
Example
Find the limit if it exists:
tan x
lim
x 0
x
sin x
1
lim
 lim
 11  1
x 0
x

0
x
cos x
Example
Find the limit if it exists:
sin(5x)
lim
x 0
x
 sin(5 x) 
 sin(5 x) 
lim 5 
 5 lim 
 5 1  5


x0
x0
5
x


 5x 
Try This
Find the limit if it exists:
3  3 cos x
lim
x 0
x
0
Lesson Close
Name 3 ways a limit may
fail to exist.
Practice
1. Sec 1.2 #1, 3, 8, 9-18,
28-38E (just find limit L),
39-42gc (graphing
calculator), 43-45