Calculus Limits. Limit of a function.. 1. One-Sided Limits...1. Infinite limits 2. Vertical Asymptotes...3. Calculating Limits Using the Limit Laws.

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1 Limi of a fucio.. Oe-Sided..... Ifiie limis Verical Asympoes... Calculaig Usig he Limi Laws.5 The Squeeze Theorem.6 The Precise Defiiio of a Limi Coiuiy.8 Iermediae Value Theorem..9 Refereces.. 9 0

2 Limi of a fucio Lim f ( ) L if we ca make he values of f() arbirarily close o L by akig o be sufficiely close > a o a bu o equal o a. Eample: Le f() 9 Soluio: Make a able o see he behavior. Discuss he behavior of he values of f() whe is close o 0. 9 ± ± ± ± As approaches 0, he values of he fucio seem o approach lim Oe-Sided Defiiio: We wrie Lim f ( ) L ad say he lef-had limi of f() as approaches a is equal o L if we ca make he values of f() arbirarily close o L by akig o be sufficiely close o a ad less ha a i.e approaches a from he lef.

3 Defiiio: We wrie Lim f ( ) L ad say he righ-had limi of f() as approaches a is equal o L if we ca make he values of f() arbirarily close o L by akig o be sufficiely close o a ad greaer ha a i.e approaches a from he righ Lim f ( ) L if ad oly if Lim f ( ) L ad Lim f ( ) L Ifiie limis Defiiio: Le f be a fucio defied o boh sides of a. The Lim f () meas ha he values of f() ca be made arbirarily large by akig sufficiely close o a, bu o equal o a.

4 Defiiio: Le f be a fucio defied o boh sides of a. The Lim f () meas ha he values of f() ca be made arbirarily large egaive by akig sufficiely close o a, bu o equal o a. Verical Asympoes Defiiio: The lie a is called a verical asympoe of he curve y f() if a leas oe of he followig saemes is rue: Lim f () Lim f () Lim f () Lim f () Lim f () Lim f ()

5 Eample: Fid he verical asympoes of f() a si Soluio: Because a cos There are poeial verical asympoes where cos 0. I fac, sice cos 0 as ( π ) ad cos 0 as ( ) π, whereas si is posiive whe is ear π, we have lim a ad lim a. This shows ha he lie π is a verical asympoe. ( π (π ) ) Similar reasoig shows ha he lies are all verical asympoes of f() a. ( ) π, (odd muliples of π ),where is a ieger, 4

6 Calculaig Usig he Limi Laws Suppose ha c is a cosa ad he limis Lim f () ad Lim g() eis. The. Lim[ f ( ) g( )] lim f ( ) lim g( ). Lim[ f ( ) g( )] lim f ( ) lim g( ). Lim[ cf ( )] c lim f ( ) 4. Lim[ f ( ) g( )] lim f ( ).lim g( ) f ( ) lim f ( ) Lim[ g( ) lim g( ) 5. ] if lim g( ) 0 Lim [ f ( )] [lim f ( )] 6. where is a posiive ieger. 7. Lim c c 8. Lim a Lim a 9., where is a posiive ieger. 0. Lim a, where is a posiive ieger.. Lim f ( ) lim f ( ), where is a posiive ieger. Evaluae lim 5. 5

7 Soluio: lim 5 lim( lim(5 ) lim ) lim lim 5 lim ( ) ( ) 5 ( ) lim Eample. Fid Soluio: Le f() lim. We ca fid he limi by subsiuig because f() is defied. Nor ca we apply he quoie rule because he limi of he deomiaor is 0. Isead we facor he umeraor as a differece of squares: ( )( ). The umeraor ad deomiaor have a commo facor of -. Whe we ake he limi as approaches, we have so 0. Therefore, we ca cacel he commo facor ad compue he limi as follows ( )( ) lim lim lim( ) The Squeeze Theorem f ( ) g( ) h( ) Lim f () Lim h ) L If Lim g( ) L. whe is ear a ad ( he 6

8 Eample. Show ha lim si 0. 0 Soluio: Firs oe ha we cao use lim si 0 lim 0 o eis. However sice si, Muliply all sides by We kow ha si. lim 0 ad lim( ) Takig f ( ) lim si 0 0 The Precise Defiiio of a Limi.lim si because limsi 0 0 0, g( ) si( ), h( ) i he squeeze heorem, we obai does Le f be a fucio o some ope ierval ha coais he umber a, ecep possibly a a iself. The we say ha he limi of f() as approaches a is L, ad we wrie Lim f ( ) L if for every umber ε > 0 here is a umber δ > 0 such ha f ( ) L < ε wheever 0 < a < δ. Eample prove ha lim(4 5) 7. Soluio. Prelimiary aalysis of he problem (guessig a value for δ ). Le ε be a give posiive umber. We wa o fid a umber δ such ha ( 4 5) 7 < ε wheever 0 < < δ 7

9 Bu ( 4 5) 7 4 4( ) 4. Therefore, we wa 4 < ε wheever ε 0 < < δ ha is, < wheever 0 < < δ. This suggess ha we should choose 4 δ ε. 4. Proof (showig ha his δ works). Give ε > 0, choose ε ( 4 5) 7 4 4( ) 4 < 4δ 4( ) ε 4 ( 4 5) 7 < ε wheever 0 < < δ Thus Therefore by defiiio of a limi lim(4 5) 7. Coiuiy A fucio f is coiuous a a umber a if lim f ( ) f ( a) δ ε. If < < δ 4 0, he Noice ha he defiiio implicily requires hree higs if f is coiuous a a:. f(a) is defied (ha is, a is i he domai of f). lim f ( ) eiss.. lim f ( ) f ( a) A fucio f is coiuous o a ierval if i is coiuous a every umber i he ierval. Eample: Show ha he fucio f ( ) is coiuous o he ierval [-,]. Soluio: If - < a <, he usig he Limi Laws, we have Lim f ( ) a a lim( ) 8

10 lim a lim( ) a f(a) Therefore f is coiuous o [-.]. If f ad g are coiuous a a ad c is a cosa, he he followig fucios are also coiuous a a:. f g. f g. cf 4 fg 5. f/g if g(a) 0 Ay polyomial is coiuous everywhere, ha is i coiuous o R (, ) Ay raioal fucio is coiuous wherever i is defied, ha is, i is coiuous o is domai. Iermediae Value Theorem Suppose ha f is coiuous o he closed ierval [a,b] ad le N be ay umber bewee f(a) ad f(b), where The here eiss a umber c i (a,b) such ha f(c) N. f ( a) f ( b). Eample. Usig he Iermediae Value Theorem, le s Show ha here is a roo of he equaio bewee ad Soluio: le f(). We are lookig for a soluio of he give equaio, ha is, a umber c bewee ad such ha f(c) 0. Therefore, we ake a, b, N0. We have f() < 0 f() 4 6 > 0 Thus, f() < 0 < f(); ha is, N 0 is a umber bewee f() ad f(). Now f is coiuous sice i is a polyomial, so he Iermediae Value Theorem says here is a umber c bewee ad such ha f(c) I oher words, he equaio has a leas oe roo c i he ierval (,). Refereces: Calculus, James Sewar 5 h ediio 9