Bab III Limit-Continu

8/3/2019 Bab III Limit-Continu

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3. LIMIT AND CONTINUITY


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3.1 Limit Function at One Point

An intuitive introduction

Let

11)(

2

= xxxf

This function is not defined at x=1 (0/0), it still makes sense to ask

what happens to values of f(x) as x moves along x-axis toward 0.

The following table shows a succession of values of f(x) correspondingto succession of values of x that move toward 0

x

f(x)

0.9 0.99 0.999 1.11.011.0010.9999 1.00011

21.9 1.99 1.999 1.9999 2.0001 2.001 2.01 2.1


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1

2

x x

f(x)

f(x)

geometrically

The result in the table suggest thatthe values of f(x) approach 1

We denote this by writing

21

1lim

2

1 =

x

x

x

we read : the limit of as x approaches 1 is equal to 2

1

12

x

x

Definition (intuitive) . It mean, the value of f(x) approaches thenumber Las x approaches c

Lxfcx

=

)(lim


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Lxfcx

=

)(lim

L

c

L

L

+L

there is such that0>

c

L


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)(lim xfcx +

The Limit from the right and left(one-sided limit)

cx

Ifxapproaches cfrom the left side (fromnumber on which is smaller than c), weget the left-hand limit, denoted by

)(lim xfcx

c x

lim ( ) lim ( ) lim ( )andx c x c x c

f x L f x L f x L +

= = =

The relationship among two-side limit an one-side limit

If )(lim xfcx

+)(lim xf

cx

then does not exist)(lim xfcx

If x approaches c from the right side (fromnumber on which is greater than c), we

get the right-hand limit, denoted by


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+


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)(lim0

xfx

0lim 20 == xx

)(lim0

xfx +

0lim0

==+x

x

0)(lim0

=

xfx

)(lim1

xfx

1lim

1==

x

x

)(lim1 xfx + 32lim2

1 =+= + xx

)(lim1

xfx

)(lim2 xfx 62lim

2

2=+=

xx

does not exist

c. The formula of f(x) does not change at, thus

b. Because the formula of f(x) change at x = 1 then first findingthe one-sided limit at point x = 1.


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d.

For x 02)( xxf =

Graph: parabola

For 0


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2. Find the value of c so that


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Problems

>+

+=

1,2

1,1)(

2

2

xxx

xxxf

)(lim1

xfx

x

f x

+1lim ( )

x

f x

1

lim ( )

xxxg 32)( =

x

g x

2lim ( )

x

g x

+2

lim ( )x

g x2

lim ( )

22)(

=

xxxf

x

f x

2

lim ( )x

f x

+2

lim ( )x

f x2

lim ( )

1. Let

a.Evaluate and

b. Does exist? if it limit exist evaluate

2. Let , evaluate

3. Let , evaluate

a. b. c.

a. b. c.

B.

x

f x

1

lim ( )


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lim ( ) lim ( )and x a x a

f x L g x G

= =

[ ] GLxgxfxgxfaxaxax

==

)(lim)(lim)()(lim

TheoremLet

then

[ ] LGxgxfxgxfaxaxax

==

)(lim)(lim)()(lim

0,)(lim

)(lim

)(

)(lim ==

Gbila

G

L

xg

xf

xg

xf

ax

ax

ax

2.

3.

4.

n

ax

n

ax xfxf ))(lim())((lim =

,n positive integer

nn

ax

n

axLxfxf ==

)(lim)(lim5. provided if n is even

1.

0L


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222 )1(1

1sin)1()1(

x

xxx

)()()( xhxgxf

lim ( ) , lim ( ) x c x c

f x L h x L

= =

Lxgcx

=

)(lim

1

1

sin)1(lim

2

1

xxx

The Pinching theorem

Let for all x in some open interval containing the point c and

then

Example Evaluate

Because 1)1

1sin(1

x

and 0)1(lim 21

=

xx

0)1(lim, 21

=

xx

0

1

1sin)1(lim 2

1=

x

xx

then


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Limits of Trigonometric Functions

1sin

lim.10

= x

x

x

1coslim.2 0=

xx

1tan

lim.30

= x

x

x

Example

2.

2

2tan5

4.4

4sin3

lim2tan5

4sin3lim

00

x

xx

x

xx

xx

xx

+=

+

2.

2

2tanlim5

4.4

4sinlim3

0

0

x

xx

x

x

x

+=

3

7

2.2

2tanlim5

4.4

4sinlim3

02

04=

+=

x

xx

x

x

x

x 0 if only if 4x 0


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Problems

tt

t sin1coslim

2

0 +

t

tt

t sec2

sincotlim

0

t

t

t 2

3tanlim

2

0

tt

tt

t sec

43sinlim

0

+

Evaluate

1.

2.

3.

4.

x

x

x 2sin

tanlim

05.


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Example Evaluate

1

1lim

2

1

+ x

x

x

a.1

1lim

2

2

1

+ x

x

x x

x

x sinlim

+b. c.

Solution

a. 021lim 21

>=+x

x,g(x)=x-1 approaches 0 from the downward

thus

=

+ 1

1

lim

2

1 x

x

x

b. 021lim2

1>=+

x

x

approaches 0 from the upward1)( 2 = xxg

thus

+=

+ 1

1lim

2

2

1 x

x

x


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c.

0lim >=+

xx

and

f(x)=sinx

x

If x approaches from the right then sin(x) approaches 0 fromdownward

=+ xx

x sinlim

thus

Because


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Limits at Infinity

Lxfx

=

)(lima. if >> |)(|00 LxfMxM

or f(x) approaches L if x increase without boundL

x

Example Evaluate

42

52lim

2

2

+

++

x

xx

x

Solution

)2(

)1(lim

2

2

42

522

x

xx

x x

x

+

++=

42

52lim

2

2

+

++ x

xx

x

2

2

42

521

lim

x

xxx

+

++=

= 1/2


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Lxfx

=

)(lim if


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Example Evaluate

xxxx

+++

3lim 2

Solution:

If x , It is form ( )

xxxx

+++

3lim 2 )

3

3(3lim

2

22

xxx

xxxxxx

x

++

+++++=

xxx

xxx

x ++

++=

3

3lim

2

22

xxx

x

x ++

+=

3

3lim

2

xx

x

xx

x

x ++

+

= )1(

)1(

lim2

312

3

||2 xx =

xxx

xx

x

x +++=

2

31

3

1)1(lim

2

1

)11(

1lim

2

31

3

=+++

+=

xx

x

x


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Problems

limx

x

x +

+

3

3

3

limx x + 2

2

3

4

)1(lim xxx

limx

x

x +1 2

1

1lim

2

+ x

x

x

limx

x x

x

+

+

2

1

.

Evaluate

1.

2.

3.

4.

5.

6.


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Continuity

A Function f(x) is said continuity at x= a if

(i) f(a) is defined

lim ( )x a f x exist(ii)

(iii) )()(lim afxfax

=

If one or more of the conditions fails to hold, then f is called discontinuous

at a and ais called apoint of discontinuity off

a

(i)

f(a) does not exist

f is not continuous at


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a

(ii)

1L

2LLeft-hand limit right-hand limit ,Thus limit does not exist at x=a

f is not continuous at x=a

(iii)

a

f(a)f(a) is defined

)(lim xfax

L exist

f is not continuous at x=a

lim ( ) ( )x a

f x f a

but


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(iv)

a

f(a)

f(a) is defined)(lim xf

axexist

)()(lim afxfax

=

f(x) is continuous x=a

Removable discontinuity

For case (i) discontinuity can be removedby define f(a) = limit of function at a

a


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ExampleDetermine whether f is continuous at x=2

2

4)(

2

=

x

xxf

=

=

2,3

2,

2

4)(

2

x

x

x

xxfa. b.


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Solution :f(x) is continuous at x=2, if

f is continuous from the left at x=2

)2()(lim2 fxfx=

aaxxfxx

+=+=

2lim)(lim22

1412)2( 2 == aaf

2 + a = 4a 1-3a = -3

a = 1

f is continuous from the right at x=2

)2()(lim2

fxfx

=+1412)2( 2 == aaf

141lim)(lim 222

==++

aaxxfxx

trivial


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1. Let

>+

=

1,22

1,1)(

2

xx

xxxf

Determine whether f is continuous at x= -1

Problems

2. Find the value of a + 2b so that f(x) will be continuous everywhere


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Continuity on a Closed Interval

A functionf(x) is said to be continuous on closed interval [a,b] if

1.f(x) is continuous on ( a,b )

2.f(x) is continuous from the right atx = a

3.f(x) is continuous from the left atx = b


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Theorem Polynomials are continuous function

A rational function is continuous everywhere except at the points

where the denominator is zero

Let , then

f(x) is continuous everywhere ifn is odd

f(x) is continuous for positive number if n is even

Example where is continuous ?

From theorem, f(x) is continuous for x-4> 0 or x > 4.

f(x) is continuous from the

right at x=4

Thus f(x) is continuous at [4, )

n xxf =)(

4)( = xxf

)4(04lim)(lim44

fxxfxx

===++


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f xx x

x( ) =

+

+

2 3

3

f xx

x( ) =

2

3

4

8

f xx

x( )

| |=

2

2

94

1)(

2

=

x

xxf

24)( xxxf =

A. Find the points of discontinuity, if any

B. where is f(x) continuous

Problems

1.

2.

3.

1.

2.


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Limit and continuity of Composition Function

Theorema

If andf(x) is continuous atL, then

Theorema

If g(x) is continuous at a,f(x) is continuous at g(a), then the composition

is continuous at a

Proof

( f is continuous at g(a) )= f(g(a)) ( g is continuous at a )

= (fog)(a)

Lxgax

=

)(lim

)())(lim())((lim Lfxgfxgfaxax

==

))(( xgf

))((lim))((lim xgfxgfaxax

=

))(lim( xgf ax=


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+

+=

43

13

cos)( 2

4

xx

xx

xf

))(()( xhgxf =

43

13

)( 2

4

+

+

= xx

xx

xh and g(x) = cosx

Example Where is continuous?

Solution:

f(x) can be written as

where

Because h(x) is continuous on R-{-4,1} and g(x) is continuous everywherethen f(x) is continuous on R-{-4,1}

Bab III Limit-Continu

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