limit kuliah ke1

8/19/2019 Limit Kuliah Ke1

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Salas, Hille, Etgen Calculus: One and Several VariablesCopyright 2007 © ohn !iley " Sons, #nc$ %ll rights reserved$

Section 2$& 'he (i)it *rocess +%n #ntuitive #ntroductiona$ 'he (i)it *rocess

b$ %rea o- a .egion /ounded by a Curvec$ 'he #dea o- a (i)it

d$ Ea)plee$ #llustration o- a (i)it-$ (i)its on Various 1unctionsg$ Ea)pleh$ OneSided (i)itsi$ Ea)ple: OneSided (i)its

3$ %nother Ea)ple4$ 1unctions %pproaching #n-inityl$ Su))ary o- (i)its that 5o 6ot Eist

Section 2$2 5e-inition o- (i)ita$ 5e-inition

b$ #llustration o- 5e-initionc$ Selection o- Epsilond$ (i)its on Open #ntervalse$ (i)it *roperties-$ Euivalent (i)it *ropertiesg$ (e-thand and .ighthand (i)its

h$ Ea)ple

Section 2$8 So)e (i)it 'heore)sa$ 'he 9niueness o- a (i)it

b$ (i)it *roperties: %rith)etic o- (i)itsc$ (i)it o- uotientsd$ (i)its that 5o 6ot Eist -or uotients

Chapter 2: (i)its and Continuity

Section 2$; Continuitya$ Continuity at a *oint

b$ 'ypes o- 5iscontinuityc$ *roperties o- Continuity

d$ Ea)plee$ Co)position 'heore)-$ Onesided Continuityg$ Continuity on #ntervals

Section 2$< 'he *inching 'heore)= 'rigono)etric (i)itsa$ 'he *inching 'heore)

b$ /asic 'rigono)etric (i)itsc$ Continuity o- the 'rigono)etric (i)its

d$ Ea)ple Section 2$> '?o /asic 'heore)sa$ 'he #nter)ediateValue 'heore)

b$ /oundedness= Etre)e Valuesc$ 'he Etre)eValue 'heore)d$ *roperties o- the '?o /asic 'heore)s


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'he (i)it *rocess

THE LIMIT PROCESS (AN INTUITIVE INTRODUCTION)

!e could begin by saying that li)its are i)portant in calculus, but that ?ould

be a )a3or understate)ent$ Without limits, calculus would not exist. Every single notion of calculus is a limit in one sense or another $

1or ea)ple:

!hat is the slope o- a curve@ #t is the li)it o-

slopes o- secant lines$

!hat is the length o- a curve@ #t is the li)it o-the lengths o- polygonal paths inscribed in thecurve$


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'he (i)it *rocess

!hat is the area o- a region bounded by a curve@ #t is the li)it o- the su) o- areaso- approi)ating rectangles$


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'he (i)it *rocess

The Idea of a Limit

!e start ?ith a nu)ber c and a -unction f de-ined at all nu)bers x near c butnot necessarily at c itsel-$ #n any case, ?hether or not f is de-ined at c and, i-so, ho? is totally irrelevant$ 6o? let L be so)e real nu)ber$ !e say that the limit of f + x as x tends to cis L and ?rite

provided that +roughly spea4ingas x approaches c, f(x) approaches L

or +so)e?hat )ore precisely provided that

f + x is close to L for all x A c which are close to c.

( )li) x c

f x L→

=


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'he (i)it *rocess

Example Set f + x B ; x < and ta4e c B 2$ %s approaches 2, ; x approaches D and ; x < approaches D < B &8$ !e conclude that

$&8+li)2

=→

x f x


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'he (i)it *rocess

Example !

Set

%s x approaches D , & x approaches F and approaches 8$ !e concludethat

#- -or that sa)e -unction ?e try to calculate

?e run into a proble)$ 'he -unction is de-ined only -or x G &$#t is there-ore not de-ined -or x near 2, and the idea o- ta4ing the li)it as x

approaches 2 )a4es no sense at all:

does not eist.

( ) & f x x= − and ta4e c B D$

& x−

( )D

li) 8 x

f x→−

=

( )2

li) x

f x→

( ) & f x x= −

( )2

li) x

f x→


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'he (i)it *rocess

Example "

1irst ?e ?or4 the nu)erator: as x approaches 8, x8 approaches 27, 2 x approaches >, and x8 2 x ; approaches 27 > ; B 2


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'he (i)it *rocess

'he curve in 1igure 2$&$; represents the graph o- a -unction f $ 'he nu)ber c ison the xais and the li)it L is on the yais$ %s x approaches c along the

xais, f + x approaches L along the yais$


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'he (i)it *rocess%s ?e have tried to e)phasiJe, in ta4ing the li)it o- a -unction f as x tends to c,it does not )atter ?hether f is de-ined at c and, i- so, ho? it is de-ined there$ 'he

only thing that )atters is the values ta4en on by f at nu)bers x near c$ 'a4e a loo4at the three cases depicted in 1igure 2$&$


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'he (i)it *rocess

Example ;Set

and let c B 8$ 6ote that the -unction f is not de-ined at 8: at 8, both nu)erator anddeno)inator are 0$ /ut that doesnKt )atter$ 1or x A 8, and there-ore for all x near 8,

( )2 F

8

x f x

x

−=

−

( ) ( )2 8 8F8

8 8

x x x x

x x

− +−= = +

− −

'here-ore, i- x is close to 8, then2 F

88

x x

x

−= +

−

is close to 8 8 B >$ !e conclude that

( )

2

8 8

F

li) li) 8 >8 x x x

x x→ →

−= + =

−


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'he (i)it *rocess

Example <

li) B &2$ x8 D

x 2 x L 2

'he -unction f + x B is unde-ined at x B 2$ /ut, as ?e said be-ore, that

doesnKt )atter$ 1or all x 2,

x8 D

x 2

A

x8 D + x 2+ x2 2 x ;

x 2 x 2B B x2 2 x ;$

'here-ore,

x8 D

x 2 x L 2li) B li) + x2 2 x ; B &2$

x L 2


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'he (i)it *rocess

Example > #- f + x B then li) f + x B ;$

8 x ;, x A 0

&0, x A 0, x L 0

#t does not )atter that -+0 B &0$ 1or A 0, and thus -or all near 0,

f + x B 8 x ; and there-ore li) f + x B li) +8 x ; B ;$ x L 0 x L 0


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'he (i)it *rocess

O#e$Sided Limit%

6u)bers x near c -all into t?o natural categories: those that lie to the le-to- c and those that lie to the right o- c$ !e ?rite

M'he le-thand li)it o- f(x) as x tends to c is L$N

to indicate that

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

!e ?rite

M'he righthand li)it o- f(x) as x tends to c is L$N

to indicate thatas x approaches c from the right, f(x) approaches L

( )li) x c

f x L−→

=

( )li) x c

f x L+→

=


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'he (i)it *rocess

Example

'a4e the -unction indicated in 1igure 2$&$7$ %s x approaches< -ro) the le-t, f + x approaches 2= there-ore

%s x approaches < -ro) the right, f + x approaches ;= there-ore

'he -ull li)it, , does not eist: consideration o- x < < ?ould -orce theli)it to be 2, but consideration o- x > < ?ould -orce the li)it to be ;$

or a full limit to exist, !oth one"sided limits have to exist and they have to !e e#ual.

( )<

li) 2 x

f x−

→

=

( )<

li) ; x

f x+→

=

( )<

li) x

f x→


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'he (i)it *rocess

Example & 1or the -unction f indicated in -igure 2$&$D,

#n this case

#t does not )atter that f +2 B 8$

Ea)ining the graph o- f near x B ;, ?e -ind that

Since these onesided li)its are di--erent,

does not eist.

( )( )

( )( )

2 2

li) < and li) < x x

f x f x− +

→ − → −

= =

( )2

li) < x

f x→−

=

( ) ( ); ;

li) 7 ?hereas li) 2 x x

f x f x− +→ →

= =

( );

li) x

f x→


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'he (i)it *rocess

Example ' Set $ 6ote that f + x B & -or x ＞ 0, and f + x B & -or x ＜ 0:

f + x B&, i- x ＞ 0

&, i- x ＜ 0$

(etKs try to apply the li)it process at di--erent nu)bers c$

#- c ＜ 0, then -or all x su--iciently close to c,

x ＜ 0 and f + x B &$ #t -ollo?s that -or c ＜ 0li) f + x B li) +& B &

x L c x L c

#- c ＞ 0, then -or all x su--iciently close to c, x ＞ 0 and f + x B &$ #t -ollo?s that

-or c ＜

0li) -+ B li) +& B &

x L c x L c

Ho?ever, the -unction has no li)it as x tends to 0:

li) f + x B & but li) f + x B &$ x L 0 x L 0

x x x f I+ =


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'he (i)it *rocess

Example !e re-er to -unction indicated in 1igure 2$&$&0 and ea)ine the behavior o- f + x -or close to 8 and close to to 7$

%s x approaches 8 -ro) the le-t or -ro) the right, f + x beco)es arbitrarily large and cannot stay close to anynu)ber L$ 'here-ore

li) f + x does not eist$

%s x approaches 7 -ro) the le-t, f + x beco)es arbitrarily large and cannotstay close to any nu)ber L$ 'here-ore

li) f + x does not eist$

'he sa)e conclusion can be reached by noting as x approaches 7 -ro) theright, f + x beco)es arbitrarily large$

x L 8

x L 7


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'he (i)it *rocess

+

Rema* 'o indicate that f + x beco)es arbitrarily large, ?e can ?rite f + xL$ 'o indicate that f + x beco)es arbitrarily large negative, ?e

can ?rite f + xL$

Consider 1igure 2$&$&0, and note that -or the -unction depictedthere the -ollo?ing state)ents hold:

as x L 8P , f + x L + and as x L 8 , f + xL.

Conseuently,as x L 8 , f + xL.

%lso,

as x L 7P , f + xL and as x L 7 , f + xL.

!e can there-ore ?riteas x L 7 , Q f + xQ L .

+


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'he (i)it *rocess

Example +

!e set f + x B

and ea)ine the behavior o- f + x +a as x tends to ;and then +b as x tends to 2$

& x 2


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'he (i)it *rocess

Example

Set f + x B& x2, x ＜ &

&I+ x &, x ＞ &$


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'he (i)it *rocess

Example !

Here ?e set f + x B sin +πI x and sho? that the -unction can have no li)it as x L 0

'he -unction is not de-ined at x B 0, as you 4no?, thatKs irrelevant$ !hat4eeps f -ro) having a li)it as x L 0 is indicated in 1igure 2$&$&8$ %s x L 0, f + x4eeps oscillating bet?een y B & and y B & and there-ore cannot re)ain close toany one nu)ber L$


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'he (i)it *rocess

Example "

(et f + x B +sin xI x$ #- ?e try to evaluate f at 0, ?e get the )eaningless ratio 0I0=- is not de-ined at B 0$ Ho?ever, f is de-ined -or all x A 0, and so ?e canconsider

li) $sin x

x x L 0

!e select nu)bers that approach 0 closely -ro) the le-t and nu)bers that

approach 0 closely -ro) the right$ 9sing a calculator, ?e evaluate f at thesenu)bers$ 'he results are tabulated in 'able 2$&$&$


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'he (i)it *rocess

'hese calculations suggest that

and there-ore that

'he graph o- f , sho?n in 1igure 2$&$&;,

supports this conclusion$ % proo- that thisli)it is indeed & is given in Section 2$


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'he (i)it *rocessS,mma- of Limit% That .ail to Exi%t

Ea)ples 7&8 illustrate various ?ays in ?hich the li)it o- a -unction f at a nu)berc )ay -ail to eist$ !e su))ariJe the typical cases here:

+i +Ea)ples 7, D$

+'he le-thand and righthand li)its o- f at c each eist, but they are not eual$

+ii f + x L as x L c , or f + x L as x L c$ , or both +Ea)ples F, &0, &&$ +'he-unction f is unbounded as x approaches c -ro) the le-t, or -ro) the right, or both$

+iii f + x Roscillates as x L c , c or c +Ea)ples &2, &8$

( ) ( )& 2 & 2li) , li) and x c x c

f x L f x L L L− +

→ →

= = ≠

-i i i - i i


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5e-inition o- (i)it

5 -i i i - (i i


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5e-inition o- (i)it

1igures 2$2$& and 2$2$2 illustrate this de-inition$

5 -i i i - (i i


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5e-inition o- (i)it

#n 1igure 2$2$8, ?e give t?o choices o- % and -or each ?e display a suitable &$ 1or a & to be suitable, all points ?ithin & o- c +?ith the possible eception o- c itsel-

)ust be ta4en by the -unction f to ?ithin T o- L$ #n part +b o- the -igure, ?e began?ith a s)aller T and had to use a s)aller &$

'he & o- 1igure 2$2$; is too large -or the givenT$ #n particular, the points )ar4ed x& and x2 inthe -igure are not ta4en by f to ?ithin T o- L$

5 -i i i - (i i


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5e-inition o- (i)it

'he li)it process can be describedentirely in ter)s o- open intervals assho?n in 1igure 2$2$


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5e-inition o- (i)it

Example

Sho? that

li) +2 x & B 8$ +1igure 2$2$> x L 2

5 -i iti - (i it


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5e-inition o- (i)it

x L &

Example !

Sho? that li) +2 8 x B


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5e-inition o- (i)itExample "

1or each nu)ber c

Example /

1or each real nu)ber c

Example 0

1or each constant '

5 -i iti - (i it


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5e-inition o- (i)it

x L 8

Example 1

Sho? that li) x2 B F$ +1igure 2$2$&&

5 -i iti - (i it


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5e-inition o- (i)it

Example &

Sho? that +1igure 2$2$&2 $2li)

;=

→ x

x

5 -i iti - (i it


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5e-inition o- (i)it

'here are several di--erent ?ays o- -or)ulating the sa)e li)it state)ent$

So)eti)es one -or)ulation is )ore convenient, so)eti)es another, #n particular, it is use-ul to recogniJe that the -ollo?ing -our state)ents areeuivalent:

5 -i iti - (i it


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5e-inition o- (i)it

x L 8

Example '

1or f + x B x2, ?e have li) x2 B F li) +8 h2 B F

li) + x2 F B 0 li) x2 F B 0$ x L 8 x L 8

h L 8

5e-inition o- (i)it


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5e-inition o- (i)it

Onesided li)its give us a si)ple ?ay o- deter)ining ?hetheror not a +t?osided li)it eists:

5e-inition o- (i)it


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5e-inition o- (i)it

Example

1or the -unction de-ined by setting

does not eist$

Poof 'he le-t and righthand li)its at 0 are as -ollo?s:

Since these onesided li)its are di--erent,

does not eist$

( ) 22 &, 0

, 0

x x f x

x x x

+ ≤=

− >

( )0

li) x

f x→

( ) ( ) ( ) ( )20 0 0 0

li) li) 2 & &, li) li) 0 x x x x

f x x f x x x− − + +→ → → →

= + = = − =

( )0

li) x

f x→

5e-inition o- (i)it


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5e-inition o- (i)it

Poof 'he le-t and righthand li)its at & are as -ollo?s:

Example +

1or the -unction de-ined by setting

g + x B

li) g + x B 2$

& x2, x ＜ &

8, x B &

; 2 x, x ＞&,

x L &

li) g + x B li) +& x2 B 2, li) g + x B li) +; 2 x B 2$

'hus, li) g + x B 2$ 6O'E: #t does not )atter that g +& A 2$ x L &

x L &2 x L & 3 x L & 3 x L &2

5e-inition o- (i)it


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5e-inition o- (i)it

(i)it 'heore)s


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(i)it 'heore)s

(i)it 'heore)s


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(i)it 'heore)s

'he -ollo?ing properties are etensions o- 'heore) 2$8$2$

(i)it 'heore)s


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(i)it 'heore)s

Example%

li) +


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(i)it 'heore)s

Example%

li) B &>, li) B , li) B B $

&

x2

&

x8 &

&

7

&

x

&

8

&

8 x L ; x L 2 x L 8

(i)it 'heore)s


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(i)it 'heore)s

Example%

22

8 2

28

8 < > < &li)

& ; & <

8 27 27li) 0& & F

x

x

x

x

x x x

→

→

− −= =

+ +

− −= =

− −

(i)it 'heore)s


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(i)it 'heore)s

Example%

1ro) 'heore) 2$8$&0 you can see that2

2& 2 0

8 7 <li) li) li)

& ; x x x x x

x x x→ → →

−

− −

%ll -ail to eist$

(i)it 'heore)s


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(i)it 'heore)s

Example

Evaluate the li)its eist:

+a li) , +b li) , +cli) $

x2 x >

x 8

+ x2 8 x ;2

x ;

x &

+2 x2 7 x


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(i)it 'heore)s

Example !

usti-y the -ollo?ing assertions$

+a li) B , +b li) B >$&I x &I2

x 2 x FU x 8 x L 2 x L F

& ;

Continuity


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Continuity

Co#ti#,it- at a Poi#t

'he basic idea is as -ollo?s: !e are given a -unction f and a nu)ber c$ !ecalculate +i- ?e can both and f +c$ #- these t?o nu)bers are eual, ?esay that f is continuous at c$ Here is the de-inition -or)ally stated$

#- the do)ain o- f contains an interval +c p, c p, then f can -ail to becontinuous at c -or only one o- t?o reasons: either

+i f has a limit as x tends to c, !ut , or

+ii f has no limit as x tends to c$

#n case +i the nu)ber c is called a remova!le discontinuity$ 'he discontinuity can be re)oved by rede-ining f at c$ #- the li)it is L, rede-ine f at c to be L$

#n case +ii the nu)ber c is called an essential discontinuity$ ou can change thevalue o- f at a billion points in any ?ay you li4e$ 'he discontinuity ?ill re)ain$

( ) ( )li) x c

f x f c→

≠

( )li) x c

f x→

Continuity


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Continuity'he -unctions sho?n have essential discontinuities at c$

'he discontinuity in 1igure 2$;$2 is, -or obviousreasons, called a jump discontinuity$

'he -unctions o- 1igure 2$;$8 have infinite discontinuities$

Continuity


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Continuity

Continuity


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Continuity

Example 'he -unction

is continuous at all real nu)bers other than 2 and 8$ ou can see this by notingthat

B 8 f gh ' ?here

f + x B Q xQ , g + x B x8 x, h+ x B x2 , ' + x B ;.

Since f, g, h, ' are every?here continuous, is continuous ecept at 2 and 8, thenu)bers at ?hich h ta4es on the value 0$ +%t those nu)bers is not de-ined$

( ) 828 ;< > x x x x

x x−= + +

− +

Continuity


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Continuity

Continuity


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Continuity

Example !

'he -unction + x B is continuous at all nu)bers greater than 8$ 'o seethis,

note that B f g , ?here

and g + x B $

6o?, ta4e any c ＞ 8$ Since g is a rational -unction and g is de-ined at c, g iscontinuous at c$ %lso, since g +c is positive and - is continuous at each positivenu)ber, f is continuous at g +c$ /y 'heore) 2$;$;, is continuous at c$

x2 &

x 8

x2 &

x 8 x x f =+

Continuity


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Continuity

Example "

'he -unction is continuous every?here ecept at B W8,

?here it is not de-ined$ 'o see this, note that B f g ' h, ?here

and observe that each o- these -unctions is being evaluated only ?here it is

continuous$ #n particular, g and h are continuous every?here, f is being evaluatedonly at nonJero nu)bers, and ' is being evaluated only at positive nu)bers$ $

&><

&+

2+−

=

x x

$&>+ ,+ ,


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Continuity

Continuity


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Continuity

Example /

5eter)ine the discontinuities, i- any, o- the -ollo?ing -unction:

f + x B

2 x &, x 0≦

&, 0 ＜ x &≦

x2 &, x ＞ &$

+1igure2$;$D

Continuity


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Continuity

Example 0

5eter)ine the discontinuities, i- any, o- the -ollo?ing -unction:

f + x B

x8, x &≦

x2 2, & ＜ x ＜ &> x, &≦ x ＜ ;

, ; ＜ x ＜ 7

7 x

Continuity


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Continuity

( )2

&

& f x

x=

−

Co#ti#,it- o# I#te4al%

% -unction f is said to be continuous on an interval i- it is continuous at each interior point o- the interval and onesidedly continuous at ?hatever endpoints the interval )aycontain$1or ea)ple:(i) 'he -unction

is continuous on M& , &N because it is continuous at each point o- +& , &,continuous -ro) the right at &, and continuous -ro) the le-t at &$'he graph o- the -unction is the se)icircle$

(ii) 'he -unction

is continuous on +& , & because it is continuous at each point o- +& , &$ #t is not continuous on M& , & because it is not continuous -ro) the right at &$ #t is not continuous on +& , &N because it is not continuous -ro) the le-t at &$

(iii) 'he -unction graphed in 1igure 2$;$D is continuous on + , &N and continuous on +& ,$ #t is not continuous on M& , because it is not continuous -ro) the right at &$

(i4) *olyno)ials, being every?here continuous, are continuous on + ,$

Continuous -unctions have special properties not shared by other -unctions$

( ) 2& f x x= −

'rigono)etric (i)its


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g

1ro) this it -ollo?s readily that



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g



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g

#n )ore general ter)s,

Example 1ind

Sol,tio#

'o calculate the -irst li)it, ?e Rpair o-- sin ; x ?ith ; x and use +2$:

'here-ore,

'he second li)it can be obtained the sa)e ?ay:

0 0

sin ; & cos2li) and li)8


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g

Example !1ind li) x cot 8 x$

x L 0



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g

Example "

1ind $

+

sin+li)

2;I

;

&

;

&

π

π

π

−

−

→ x

x

x



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g

Example /

1inding li) $ x2

sec x & x L 0

'?o /asic 'heore)s


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% -unction ?hich is continuous on an interval does not Rs4ip any values, and thusits graph is an Runbro4en curve$ 'here are no Rholes in it and no R3u)ps$ 'hisidea is epressed coherently by the intermediate"value theorem$

'?o /asic 'heore)s


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Example

!e set f + x B x2 2$ Since f +& B & ＜ 0 and f +2 B 2 ＞ 0, there eists a nu)ber c

bet?een & and 2 such that f +c B 0$ Since - increases on M&, 2N, there is only one suchnu)ber$ 'his is the nu)ber ?e call $

So -ar ?e have sho?n only that lies bet?een & and 2$ !e can locate )ore precisely by evaluating f at &$


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Example !

'he -unction f + x B 2I x ta4es on the value 2 at x B & and it ta4es on the value 2 at x B&$ Certainly 0 lies bet?een 2 and 2$ 5oes it -ollo? that - ta4es on the value 0so)e?here bet?een & and &@ 6o: the -unction is not continuous on M&, &N, and

there-ore it can does s4ip the nu)ber 0$

'?o /asic 'heore)s


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5o,#ded#e%%6 Exteme Val,e%

% -unction f is said to be !ounded or un!ounded on a set in the sense in ?hich

the set o- values ta4en on by f on the set is bounded or unbounded$ 1or ea)ple, the sine and cosine -unctions are bounded on + ,:

& G sin x G & and & G cos x G & -or all x ∈ + ,.

/oth -unctions )ap + , onto M& , &N$

'he situation is )ar4edly di--erent in the case o- thetangent$ +See 1igure 2$>$;$ 'he tangent -unction is bounded on M0 , X ;N= on M0 , X 2 it is bounded belo? but not bounded above= on +X 2 , 0N it is bounded above but not bounded belo?= on +X 2 , X 2 it is unbounded both belo?

and above$

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Example "

(et

#t is clear that g is unbounded on M0, $ +#t is unbounded above$ Ho?ever, it is

bounded on M&, $ 'he -unction )aps M0, onto M0, , and it )aps M&, onto +0,&N$

+1igure 2$>$


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1or a -unction continuous on a bounded closed interval, the eistenceo- both a )ai)u) value and a )ini)u) value is guaranteed$ 'he

-ollo?ing theore) is -unda)ental$

'?o /asic 'heore)s


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1ro) the inter)ediatevalue theore) ?e 4no? that

*continuous functions map intervals onto intervals.+

6o? that ?e have the etre)evalue theore), ?e 4no? that

*continuous functions map !ounded closed intervals Ma, !N onto!ounded closed intervals Mm, N$+

O- course, i- f is constant, then B m and the interval Mm, N collapses to a point$

limit kuliah ke1

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