Definition and finding the limit When substitution results in a 0/0 fraction, the result is called...
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Transcript of Definition and finding the limit When substitution results in a 0/0 fraction, the result is called...
Definition and finding the limit
When substitution results in a 0/0 fraction, the result is called an indeterminate form.
The limit of an indeterminate form exists, but to find it you must use a technique, such as factor and cancel.
Calculus Date: 9/26/14
Objective: SWBAT define, calculate & apply properties of limits graphically, numerically and now analytically.
Do Now –
Mini Quiz 5 minutes Take out a piece of paper. Can be a half sheet.
HW Requests: HW: pg 30 SM all
In class: worksheets ( possibilities Worksheet limits or McCafrey)
HW: Complete WorksheetsAnnouncements:
Mandatory session Sine and Cosine functions starting
with the Unit Circle
Quiz Friday
To get ahead,You have to do extra!
Mini Quiz7 minutes
2.Show your work then
1. Try Substitution, if doesn’t work
2. Try Factor and cancel and then3. Try Substitution again, if doesn’t work4. Do DNE or +/- infinity check
Techniques-Finding limits for Rational Expressions
Let’s go to the SM pg #28 #1-8 (10) HW: pg 30 SM all
- If the right and left side limit are not equal the limit does not exist - DNE
- If the right and left side limit are not equal the limit does not exist – DNE
One sided Limits- If the left side number is negative then the - If the left side number is positive then the - If the right side number is negative then the - If the left side number is negative then the
Rationalizing Technique
x
xx
11lim
0
If there is a radical in the numerator or the denominator,rationalize, simplify and cancel, then try substitution.
)11(
1
)11()11(
11
11
11.
11
x
xx
x
xx
x
x
x
x
x
Substituting we get 2
1
)110(
1
Hint: Often you can cancel a common term in the numerator and denominator when simplifying
Rationalizing Technique
xxx
x
x
x
x
x
2
1
)2)(4(
4
2
2.
4
2
Rationalize, simplify (cancel) and try substitution.
Substituting we get = 1 4
Try This
Find:
f(0)is undefined; 2 is the limit
2( )
1 1
xf x
x
0lim ( )x
f x
Find: Try This
( ) , 01 1
xf x x
x
f(0) is defined; 2 is the limit
21
0lim ( )x
f x
1, x = 0
Try ThisFind the limit if it exists:
0limx
x
x
DNE
Try ThisFind: if
1lim ( )x
f x
1lim ( ) 3x
f x
Try ThisFind 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
Example
Find the limit if it exists:3
1
1lim
1x
x
x
Try substitution
Example
Find the limit if it exists:3
1
1lim
1x
x
x
Substitution doesn’t work…does this mean the limit doesn’t exist?
Use the factor & cancellation technique
3 21 ( 1)( 1)
1 1
x x x x
x x
2 1x x and
are the same except at x=-1
Use the factor & cancellation technique
2 1x x
After factoring and cancelling, now try substituting -1 again.
= 33
1
1lim
1x
x
x
Try This
Find the limit if it exists:2
3
6lim
3x
x x
x
5
Isn’t
that
easy?
Did you think ca
lculus
was going to
be
difficu
lt?
Try ThisSolve using limit properties and substitution:2
2
4lim
3x
x x
x
6
Try ThisFind the limit if it exists:
22
2lim
4x
x
x
1
4
ExampleSometimes limits do not exist. Consider:
3
2
3lim
2x
x
x
If substitution gives a constant divided by 0, the limit does not exist (DNE)
Try This
Find the limit if it exists:2
3
6lim
3x
x x
x
The limit doesn’t existConfirm by graphing
Lesson Close
Name 3 ways a limit may fail to exist.
Exit Ticket
• In Class: SM – pg 28 #1-5
• HW: SM pg 30 #1-15
Try ThisFind the limit if it exists:
2
1
2 3lim
1x
x x
x
-5
Limit properties again
The existence or non-existence of f(x) as x approaches c has no bearing on the existence of the limit of f(x) as x approaches c.
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.
In order for a limit to exist at c, the left-hand limit must equal the right hand limit.
Limits, again!
)(lim)(lim xfxfcxcx
Lxfcx
)(lim
If the left-hand limit equals the right hand limit, then the limit exists and we write:
Watch out for piecewise functions
When finding the limit of a function it is important to let x approach a from both the right
and left. If the same value of L is approached by the function then
the limit exist and
Lxf )(limax
ExampleConsider3 1
( )1
xf x
x
for ( ,1) (1, ) and
3
1
1lim
1x
x
x
=?
Try ThisGraph and find the limit (if it exists):
3
3lim
3x x DNE
ExampleTrig functions may have limits.
2
lim(sin )x
x
Try This
2
lim(cos )x
x
2
lim(cos ) cos 02x
x
Using the Product Rule Technique
Important Idea
The functions have the same limit as x-1
Try This
Graph and
3
1
1
1
xY
x
2
2 1Y x x on the same axes. What is the
difference between these graphs?
3 1( )
1
xf x
x
Why is there a “hole” in the graph at x=1?
Analysis
Example
Consider
3 1( ) , 1
1
xf x x
x
What happens at x=1?
x .75 .9 .99 .999
f(x)
Let x get close to 1 from the left:
Try This
Consider
3 1( ) , 1
1
xf x x
x
x 1.25 1.1 1.01
1.001
f(x)
Let x get close to 1 from the right:
Try ThisWhat number does f(x) approach as x approaches 1 from the left and from the right?
3
1
1lim 3
1x
x
x
Informal Definition of Limit• If f(x) becomes arbitrarily close to a single
number L as x approaches a number c from either side, the limit of f(x), as x approaches c, is L.
Lxfcx
lim
- Definition of Limit
• Let f be a function defined on an open interval containing c (except possibly at c) and let L be a real number. The statement means that for each >0 there exists a >0 such that if then .
Lxfcx
lim
cx0 Lxf
Basic Limits
•
•
•
bbcx
lim
cxcx
lim
nn
cxcx
lim
Constant Function Limits
bbax
lim
• a and b are both constants
• This means that for any constant function f(x) = b, as x approaches any constant a, the limit will always be b.
Linear Function Limits
axax
lim
The limit of f(x) = x as x approaches any constant is the constant itself.
Exponential Function Limits
ax
axbb
lim
Just plug in a for x
Properties
Let and
Scalar multiple: Sum or difference:
Product:
Quotient: , if K0
Power:
Lxfcx
lim Kxgcx
lim
bLxbfcx
lim
KLxgxfcx
lim
LKxgxfcx
lim
K
L
xg
xfcx
lim
nn
cxLxf
lim
• Let's try a practice problem.
• Property (B) tells us we can split these apart:
• Using limit (1) and limit (2) from the basic limits, we get:
3lim2
xx
3limlim22
xx
x
1323lim
33lim
2lim
2
2
2
x
x
x
x
x
Putting it all together
• So,
5
714131
743lim
23
23
1
xxxx
This is called the Substitution method
Try ThisSolve using limit properties and substitution:
2
3lim 2 3 2x
x x
25