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Lesson 1

Indeterminate Forms

OBJECTIVES:

• to define, determine, enumerate the

different indeterminate forms of

functions;

• to apply the theorems on differentiation

in evaluating limits of indeterminate

forms of functions using L’Hopital’s Rule.

.

To evaluatethe said limit Theorems on L' Hopital' s Ruleis will be used.

longer be applied tothe second example.

Obviously,the principle applied inthe previous problems canno

0

0

0

sin( 0 )

2 ( 0 )

sin 2 ( 0 )

2x

sin 2x lim

2x

sin 2x Let usconsider evaluating the lim

x 0

x 0

= = = →

→

.

∞

4. - and

B.Secondary Forms:

and

A.PrimaryForms:

of indeterminateforms:

0 0

Kinds

.

.

y- sin3y

tany-3y

2. lim

y→ 0

(0)-sin3(0)

tan(0)-3(0)

y-sin3y

tany-3y

lim

y

0

( )

( )

( )

1 - 3cos

sec 0 3

1 - (cos3y)

sec y 3 ( 1 )

lim

y-sin3y

dx

d

tany-3y

dx

d

lim

y-sin3y

tany-3y

lim

By LHR

2 2

y 0 y 0 y 0

→ → →

1 y-sin3y

tany-3y lim y 0

∴ = →

.

( )

( )

2

4

x 4 x

ln sin2x

3. lim

π

( )

( )

2 2 x 4

ln sin2 ln sin ln sin2x (^4 ) lim 4x 0 0 4 4

π →

 π^   π         = = = π − (^)  π  π −   

( )

( )

( )

( )

2 ( 4 x)( 4 )

(cos 2 x) 2 sin2x

lim

4 x dx

d

ln sin2x dx

d

lim 4 x

ln sin2x lim

By LHR

4

(^2) x 4

x

2

4

x − −

π π π

( ) ( )

Thisisstillindetermin ate

0

0

80

2

2cot

4

8 4

4

2cot

8 4 x

2cot2x lim

4

x

 

  



−

=

 

  

 − −

= − −

⇒ →

π

π π

π

π π

.

x

2

x e

x

4. lim

→+∞

( )

→ +∞ +∞ e e

x lim x

2

x

2

[ ]

[ ]

• ∞

+∞

+∞ = = = +∞ →+∞ →+∞ →+∞ e

2

e ( 1 )

2x lim

e dx

d

x dx

d

lim e

x lim

By LHR

x x x

2

x x

2

x

[ ]

[ ]

[ ]

[ ]

x 2

2

2 2

2

x x

2

x x

2

x e d x

d

x d x

d

lim

e dx

d

x dx

d

lim e

x lim

Repeat LHR

→+∞ →+∞ → +∞

= =

[ ]

[ ]

0

2

e

2

e ( 1 )

2 ( 1 ) lim

e dx

d

2x dx

d

lim x x x x

= +∞

⇒ = = = = →+∞ →+∞ +∞

e

x

lim

x

2

x

→+∞

. lntan 3x

lncos3x

5. lim

6

x

π →

∞

∞

= π

π

= π

π

= π →

ln

ln

2

lntan

2

lncos

lntan 3

6

lncos 3

lntan3x

lncos3x lim x

0

(^66)

ln ( ) 1 0

Note:

ln ( ∞) =∞

ln ( ) 0 =−∞

[ ]

[ ]

(sec 3 x) 3 tan 3 x

1

sin 3 x 3 cos3x

1

lim

lntan3x dx

d

lncos3x dx

d

lim lntan3x

lncos3x lim

Apply LHR

2 6

x 6

x 6

x

−

= = → → → π π π

2

(^2 )

2 x 2 x x (^6 6 6 )

sin 3x

-3tan3x tan 3x (^) cos 3x lim lim lim 1 sec 3x^1 3 sec 3x tan3x cos 3x

π π π → → →

  ⇒ = (^)  − (^) = −   (^)      

( )

2 2

x 6

lim sin3x sin3 1 →π^6

 π ⇒ = − (^)   = −  

x 6

ln cos 3x lim 1 →π ln tan 3x

Therefore = −