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Lesson 2

Other Indeterminate

Forms

• to define and enumerate other

indeterminate forms (secondary forms) of

functions;

• to reduce secondary forms to primary

forms;

• to apply the theorems on differentiation in

evaluating limits of indeterminate forms of

functions using L’Hopital’s Rule.

OBJECTIVES:

The INDETERMINATE FORMS 0 • ∞ and ∞ - ∞

( ) ( ) ( )

[ ( ) ( )]

[ ( ) ( )] ( ) ( )

Then apply L' Hopital' s Rule.

or.

0

0 equivalent quotient whoselimit whenevaluated may resultto

Thelimit could beevaluated bytrnasforming thedifferenceintoan

That is lim f x g x lim f x - lim g x -.

thenlim f x g x is said tobeindeterminateof the form -.

B. If lim f x , and lim g x whichareboth positive , thenthe

x a x a x a

x a

x a x a

∞

∞

⇒ − = =∞ ∞

− ∞ ∞

=∞ =∞

→ → →

→

→ →

.

∞ The INDETERMINATE FORMS 0 ,∞ ,and 1

0 0

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) (^) [ ( )]

( )

for the function,thenapplythe propertiesof logarithmand then LHR.

Theseindeterminate formsmaybeevaluated byletting avariable y

assumed the indeterminate forms 0 , , 1 ,respectively.

or as xapproaches or - thentheexpressionlim f x

lim f x 1 , and lim g x

lim f x , and lim g x 0 , or

lim f x 0 , and lim g x 0 , or

Giventwo functions f x and g x ,and if :

Defininition :

0 0

g x

x a

x a x a

x a x a

x a x a

∞

→

→ →

→ →

→ →

∞

+∞ ∞

• = = ∞
• = ∞ =
• = =

2. lim.^ [ x lnx] x→ 0

[ ] (^) = ( ) (^) = ( (^) −∞) →

lim x lnx 0 ln 0 0 x 0

[ ] ∞

−∞ = = = → →

0

1

ln 0

x

1

lnx lim x lnx lim

Transformingthegiven function toanequivalent function

x 0 x 0

lim [ x lnx] 0 x 0

→

[ ]

lim( x) 0

x

x

lim

x

dx

d

lnx dx

d

lim

x

lnx lim

ApplyLHR

x 0

2

x 0 x 0 x 0

→ → → →

. (^)  

  



−

− → (^) x 1

1

lnx

1

3. lim x 1

= − =∞− ∞ −

= −  

  



−

− → (^0)

1

0

1

1 1

1

ln 1

1

x 1

1

lnx

1 lim x 1

0

0

(1-1)ln 1

( 1 1 ) ln 1

(x-1)lnx

(x 1 ) lnx lim x 1

1

lnx

1 lim

Trnsformingtoasimple fraction

x 1 x 1

=

− −

− −

 

  



−

− → →

[ ]

[ ] ( ) ( 1 ) (ln x)( 1 )

x

1 x 1

( 1 ) x

1 1

lim

(x-1)lnx dx

d

(x 1 ) lnx dx

d

lim x 1

1

lnx

1 lim

Apply LHR

x 1 x 1 x 1 ^ + 

  

 −

 

  

 −

=

− −

 

  



−

− → → →

0

0

1 1 1

1 1

1

= − +

−

⇒ → ( )ln(1)

x

x- 1 xlnx

x

x- 1

lim x

.

= − = − = ∞ ∞  



 

 − →

x x sec x (sec )

1 lim x 2 0

1

0

1

0 0

1

0

1

2

1

0 2

0

0

0

1 cos 2 ( 0 )

x

1 - cos2x lim x

cos 2 x

x

1 lim x sec 2 x

1

x

1 lim

Trnsforming totheequivalent function

x 0 2 2 x 0 2 2 x 0 2

=

−

 



 



 



 

 = −  



 

 − → → →

 



 

 − x→ (^) x sec 2 x

1

x

1

4. lim 2 2 0

.

[ ]

[ ]

( )( )

sin 2 ( 0 )

2 x

sin 2 x 2 lim

x dx

d

1 - cos2x dx

d

lim x

1 - cos2x lim

Apply LHR

x 0 2 x 0 2 x 0

→ → →

( )

( )

( )

( ) lim 2 cos 2 x 2 cos 0 2 1

cos 2 x ( 2 )( 1 ) lim

x dx

d

sin 2 x dx

d

lim x

sin 2 x lim

ApplyLHRagain

x 0 x 0 x 0 x 0

→ → → →

x sec 2 x

x

lim

2 2 x 0

→

. x^1

1

x 1

6. lim(x)

−

→ +

− −^ ∞ →

= = =

lim (x) ( 1 ) ( 1 )^0 ( 1 )

1 1 1

1 x 1

1

x 1

x 1

1

Let y =(x) −

x 1

lnx lnx x 1

1 ln y ln(x)x^1

1

−

= −

= − =

0

0

1 1

ln 1

x 1

lnx limlny lim

Applyingthelimitonbothsidesasx 1

x 1 x 1

= −

= −

=

→

→+^ → +

1

1

x

1 lim 1

1 x

1

lim

x 1 dx

d

lnx dx

d

lim x 1

lnx lim

ApplyLHRontherightmember

x 1 x 1 x 1 x 1

= = =

−

= − →+^ →+ →+ →+

.

( )

lim ( )x e 2. 71

lim y e but y x

limlny 1 ,taketheinverse functionof bothsides x 1

lnx lim

Thus

x 1

1

x 1

x 1

1 1

x 1

x 1 x 1

− →

− →

→ →

2

2

x 0 x 0 x 0

x

( csc x)( 1 )

cotx

lim

x

dx

d

(lncotx)

dx

d

lim

x

lncotx

lim

ApplyLHRonrightmember

→ +^ → + → +

2 sinxcos x

2 x lim

x

1

sinxcosx

1

lim

x

1

sin x

1

cosx

sinx

lim

2

x 0 2

x 0 2

2

x (^0) ⋅

⋅

−

=

−

 

  



= → +^ → + → +

0

0

sin( 0 )

2 ( 0 )

sin 2 x

2 x lim

2

x 0

= = = → +

cos 2 x

2 x

lim

(cos 2 x)( 2 )

4 x

lim

(sin 2 x)

dx

d

( 2 x )

dx

d

lim

sin 2 x

2 x

lim

ApplyLHRagain

x 0 x 0

2

x 0

2

x → 0 +^ → + → + → +

= = = 0 1

0

cos( 0 )

2 ( 0 ) = = =

Since y ( cot x) then lim(cot x) 1

lim y e 1

lim lny 0 , taketehinverse function of bothsides

x

lncotx

lim

Hence

x

x 0

x

0

x 0

x 0 x 0

→

→

→ →