1003 Limits pt.3
Limits at Infinity and End Behavior
AP CALCULUS
REVIEW:
ALGEBRA is a ________________________
Function
Evaluates
machine that ___________________
a function
___________
a point.
at
Limit
CALCULUS is a ________________________
Describes the behavior of
machine that ___________________________
a function ___________
a point
near
END BEHAVIOR
lim
x
LIMITS AT INFINITY ( x   )
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 ↑↑ x3↑↓
Trigonometric: cyclical
Exponential:
Logarithmic:
NOW primarily RATIONAL and COMPOSITE.
END Behavior: Limit
Layman’s Description:
( x   )
If x>m then 𝑓 𝑥 − 𝐿 < 𝜀
𝜀
m
Notation:
Closer to L than ε
lim 𝑓 𝑥 = 𝐿
𝑥→∞
Horizontal Asymptotes:
If it has a limit = L then the HA y=L
Note:
GNAW: Graphing
2
3
x
EX: f ( x) 
x2  1
Lim f ( x)  3
x
Lim f ( x)  3
x
EX: f ( x) 
2x
x2  1
Lim f ( x)  2
x
𝑥2
= 𝑥
Gives 2 HA
Lim f ( x)  -2
x
If you cover the
middle what happens?
GNAW: Algebraic
Method: DIRECT SUBSTITUTION gives a second INDETERMINANT FORM


Theorem:
Method:
#
lim # = 0
𝑥→∞ 𝑥
Divide by largest degree in denominator
End Behavior Models
EX: (with Theorem)
2𝑥 1
−
𝑥 𝑥
𝒙 1
𝑥→∞ 𝒙+ 𝑥
lim
2x 1
Lim
=
x  x  1
0 10
2−𝑥
2
=
=
0
0 1
1+𝑥
1
lim 𝑓 𝑥 = 2
𝑥→∞
End behavior HA y=2
2x  5
Lim 2
x  3 x  1
3
3
0
0= lim
2𝑥 𝟓
+
𝑥2 𝒙𝟐
𝟐
𝟑𝒙 1
𝑥→∞ 𝟐 + 2
𝒙
𝑥
lim
𝑥→∞
0
2
0= 𝑥
5
2𝑥+ 2
𝑥
1
3+ 2
𝑥
3
lim 𝑓 𝑥 = 𝐷𝑁𝐸
𝑥→∞
lim 𝑓 𝑥 = ∞
𝑥→∞
End behavior acts like y =
2
𝑥
3
Leading term determines end
behavior. Even exp both ↑ odd exp
↑↓
End Behavior Models
Leading term test (reduce leading term)
Summary: ________________________________________
A). If degree on bottom is largest limit = 0
B). If the degrees are the same then the limit =
reduced fraction
C). If the degrees on top is larger limit DNE but EB acts
like reduced power function
2x  5
Lim 2
x  3 x  1
lim 𝑓 𝑥 =
𝑥→∞
2
=0
3𝑥
E.B. HA y=0
2 x2  5
Lim 2
x  3 x  1
lim 𝑓 𝑥 =
𝑥→∞
2
3
EB y=
2 2
=
3 3
2 x3  5
Lim 2
x  3 x  1
lim 𝑓 𝑥 =
𝑥→∞
2𝑥
= 𝐷𝑁𝐸
3
2
3
EB acts like y = 𝑥
Continuity
General Idea:
General Idea: ________________________________________
Can you draw without picking up your pencil
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)
Has a point
f(a) exists
b)
Has a limit
lim 𝑓 𝑥 𝑒𝑥𝑖𝑠𝑡𝑠
c)
Limit = value
𝑥→𝑎
lim 𝑓 𝑥 = 𝑓(𝑎)
𝑥→𝑎
Continuity Theorems
Interior Point: A function y  f  x  is continuous at an interior point c of its
domain if
lim
f x  f c 
xc  
ONE-SIDED CONTINUITY
Endpoint: A function y  f  x  is continuous at a left endpoint a of its domain
if lim f  x   f  a 
x a
or
continuous at a right endpoint b if
lim f  x   f  b  .
x 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
a)
b)
No value
f(a) DNE
hole
No limit
lim 𝑓 𝑥 𝐷𝑁𝐸
𝑥→𝑎
c)
Vertical
asymptote
Limit does not equal
value
Limit ≠ value
jump
Discontinuity: cont.
Method:
0
ℎ𝑜𝑙𝑒
0
(a). Test the value =
1
0
Vertical
Asymptote
Lim DNE
Jump
= cont.
≠ hiccup
(b). Test the limit
Look for f(a) =
lim 𝑓 𝑥 =
lim+ 𝑓 𝑥 =
𝑥→𝑎−
𝑥→𝑎
lim 𝑓(𝑥)
(c). Test f(a) = 𝑥→𝑎
Removable or
f(a) =
Essential Discontinuities
Holes and hiccups are removable
Jumps and Vertical Asymptotes are essential
lim 𝑓(𝑥)
𝑥→𝑎
Examples:
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?
EX:
0
x 2
=
f ( x) 
0
x4
x≠ 4
0
lim 𝑓 𝑥 =
𝑥→4
0
Hole discontinuous because f(x) has no value
It is removable
removable or essential?
Examples: cont.
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?
1
f ( x) 
( x  3) 2
x≠3
lim
𝑥→3
1
𝑥−3
=
2
1
0
VA discontinuous because no value
It is essential
removable or essential?
Examples: cont.
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?
3  x, x  1
f ( x)  
3  x, x  1
Step 1: Value must look at 4 equation
f(1) = 4
Step 2: Limit
lim 3 + 𝑥 = 4
𝑥→1−
lim 3 − 𝑥 = 2
𝑥→1+
lim 𝑓 𝑥 = 𝐷𝑁𝐸 2 ≠ 4
𝑥→1
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
Hole discont.
No value
VA discont.
Because no value
no limit
Hiccup discont.
Because limit ≠
value
x =1
Continuous limit
= value
x=2
VA discont. No
limit
x=3
Jump discont.
Because limit
DNE
Algebraic Method
 3x  2 x  2
f ( x)   2
3x  4 x  2
Look at function with equal
a. Value: f(2) = 8
lim 𝑓 𝑥 = 8
b. Limit:
𝑥→2−
lim 𝑓 𝑥 = 8
𝑥→2+
lim 𝑓 𝑥 = 8
𝑥→2
c. Limit = value: 8=8
Limit = Value ∴ 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑠
Algebraic Method

 1- x 2
x 1
 2
f ( x)   x - 2 1  x  3
 x2  9

x3
 x  3
At x=1
a. Value: f(1) = -1
lim 𝑓
b. Limit: 𝑥→1−
𝑥 = 1 − 𝑥2 = 0
At x=3
0
a. Value: x=3 f(3) =
0
b.
Limit
lim 𝑓 𝑥 = 𝑥2 − 2 = −1
𝑥→1+
lim 𝑓 𝑥 = 𝐷𝑁𝐸
𝑥→1
c.
Jump discontinuity
because limit DNE,
essential
Hole discontinuity
c. because no value,
removable
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
f(b)
If f© is between f(a) and
f(b) there exists a c
between a and b
f(c)
f(a)
** Existence Theorem
a
EX: Verify the I.V.T. for f(c)
f ( x)  x 2
on
c
b
Then find c.
1, 2
f (c )  3
f(1) =1
f(2) = 4
Since 3 is between 1 and 4. There exists a c between 1 and 2 such
that f(c) =3 x2=3 x=±1.732
Consequences: cont.
I.V.T - Zero Locator Corollary
Intermediate Value Theorem
EX: Show that the function has a ZERO on the interval [0,1].
f ( x)  x 3  2 x  1
f(0) = -1
f(1) = 2
Since 0 is between -1 and 2 there exists a c between
0 and 1 such that f(c) = c
CALCULUS AND THE CALCULATOR:
The calculator looks for a SIGN CHANGE between Left Bound and
Right Bound
Consequences: cont.
I.V.T - Sign on an Interval - Corollary
(Number Line Analysis)
EX:
We know where the
zeroes are located
( x  1)( x  2)( x  4)  0
-5 -4 -3 -2
-1 0
1 2
3
4
5
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.
y
y












x
x









