Sections 2.2 & 2.3 Limit of a Function and Limit Laws

Transcription

1 Mah 80 Secions. &. Limi of a Funcion and Limi Laws In secion. we saw how is arise when we wan o find he angen o a curve or he velociy of an objec. Now we urn our aenion o is in general and mehods for compuing hem. We will begin wih an informal definiion of i. Eample Le s invesigae he behavior of he funcion for values of near. Similarly, we define he noion of a i for an arbirary funcion. Definiion If f is a funcion defined on an open inerval abou a, ecep possibly a a iself, we wrie f a ( ) = L f, as approaches a, equals L and say he i of ( ) if we can make he values of f ( ) sufficienly close o a (on eiher side of a ) bu no equal o a. arbirarily close o L ( as close o L as we like) by aking Roughly speaking, his says ha he values of f ( ) number a (from eiher side of a ) bu a. Noes: - In finding he i of f ( ) as a we never consider = a. - f ( ) need no even be defined when = a - The only hing ha maers is how f is defined near a become closer and closer o he number L as approaches he Eample Here we have hree funcions. This eample illusraes ha he i of a funcion does no depend on how he funcion is defined a he poin being approached.

2 Eercise For he funcion g( ) graphed here, find he following is or eplain why hey do no eis. a) g( ) c) g( ) b) g( ) d) g( ).5 Eample Here are some ways ha is can fail o eis. Theorem Limi Laws (.) This heorem shows us how o calculae is of funcions ha are arihmeic combinaions of funcions having known is. If ( ) and ( ) f a a g eis, hen. Sum Rule: ( ) ( ). Produc Rule: ( ) ( ) ( f ± g ) = f ( ) ± g( ) a a a ( f g ) = f ( ) g( ) a a a. Quoien Rule: ( ) ( ) ( ) ( ) f f a =,g( ) 0 a g g a a 4. Consan Muliple Rule: ( ) ( ) ( ) c f = c f, c R a a n 5. Power Rule: f ( ) f ( ) a a 6. Roo Rule: n f ( ) n f ( ) n =, where n is a posiive ineger a a =, where n is a posiive ineger. f > 0 a If n is even, we assume ha ( )

3 Special Limis (.) The Consan Funcion The Ideniy Funcion c = c a = a a Direc Subsiuion Propery If f ( ) is a polynomial or a raional funcion and a is in he domain of f, hen ( ) = ( ) f f a a No all is can be evaluaed by direc subsiuion. Noe: Special Cases hese are cases where he Limi Laws canno be applied. 0,,0,,,0, Eercise Use he Limi Laws and he graphs of f and g o evaluae he following is, if hey eis: a) f ( ) + 5g( ) 4 f b) f ( ) g( ) -5 g - ( ) ( ) f c) g Eercise Evaluae he following is: a) ( 4) 5 + d) (#45/.) sec 0 g) 0 cos5 + 0,000 b) + 5 e) (#/.) ( 5 ) + c) (#4/.) ( sin ) f) ( z 8) 0 z 0

4 Soluions 4

5 Noe: Funcions wih he Direc Subsiuion Propery are called coninuous a c and will be sudied in Secion.5. However, no all is can be evaluaed by direc subsiuion. Einaing Zero Denominaors Algebraically (.) Some of he recommended mehods are : simplifying he raional epression, raionalizing he numeraor or denominaor, subsiuion. Eercise 4 Find he following is: a) + b) + 4+ c) 5y + 8y y y y d) u u u 4 e) + f) + g) + h) ( h) + 9 h 0 h i) j) 4 4 k) l) h 5h + 4 h m) p) h 0 5h + 4 h 6 n) r) 0 + c o) 0 + s) Answers: a) ; b) -/; c) -/; d) 4/; e) /; f) ; g) /; h) 6; i) /6; j) 6; k) -/; l) ; m) 5/4; n) -; o) ; p) /; r) c/; s) / Eercise 5 Evaluae ( + ) ( ) h 0 f h f h for he given value of and funcion f. a) f ( ) =, = b) f ( ) =, = 7 Answers: a) -4; b) 7. Eercise 6 Find g( ) if g( ) + if =. π if = 5

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7 The ne wo heorems give wo addiional properies of is. The Squeeze Theorem ( The Sandwich Theorem) (.) Hypohesis: () g( ) f ( ) h( ) for all in some open inerval conaining a, ecep possibly a a and g = h = L () ( ) ( ) a a Conclusion: f a ( ) = L Theorem Hypohesis: () f ( ) g( ) for all in some open inerval conaining a, ecep possibly a a and () f a ( ) and g( ) boh eis a Conclusion: f ( ) g( ) a a Eercise 7 If ( ) f for 5 5, find ( ) 0 f. Eercise 8 Find he following is: a) 0 sin b) cos( 0π ) 0 c) sin 0 d) 0 + π sin 4 Eercise 9 If f ( ) for, and ( ) auomaically know ( ) 4 f for < and >, a wha poins c do you f? Wha can you say abou he value of he i a hese poins? c Eercise 0 Is here a number a such ha i. + a+ a+ + eis? Of so, find he value of a and he value of he Eercise Evaluae

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