AP CALCULUS
1003 Limits pt.3
Limits at Infinity and End Behavior
REVIEW:
ALGEBRA is a ________________________ machine that ___________________ a function ___________ a point.
CALCULUS is a ________________________ machine that ___________________________ a function ___________ a point
Function Evaluates
at
Limit
Describes the behavior of
near
END BEHAVIOR
lim x
LIMITS AT INFINITY
Part 2: End Behavior
GENERAL IDEA: The behavior of a function as x gets very large ( in a positive or negative direction)
ALREADY KNOW many functions:
Polynomial: x2
Trigonometric: cyclical
Exponential:
Logarithmic:
NOW primarily RATIONAL and COMPOSITE.
( )x
x3
END Behavior: Limit
Laymanβs Description:
Notation:
Horizontal Asymptotes:
Note:
( )x
m
If x>m then
Closer to L than Ξ΅
limπ₯ββ
π (π₯ )=πΏ
If it has a limit = L then the HA y=L
π
GNAW: Graphing
EX:
EX:
2
2
3( )
1
xf x
x
2
2( )
1
xf x
x
( )
( )x
x
Lim f x
Lim f x
( )
( )x
x
Lim f x
Lim f x
If you cover the middle what happens?
Gives 2 HA
2
-2
3
3
GNAW: Algebraic
Method: DIRECT SUBSTITUTION gives a second INDETERMINANT FORM
Theorem: limπ₯ββ
ΒΏπ₯ΒΏ
ΒΏ=0
Method: Divide by largest degree in denominator
End Behavior Models
EX: (with Theorem)
2 1
1x
xLim
x
3
2
2 5
3 1x
xLim
x
=
=
limπ₯ββ
π (π₯ )=2End behavior HA y=2
=
limπ₯ββ
π (π₯ )=π·ππΈ
limπ₯ββ
π (π₯ )=βEnd behavior acts like y =
0
0
00
0
0
0
0
End Behavior Models
2
2
2 5
3 1x
xLim
x
Summary: ________________________________________
A).
B).
C).
3
2
2 5
3 1x
xLim
x
2
2 5
3 1x
xLim
x
Leading term determines end behavior. Even exp bothodd exp
Leading term test (reduce leading term)
If degree on bottom is largest limit = 0
If the degrees are the same then the limit = reduced fraction
If the degrees on top is larger limit DNE but EB acts like reduced power function
limπ₯ββ
π (π₯ )= 23 π₯
=0
E.B. HA y=0
limπ₯ββ
π (π₯ )=23=
23
EB y=
limπ₯ββ
π (π₯ )=2π₯3
=π·ππΈ
EB acts like y =
Continuity
General Idea:
General Idea: ________________________________________
We already know the continuity of many functions:
Polynomial (Power), Rational, Radical, Exponential, Trigonometric, and Logarithmic functions
DEFN: A function is continuous on an interval if it is continuous at each point in the interval.
DEFN: A function is continuous at a point IFF
a)
b)
c)
Can you draw without picking up your pencil
Has a point f(a) exists
Has a limit limπ₯βπ
π (π₯ )ππ₯ππ π‘π
Limit = value limπ₯βπ
π (π₯ )= π (π)
Continuity Theorems
Interior Point: A function is continuous at an interior point of its
domain if
ONE-SIDED CONTINUITY
Endpoint: A function is continuous at a left endpoint of i
limx c
y f x c
y f x a
f x f c
ts domain
if lim
or
continuous at a right endpoint if lim .
x a
x b
f x f a
b f x f b
Continuity on a CLOSED INTERVAL.
Theorem: A function is Continuous on a closed interval if it is continuous at every point in the open interval and continuous from one side at the end points.
Example :The graph over the closed interval [-2,4] is given.
From the right
From the left
Discontinuity
No valuef(a) DNE hole
jump
No limit
Limit does not equal valueLimit β value
Vertical asymptotea)
c)
b)
Discontinuity: cont.
Method:
(a).
(b).
(c).
Removable or Essential Discontinuities
Test the value = Look for f(a) =
Test the limitlimπ₯βπβ
π (π₯ )=ΒΏΒΏlim
π₯βπ+ΒΏ π (π₯ )=ΒΏ ΒΏΒΏΒΏ
Holes and hiccups are removableJumps and Vertical Asymptotes are essential
Test f(a) = limπ₯βπ
π (π₯) limπ₯βπ
π (π₯)f(a) =
00hπππ
Vertical Asymptote
Lim DNEJump
= cont.β hiccup
Examples:
EX:2
( )4
xf x
x
Identify the x-values (if any) at which f(x)is not continuous. Identify the reason for the discontinuity and the type of discontinuity. Is the discontinuity removable or essential?
removable or essential?
00
=
xβ 4
limπ₯β 4
π (π₯ )=00
Hole discontinuous because f(x) has no valueIt is removable
Examples: cont.
2
1( )
( 3)f x
x
Identify the x-values (if any) at which f(x)is not continuous. Identify the reason for the discontinuity and the type of discontinuity. Is the discontinuity removable or essential?
removable or essential?
xβ 3
limπ₯β 3
1(π₯β3 )2
=10
VA discontinuous because no value
It is essential
Examples: cont.
3 , 1( )
3 , 1
x xf x
x x
Identify the x-values (if any) at which f(x)is not continuous. Identify the reason for the discontinuity and the type of discontinuity. Is the discontinuity removable or essential?
Step 1: Value must look at 4 equationf(1) = 4
Step 2: Limitlimπ₯β1β
3+π₯=4
limπ₯β1+ΒΏ 3βπ₯=2ΒΏ
ΒΏ
limπ₯β 1
π (π₯ )=π·ππΈ 2β 4
It is a jump discontinuity(essential) because limit does not exist
Graph:
Determine the continuity at each point. Give the reason and the type of discontinuity.
x = -3
x = -2
x = 0
x =1
x = 2
x = 3
Hole discont. No value
VA discont. Because no value no limit
Hiccup discont. Because limit β value
Continuous limit = value
VA discont. No limit
Jump discont. Because limit DNE
Algebraic Method
2
3 2 2( )
3 4 2
x xf x
x x
a.
b.
c.
Value: f(2) = 8
Look at function with equal
Limit:
limπ₯β 2β
π (π₯ )=8
limπ₯β2+ΒΏ π (π₯ )=8ΒΏ
ΒΏ
limπ₯β 2
π (π₯ )=8
Limit = value: 8=8
Limit = Value
Algebraic Method 2
2
2
1- 1
( ) - 2 1 3
9 3
3
x x
f x x x
xx
x
At x=1a.
b.
c.
At x=3a.
b. Limit
c.
Value:
Limit: limπ₯β 1βπ (π₯ )=1βπ₯ 2=0
limπ₯β 1+ΒΏ π (π₯ )=π₯ 2β2=β 1ΒΏ
ΒΏ
limπ₯β 1
π (π₯ )=π·ππΈ
Jump discontinuity because limit DNE, essential
Value: x=3 f(3) =
Hole discontinuity because no value, removable
No furth
er test n
ecessary
f(1) = -1
Rules for Finding Horizontal Asymptotes
1. If degree of numerator < degree of denominator, horizontal asymptote is the line y=0 (x axis)
2. If degree of numerator = degree of denominator, horizontal asymptote is the line y = ratio of leading coefficients.
3. If degree of numerator > degree of denominator, there is no horizontal asymptote, but possibly has an oblique or slant asymptote.
Consequences of Continuity:
A. INTERMEDIATE VALUE THEOREM
** Existence Theorem
EX: Verify the I.V.T. for f(c) Then find c.
on 2( )f x x 1,2 ( ) 3f c
If fΒ© is between f(a) and f(b) there exists a c between a and b
ca b
f(a)
f(c)
f(b)
f(1) =1f(2) = 4Since 3 is between 1 and 4. There exists a c between 1 and 2 such that f(c) =3 x2=3 x=Β±1.732
Consequences: cont.
EX: Show that the function has a ZERO on the interval [0,1].3( ) 2 1f x x x
I.V.T - Zero Locator Corollary
CALCULUS AND THE CALCULATOR:
The calculator looks for a SIGN CHANGE between Left Bound and Right Bound
f(0) = -1f(1) = 2Since 0 is between -1 and 2 there exists a c between 0 and 1 such that f(c) = c
Intermediate Value Theorem
Consequences: cont.
EX: ( 1)( 2)( 4) 0x x x
I.V.T - Sign on an Interval - Corollary(Number Line Analysis)
0 1 2 3 4 5-1-2-3-4-5
We know where the zeroes are located
Choose a point between them to determine whether the graph is positive or negative
Consequences of Continuity:
B. EXTREME VALUE THEOREM
On every closed interval there exists an absolute maximum value and minimum value.
x
y
x
y