AP CALCULUS 1003 Limits pt.3 Limits at Infinity and End Behavior.

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

Layman’s Description:

Notation:

Horizontal Asymptotes:

Note:

( )x

m

If x>m then

Closer to L than ε

lim𝑥→∞

𝑓 (𝑥 )=𝐿

If it has a limit = L then the HA y=L

𝜀

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: lim𝑥→∞

¿𝑥¿

¿=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

=

=

lim𝑥→∞

𝑓 (𝑥 )=2End behavior HA y=2

=

lim𝑥→∞

𝑓 (𝑥 )=𝐷𝑁𝐸

lim𝑥→∞

𝑓 (𝑥 )=∞End 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

lim𝑥→∞

𝑓 (𝑥 )= 23 𝑥

=0

E.B. HA y=0

lim𝑥→∞

𝑓 (𝑥 )=23=

23

EB y=

lim𝑥→∞

𝑓 (𝑥 )=2𝑥3

=𝐷𝑁𝐸

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 lim𝑥→𝑎

𝑓 (𝑥 )𝑒𝑥𝑖𝑠𝑡𝑠

Limit = value lim𝑥→𝑎

𝑓 (𝑥 )= 𝑓 (𝑎)

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 ≠ value

Vertical asymptotea)

c)

b)

Discontinuity: cont.

Method:

(a).

(b).

(c).

Removable or Essential Discontinuities

Test the value = Look for f(a) =

Test the limitlim𝑥→𝑎−

𝑓 (𝑥 )=¿¿lim

𝑥→𝑎+¿ 𝑓 (𝑥 )=¿ ¿¿¿

Holes and hiccups are removableJumps and Vertical Asymptotes are essential

Test f(a) = lim𝑥→𝑎

𝑓 (𝑥) lim𝑥→𝑎

𝑓 (𝑥)f(a) =

00h𝑜𝑙𝑒

Vertical Asymptote

Lim DNEJump

= cont.≠ 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

=

x≠ 4

lim𝑥→ 4

𝑓 (𝑥 )=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?

x≠3

lim𝑥→ 3

1(𝑥−3 )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: Limitlim𝑥→1−

3+𝑥=4

lim𝑥→1+¿ 3−𝑥=2¿

¿

lim𝑥→ 1

𝑓 (𝑥 )=𝐷𝑁𝐸 2≠ 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 ≠ 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:

lim𝑥→ 2−

𝑓 (𝑥 )=8

lim𝑥→2+¿ 𝑓 (𝑥 )=8¿

¿

lim𝑥→ 2

𝑓 (𝑥 )=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: lim𝑥→ 1−𝑓 (𝑥 )=1−𝑥 2=0

lim𝑥→ 1+¿ 𝑓 (𝑥 )=𝑥 2−2=− 1¿

¿

lim𝑥→ 1

𝑓 (𝑥 )=𝐷𝑁𝐸

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 f© 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=±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