Class Summary. 2.2 The Limit of a Function 2.3 Calculating Limits Using the Limit Laws

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1 Class Summary. The Limi of a Funcion.3 Calculaing Limis Using he Limi Laws Definiion of a Limi Given a funcion f defined for all near a, ecep possibly a a. We say ha he i of f( ), as approaches a, equals L If we can make he values of f( ) arbirarily close o L (as close o L as we like) by aking o be sufficienly close o a, bu no equal o a. In his case, we wrie Eample : Invesigae ( ) Soluion:. f( ) = L. The graph above suggess ha he values of approaches 5 as approaches. We guess ha ( ) = 5. Eample : Invesigae Soluion: Le's ake a look a he values of Since 93 for several values of close o is even, we can ake only posiive values of.

2 As approaches 0, he value of he funcion seem o approach , so we guess ha =. 6 Eample : Invesigae Soluion: 0. Suppose ha Since value of 0 = L, for some number L (in his case, we say 0 is always as approaches 0 from he righ, L =. Bu he eiss.). is always as approaches 0 from he lef, so L =. Here we ge a conradicion. Therefore, 0 does no eis.

3 One-sided is We say ha he i of f( ) as approaches a from he lef (righ) is equal o L and wrie f( ) = L ( f( ) = L ) if we can make he values of f( ) arbirarily close o L by aking o be sufficienly close o a and < a ( a< ). By comparing he definiion of is wih he definiions of one-sided is, we see ha f( ) = L f( ) = L and f( ) = L. Infinie Limis Le f be a funcion defined on boh sides of a, ecep possibly a a. Then we say he i of f( ) is (negaive) infiniy as approaches a, and wrie f ( ) = ( ) if he values of f( ) can be made as large (negaive) as we like by aking close enough o a, bu no equal o a. Eample : Invesigae. 0 Soluion: We see from he able above ha as ges close o 0, wrie =. 0 ges very large. We

4 Le's ake a look a he graph of. Verical Asympoes The line = a is called a verical asympoe of he curve y = f( ) if a leas one of he following saemens is rue :. f( ) =. f( ) = 3. f( ) = 4. f( ) = For eamples :. = and = = and =. verical asympoe : =.

5 3. an ( ) = and an ( ) π π = 4. ln ( ) 0 verical asympoes : =. ( n ) = π, where n. verical asympoe : = 0. Limi Laws Suppose ha c is a consan, f( ) = L and g ( ) = L. Then f( ) ± g ( ) = L± L.. ( ). cf ( ) = cl. f( ) g ( ) = L L. 3. ( ) 4. ( f ) L ( ) n n =, where n is a posiive ineger.

6 5. = g ( ) L 6. f( ) L = g ( ) L, if L 0., if L 0. Noe ha Suppose c is a consan, wih c = c, = a and i laws, we ge for any a a polynomial P, ( ) P ( ) = Pa ( ). If P ( ) and Q ( ) are polynomials, hen P ( ) Pa ( ) =, if Qa ( ) 0. Q ( ) Qa ( ) Quesion: Wha if Qa ( ) = 0? Problems o hink abou :. Suppose ha boh f ( ) a. Suppose ha boh f ( ) and ( f( ) g ( )) and ( f( ) g ( )) eis, does g ( ) eis, does g ( ) a eis? eis? f( ) 3. Suppose ha g ( ) If so, wha is he i? eiss and g ( ) = 0, does f ( ) a eis? 4. Suppose ha c is a consan, f( ) = L and g ( ) =. Wha can we say abou he following is? ( f g) ( ) ( ), cg ( ) a, ( f( ) g ( )) and f( ). g ( ) 5. Suppose ha c is a consan, f( ) = and g ( ) =, a Wha can we say abou he following is? ( f g), ( f( ) g ( )) ( ) ( ) f( ). g ( ), cg( ), ( f( ) g ( )) and

7 If f ( ) = and g ( ) =, hen a a. ( ) f g ( ) =.. Le c be a real number, hen ( cf ) ( ) = 0, if c > 0, if c = 0, if c < ( f( ) g ( )) For eample : is an indeerminae form. (csc co ) 0 cos cos = ( ) = 0 sin sin 0 sin cos = ( ) 0 sin 0 = = 0. () For all 0, we can divide boh numeraor and denominaor by. Theorem If f( ) g ( ) when is near a (ecep possibly a a ) and boh f( ) and g ( ) a eis, hen f( ) g ( ). Quesion : Suppose ha f( ) < g ( ) when is near a (ecep possibly a a ) and boh f( ) and g ( ) a eis, is i rue ha f( ) < g ( )?

8 The Squeeze Theorem If h ( ) f( ) g ( ) when is near a (ecep possibly a a ) and h ( ) = g ( ) = L, hen f ( ) = L. a a Eample : Show ha Soluion : Since sin I is easy o show ha sin = 0. 0 for all 0, we have sin 0 0 = and ( ) 0, for all 0. = 0. By he Squeeze Theorem, we ge sin = 0. 0 Quesion :. Suppose ha f( ) = L, is i rue ha f ( ) = L?. Suppose ha f ( ) eiss, is i rue ha f( ) also eiss? 3. Suppose ha f ( ) = 0, is i rue ha f( ) = 0?