Index FAQ Functional limits Intuitive notion Definition How to calculate Precize definition Continuous functions.

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Functional limits

Intuitive notion Definition

How to calculatePrecize definition

Continuous functions

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Intuitive meaning of the limit of a function

You write , which means that as x “approaches” c, the function f( x) “approaches” the real number L

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Video help: http://www.calculus-help.com/tutorials/

Lesson 1: What Is a Limit?

Lesson 2: When Does a Limit Exist?

Lesson 3: How do you evaluate limits? Worked out EXAMPLES:

http://www.sosmath.com/calculus/limcon/limcon04/limcon04.html http://tutorial.math.lamar.edu/Classes/CalcI/LimitsAtInfinityII.aspx


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The limit of a function f( x) is a number, what the function intends to take, what we can observe om the graph of it

in other word: the number to which the functional values approach either in the infinity, or negative infinity, or at a certain point, which NOT necessarily belongs to the domain of the function.

The limit might or might not be equal to the functional value at that point in which the limit is taken

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Limit of the Function Note: we can approach a limit from

• left … right …both sides Function may or may not exist at that point At a

• right hand limit, no left• function not defined

At b • left handed limit, no right• function defined

a b

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A)x(flimxx

0

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You write :

L)x(flimx

c

which means that as x “approaches” c, the function f( x) “approaches” the real number L

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Can be observed on a graph.Observing a Limit

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Observing a Limit Can be observed on a graph.

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Non Existent Limits f(x) grows without bound

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Lxfcx

)(lim Lxfcx

)(lim

A)x(flimxx

0

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What is the number this function does intend to take?

-In the infinity?-In the negative infinity?-At zero? - At x=1 from the right? - At x=1 from the left?-At 2?

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What is the number this function does intend to take? -In the infinity:

-In the negative infinity?

-At zero?

-At x=1 from the right?

At x=1 from the left?

At x=2, substituting 2:

1)x(flimx

1)x(flimx

0)x(flim0x

)x(flim01x

)x(flim01x

3/2)x(flim2x


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A)x(flimxx

0

What function could it be?

Index FAQ Copyright (c) 2003 Brooks/Cole, a division of

Thomson Learning, Inc.

Computing Limits: substitution

Ex.

Ex.

2

3lim 1x

x

2

3 3lim lim1x xx

2

3 3

2

lim lim1

3 1 10x xx

1

2 1lim3 5x

xx

1

1

lim 2 1

lim 3 5x

x

x

x

1 1

1 1

2lim lim1

3lim lim5x x

x x

x

x

2 1 1

3 5 8

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Non Existent Limits



One-Sided Limit of a FunctionThe right-hand limit of f (x), as x approaches a, equals L

written:

if we can make the value f (x) arbitrarily close to L by taking x to be sufficiently close to the right of a.

lim ( )x a

f x L

a

L( )y f x



One-Sided Limit of a FunctionThe right-hand limit of f (x), as x approaches a, equals L

written:

if we can make the value f (x) arbitrarily close to L by taking x to be sufficiently close to the right of a.

lim ( )x a

f x L

a

L( )y f x



One-Sided Limit of a FunctionThe left-hand limit of f (x), as x approaches a, equals M

written:

if we can make the value f (x) arbitrarily close to L by taking x to be sufficiently close to the left of a.

lim ( )x a

f x M

a

M

( )y f x

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One sided limitsNumbers x near c fall into two natural categories: those that lie to the left of c and those that lie to the right of c. We write

[The left-hand limit of f(x) as x tends to c is L.]to indicate that

as x approaches c from the left, f(x) approaches L.

We write

[The right-hand limit of f(x) as x tends to c is L.]

to indicate that

as x approaches c from the right, f(x) approaches L

limx c

f x L

limx c

f x L

For a full limit to exist, both one-sided limits have to exist and they have to be equal.



One-Sided Limit of a Function2 if 3( )

2 if 3x xf xx x

Ex. Given

3lim ( )x

f x

3 3lim ( ) lim 2 6x x

f x x

2

3 3lim ( ) lim 9x x

f x x

Find

Find 3

lim ( )x

f x

The limit does not exists at 3, but it exists from the left, and from the right

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1, if x ＞ 0

−1, if x ＜ 0.

Let’s try to apply the limit process at different numbers c.

If c ＜ 0, then for all x sufficiently close to c,

x ＜ 0 and f(x) = −1. It follows that for c ＜ 0lim f(x) = lim (−1) = −1x → c x → c

If c ＞ 0, then for all x sufficiently close to c, x ＞ 0 and f(x) = 1. It follows that for c ＜ 0

lim f(x) = lim (1) = 1x → c x → c

However, the function has no limit as x tends to 0:

lim f(x) = −1 but lim f(x) = 1.x → 0- x → 0+

x/x)x(f

Example

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Computing Limits We saw already the first step:

substitution If fails: try to factorize the terms, then

simplify

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Example

2

x 2

x x 6 0lim Which is undefined!x 2 0

2

x 2 x 2 x 2

x x 6 (x 3)(x 2)lim lim lim (x 3) 5x 2 x 2

2x x 6NOTE : f ( x ) graphs as a straight line.x 2

Graph it.

Substitotion failed, but the limit exist!!!!

What happens at x = 2?

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?04

4)2(22

42lim 222

xx

x

Good job if you saw this as “limit does not exist” indicating a vertical asymptote at x = -2.

?00

4)2(22

42lim 222

xx

xThis limit is indeterminate. With some algebraic manipulation, the zero factors could cancel and reveal a real number as a limit. In this case, factoring leads to……

41

21lim

)2)(2(2lim

42lim

2

222

x

xxx

xx

x

xx The limit exists as x approaches 2 even though the function does not exist. In the first case, zero in the denominator led to a vertical asymptote; in the second case the zeros cancelled out and the limit reveals a hole in the graph at (2, ¼).

x

y

42)( 2

xxxf

Examples



Computing LimitsEx.

2

3 if 2lim ( ) where ( )

1 if 2x

x xf x f x

x

6

-2

2 2lim ( ) = lim 3x x

f x x

23 lim

3( 2) 6x

x

Note: f (-2) = 1

is not involved

The limiot exists at -2 because the left and right hand limits are equal

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Computing Limits We saw already the first step:

substitution If fails: try to factorize the terms, then

simplify If fails: try the „conjugate”

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Using ConjugatesUsing Conjugates

0

3 2lim 7x

xx

7 7

7 7

( 3 2) 3 2 ( 3) 4lim lim( 7) 3 2 ( 7) 3 2

7 1lim lim( 7) 3 2 3 2

Now, substitution is possible, and the answer is1 1 1

47-3 2 4 2

x x

x x

x x xx x x x

xx x x

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sin xf xx

sinlimx

xx

Find:

1 sin 1x

so for :0x 1 sin 1xx x x

1 sin 1lim lim limx x x

xx x x

sin0 lim 0x

xx

by the sandwich/squeeze/pinching theorem:

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Well-known limits 1. 2.

0 x

xsinlimx 1

xxsinlim

0x

3. 0

x

1xcoslim0x

4. 1

x1elim

0x

x

elnx

1alim0x

x

5. e)x11(lim

x

x

e))x(f

a1(lim)x(f

f(x)a

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Well-known...

If r>11lim 0.rx x

1lim 0.rx x

Can you tell, what if r<1?

6.

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Squeezing/pinching/sandwichtheorem for functions

Suppose that

g(x)h(x)f(x), in the neirbourhood of x=c, (not necesserily at c though)

and

lim g(x)=limf(x)=L at x=c

Then

lim h(x) exists at c, and

lim(h(x))=L

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Proof2

0 x

We are considering the area of triangle OAB, circle section OAB and triangle OAD

OADOABOAB AAA

tionsec

222sin xtgxx

xtgxx sin0sin:/ x

xxx

cos1

sin1

xxx cossin1 1

xxsinxcos:isthat

1x

xsinlim0x

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The limit of sin(x)/x as x goes to 0 is proof

11limand1xcoslim0x0x

Since

applying squeeze theorem

1x

xsinlim00x

x)xsin(

xxsinBecause

1x

xsinlim0x

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Proof of 0

x1xcoslim

0x

)1x

xsinlimthatusedwe(020lim1

)1x(cos)x(sinlim1

)1x(cosx)x(sinlim

)1x(cosx)x(cos1lim

)1x(cosx)xcos1)(xcos1(lim

xxcos1lim

00x0x

0x0x0x

0x0x

2 2

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Applying well-known limitsExample 1Find

SolutionTo calculate the first limit, we “pair off” sin 4x with 4x and use (2.5.6):

Therefore,

The second limit can be obtained the same way:

0 0

sin 4 1 cos2lim and lim3 5x x

x xx x

0 0 0

sin 4 4 sin 4 4 sin 4 4 4lim lim lim 13 3 4 3 4 3 3x x x

x x xx x x

0 0 0

1 cos2 2 1 cos2 2 1 cos2 2lim lim lim 0 05 5 2 5 2 5x x x

x x xx x x



Limit of a Function

The limit of f (x), as x approaches a, equals L written:

if we can make the value f (x) arbitrarily close to L by taking x to be sufficiently close to a.

lim ( )x a

f x L

a

L( )y f x

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Formal Definition of a LimitIf for any (as close as you want to get to L)there exists a (we can get as close as

necessary to c), such as:

Then the limit exits:

( )f x L when x c

lim ( )x cf x L

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Formal Definition of a Limit The

For any ε (as close asyou want to get to L)

There exists a (we can get as close as necessary to c )

lim ( )x cf x L

L •

c

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Specified Epsilon, Required Delta

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Formal Definition of a LimitIf for any (as close as you want to get to L)there exists a (we can get as close as

necessary to c), such as:

Then the limit exits:

( )f x L when x c

lim ( )x cf x L

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Finding the Required Consider showing

|f(x) – L| = |2x – 7 – 1| = |2x – 8| < We seek a such that when |x – 4| < |2x – 8|< for any we choose It can be seen that the we need is

4lim(2 7) 1x

x

2

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No limitExample Here we set f(x) = sin (π/ x) and show that the function can have no limit as x → 0

The function is not defined at x = 0, as you know, that’s irrelevant. What keeps f from having a limit as x → 0 is indicated in the Figure above

As x → 0, f(x) keeps oscillating between y = 1 and y = –1 and therefore cannot remain close to any one number L.

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Finite limit in the infinityDefinition: The limit of a function in the infinity is A if for arbitrary >0 there exists a positive number K , such that if x>K, then f(x)-A<

Example: f(x)=sinx/x+A

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Infinit limit at a (finite) pointDefinition: Let x0 is a point of the domain of the definition of function f. The limiting value of f at x0 is (positive) infinity, if for all K>0 there exists a >0 such that if x-x0 < then f(x)>K

Example: 1/x at 0

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Limiting value – defitions HAND IN!!

Based on the previous defitinions, define the following:

-Limit in the infinity is infinity/negative infinity-Limit in the negatíve infinity is + infinity/-infinity-Limim in the negative infintiy is a number A

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ContinuityContinuity at a Point

The basic idea is as follows: We are given a function f and a number c. We calculate (if we can) both and f (c). If these two numbers are equal, we say that f is continuous at c. Here is the definition formally stated.

limx cf x f c

limx cf x



Continuity of a FunctionA function f is continuous at the point x = a if the following are true:

) ( ) is definedi f a) lim ( ) existsx a

ii f x

) lim ( ) ( )x a

iii f x f a

a

f(a)



Continuous Functions

A polynomial function y = P(x) is continuous at every point x.

A rational function is continuous at every point x in its domain.

All elementary functions and their inverses are continuous

( )( ) ( )p xR x q x

If f and g are continuous at x = a, then

, , and ( ) 0 are continuous

at

ff g fg g agx a

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f (x) = (x2 – 9)/(x + 3) at x = -3

2

x 3

x 9lim f ( 3)x 3

c.

2

x 3

x 9limx 3

b. - 6

The limit exist!

f (-3) = 0/0a. Is undefined!

Therefore the function is not continuous at x = -3.

-3

-6

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2

1

1f x

x

A function f is said to be continuous on an interval if it is continuous at each interiorpoint of the interval and one-sidedly continuous at whatever endpoints the interval maycontain.For example:

The function

is continuous on [−1, 1] because it is continuous at each point of (−1, 1), continuous from the right at −1, and continuous from the left at 1. The graph of the function is the semicircle.

.

21f x x

Continuity on Intervals

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Classification of points of discontinuity:first and second kind

Infinite (dicont. of second kind): at least one of the one sides limits does not exists, or functional values tend to the infinity/-infinity

Removable: limit from the right and from the left exist, and equal

Jump: Limits exist from both side, but not equal

Discontinuity of first kind: removable, jump

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Important Theorems about continuous functions

Extreme Value Theorem Weirerstrass, r ETV

Intermediate Value Theorem - ITV

Some applications



Intermediate Value TheoremIf f is a continuous function on a closed interval [a, b] and L is any number between f (a) and f (b), then there is at least one number c in [a, b] such that f(c) = L.

( )y f x

a b

f (a)

f (b)

L

c

f (c) =



Intermediate Value Theorem

2Given ( ) 3 2 5,Show that ( ) 0 on 1,2 .

f x x xf x

Ex.

(1) 4 0(2) 3 0ff

f (x) is continuous (polynomial) and since f (1) < 0 and f (2) > 0, by the Intermediate Value Theorem there exists a c on [1, 2] such that f (c) = 0.

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Limitations of IVT

The IVT is a powerful tool, but it has its limitations. To illustrate, suppose that d(t) represents the decibel level of Pork Chop's motorcycle engine, and suppose

d(0) = 100 and d(10) = 35, where t is measured in seconds. d is a continuous function. By IVT in the ten second interval between time t=0 and time t=35 Pork

Chop's decibel level reached every value between 35 and 100. It does NOT say anything about: When or how many times (other than at least once) a particular

decibel was attained. Whether or not decibel levels bigger than 100 or less than 35 were

reached.

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Definition of absolute extrema Suppose that f is a function defined on a

domain D containing c. Then Absolute maximum value at c if

f(c) f(x) for all x D Absolute minimum value at c if

f(c) f(x) for all x D

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Extreme value theoremWEIERSTRASS OR EVT

Can find absolute extrema under certain hypotheses:

If f is continuous on a closed interval [a,b], with - < a < b < , then f has an absolute maximum M and an absolute minimum m on [a,b]

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Example

No maximum or minimum value on the domain. However, on [-3,3], it has both.

Question: does function f fullfil EVT?

20,)( 2 xxxf

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Conclusions about hypotheses

Conclude that hypothesis that interval be closed, [a,b], essential

Conclusion that f is continuous also essential:

32,3

20,2)(

xxxx

xf

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Examples fulfilling hypotheses

f(x) = 2 - 3x where -5 < x < 8

g(x) = sin(x) where 0 < x < 2

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Limitations of Extreme Value Theorem

Polynomial f(x)=x5 - 3x2 + 13 is continuous everywhere

Must have absolute max, min on [-1, 10] by theorem

Theorem doesn’t say where these occur Extreme value theorem just an “existence

theorem” Learn tools for finding extrema later using the

derivative

Index FAQ Functional limits Intuitive notion Definition How to calculate Precize definition Continuous functions.

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