INDETERMINATE FORMS

OBJECTIVES:• define, determine, enumerate the

different indeterminate forms of functions;

• apply the theorems on differentiation in evaluating limits of indeterminate forms of functions using L’Hopital’s Rule.

.

( ) ( ) ( ) 2313-x lim1x

3x1-x lim

1x3x4x

lim

:follows as numerator the factor we exist, to limit

the for and form, ateindetermin an is limit the 00

113)1(4)1(

1x3x4x

lim

1x

3x4x lim of itlim the Evaluate :callRe

1x1x

2

1x

22

1x

2

1x

−=−==−

−=−

+−

=−

+−=−

+−−

+−

→→→

→

→

21x

3x4x lim ,thus

2

1x−=

−+−

→

used. be will Rule sHopital'L' on Theorems limit

said the evaluate To example. second the to applied be longer

no can problems previous the in applied principle the Obviously,

00

0

)0sin()0(2

)0(2sin2x

2x sin lim

2x

2x sin lim the evaluating consider us Let

0x

0x

===→

→

.

∞∞∞∞

∞⋅

∞∞

1 , ,0 5.

and - 4.

0 3.

:Forms Secondary B.

2.

and 00

1.

:Forms Primary A.

:forms ateindeterminof Kinds

00

.

Theorem 3.6.1 (p. 220) L'Hôpital's Rule for Form 0/0

.

Applying L'Hôpital's Rule (p. 220)

Theorem 3.6.2 (p. 222) L'Hopital's Rule for Form ∞/∞

.

2x2x sin

lim .10x→

EXAMPLE:Evaluate the following limits.

( )( )

( )00

0

0sin02

02sin2x

2x sin lim

0x===

→

( )

( )( )

( )

( )10 cos

202cos2

122x2cos

lim2x

dxd

2x sindxd

lim2x

2x sin lim

:Rule s'Hopital'L gsinu By

0x0x0x

===

==→→→

12x

2x sin lim

0x=∴

→

.3y sin-y3y-y tan

lim .20y→

( ) ( )( ) ( ) 0

0

0000

0sin3-003-0tan

3y sin-y3y-y tan

lim0y

=−−==

→

( )

( )( )

( )

( )1

22

3131

0 3cos-130 sec

33y cos-113ysec

lim3y sin-y

dxd

3y-y tandxd

lim3y sin-y3y-y tan

lim

:LHR By

2

2

0y0y0y

=−−=

−−=−=

−==→→→

13y sin-y

3y-y tan lim

0y=∴

→

.

( )( ) 2

4x x4

2x sin ln lim .3

−ππ→

( )( )

( )

( )

( )( ) ( )4x42

2x2cos2x sin

1

limx4

dxd

2x sin lndxd

limx42x sin ln

lim

:LHR By

4x2

4x

2

4x −−π

=−π

=−π π

→π

→π

→

( ) ( )

.ateminerdetin still is This

00

082

2cot

448

42 cot 2

x482x 2cot

lim4

x

−

π

=

π−π−

π

=−π−π

→

( )( ) 0

00

2sinln

44

42 sinln

x42x sin ln

lim 22

4x

=

π

=

π−π

π

=−ππ

→

( ) 01 ln

:Note

=( ) ∞=∞ ln( ) −∞=0 ln

.

[ ]

( )[ ]( )

32x2csc4

lim)4(8

2x2csc2 lim

x48dxd

2x cot 2dxd

lim

:LHR peatRe

2

4x

2

4x

4x

−=−−

−=−π−

π→π→π→

( )81

181

42 csc

81

x2csc81

lim 2

2

2

4x

−=−=

π−=−⇒

π→

( )( ) 8

1

x4

sin2x ln lim

2

4x

−=−

∴→ ππ

.

x

2

x ex

lim .4+∞→

( )∞+

∞=∞+=⇒∞++∞→ ee

x lim

2

x

2

x

[ ]

[ ] ( )( )

∞+∞+=∞+===

∞++∞→+∞→+∞→ e2

1e2x

lime

dxd

xdxd

limex

lim

:LHR By

xxx

2

xx

2

x

[ ]

[ ]( )( ) 0

2e2

1e12

lime

dxd

2xdxd

lim

:LHR peatRe

xxx

x=

∞+====⇒

∞++∞→+∞→

0e

x lim

x

2

x=∴

+∞→

.3x tan ln3x cos ln

lim .56

xπ

→

( )( ) ∞

∞=∞

=π

π

=

π

π

=⇒π→

- ln

0 ln

2 tan ln

2 cos ln

63 tan ln

63 cos ln

3x tan ln3x cos ln

lim6

x

( ) 01 ln

:Note

=( ) ∞=∞ ln( ) −∞=0 ln

[ ]

[ ]

( )

( )3x3secx3tan

1

3x3sincos3x

1

lim3x tan ln

dxd

3x cos lndxd

lim3x tan ln3x cos ln

lim

:LHR Apply

26

x6

x6

x

−==

π→

π→

π→

x3cos1

3xcosx3sin

limx3secx3tan

lim3x3sec

x3tan1

3x tan3 lim

2

2

2

6x

2

2

6x26

xπ→π→π→

−=

−=

•

−

( ) 16

3sinx3sin lim2

2

6x

−=

π−=⇒

π→

13x tan ln

3x cos ln lim

6x

−=∴→π

.

( ) ( ) ( )( )

( ) ( )

( ) ( ) ( )

( )

applies. Rule sHopital'L' case theof

either In . or 00

to result may which evaluated is limit the then

xg1xf

lim xgxf lim

, Hence one. equivalent an to dtransforme

is products their limit, such evaluate To limit. its approaches

x as 0 or 0 form the having undefined is xg and xf

of product the unsigned), or signed be could (which 0xg lim and

0xf lim that such functions abledifferenti two are xg and xfIf .A

:onDefininiti

axax

ax

ax

∞∞

=•

•∞∞•

=

=

→→

→

→

∞∞∞• - and 0 FORMS ATEINDETERMIN The

.

( ) ( ) ( )( ) ( )[ ]

( ) ( )[ ] ( ) ( )

Rule. sHopital'L' apply Then . or 00

to result may

evaluated when limit whose quotient equivalent an into

difference the ngtransformi by evaluated be could limit The

. xg lim xf lim xgxf lim is That .-

form theof ateindetermin be to said is xgxf lim the

then , positive both are which xg lim and , xf limIf .B

axaxax

ax

axax

∞∞

∞−∞=−=−⇒∞∞

−

∞=∞=

→→→

→

→→

∞∞∞• - and 0 FORMS ATEINDETERMIN The

.

( ) ( )( ) ( )( ) ( )( ) ( )

( ) ( )[ ] ( )

LHR. then and logarithm

of properties the apply then function, the for y variable

a letting by evaluated be may forms ateindetermin These

ly.respective ,1 , ,0 forms ateminerdetin the assumed

xf lim expression the then or approaches x as or

xg lim and ,1xf lim

or ,0xg lim and ,xf lim

or ,0xg lim and ,0xf lim

:if and ,xg and xf functions two Given

:Definition

00

xg

ax

axax

axax

axax

∞

→

→→

→→

→→

∞

∞−∞+

∞==•

=∞=•

==•

∞∞ 1 and , ,0 FORMS ATEINDETERMIN The 00

.

EXAMPLE:Evaluate the following limits:

[ ]2x csc x lim .10x→

[ ] [ ] ∞•==→

00csc02x csc x lim0x

[ ]00

0sin0

sin2xx

lim2x csc x lim

:function rational equivalent an to function the gminTransfor

0x0x===

→→

[ ]

[ ] ( ) ( )

[ ]21

2x csc xlim 21

)0(cos21

2cos2x1

lim2cos2x

1 lim

sin2xdxd

xdxd

limsin2x

x lim

:LHR Apply

0x

0x0x0x0x

=∴⇒==

===

→

→→→→

.

[ ]x ln x lim .20x→

[ ] ( ) ( )∞−==→

00 ln 0x ln x lim0x

[ ]∞∞−===

→→

010 ln

x1x ln

limx ln x lim

:function equivalent an to function given the gminTransfor

0x0x

[ ] 0x ln x lim 0x

=∴→

[ ] ( )( ) 0xlim

x1

1x1

lim

x1

dxd

x lndxd

lim

x1x ln

lim

:LHR Apply

0x

2

0x0x0x=−=−=

=→→→→

.

−−

→ 1x1

x ln1

lim .31x

∞−∞=−=−

−=

−−

→ 01

01

111

1 ln1

1x1

x ln1

lim1x

( )( )

( )( ) 0

01 ln 1-11 ln11

x ln 1-xx ln1x

lim1x

1x ln

1 lim

:fraction simple a ot gminTransfor

1x1x=−−=−−=

−−

→→

( )[ ]

( )[ ]

( )

( ) ( ) ( ) ( )1x ln1x1

1x

1x1

1lim

x ln 1-xdxd

x ln1xdxd

lim1x

1x ln

1 lim

:LHR Apply

1x1x1x

+

−

−

=−−

=

−−

→→→

( ) ( ) 00

1 ln 11111

x ln x1-x1x

lim

xx ln x1-x

x1-x

lim1x1x

=+−

−=+−=+⇒

→→

.

( ) [ ]

[ ] ( ) ( ) ( )1xln1x1

x1

1 lim

x ln x1-xdxd

1-xdxd

limx ln x1-x

1-x lim

:LHR again Apply

1x1x1x

++=

+=

+⇒

→→→

( ) 21

1 ln21

x ln21

lim1x

=+

=+

⇒→

2

1

1x

1

x ln

1 lim

1x=

−−∴

→

.

−

→ x2secx1

x 1

lim .4220x

( ) ∞∞=−=−=

−

→-

01

01

0sec01

01

x2secx1

x1

lim220x

( )00

0

02cos1x

cos2x-1 lim

xx2cos

x1

limx2secx

1x1

lim

: function equivalent the to gminTransfor

20x220x220x=−=

=

−=

−

→→→

[ ]

[ ]( ) ( ) ( )

00

002sin

x22x2sin

limx

dxd

cos2x-1dxd

limx

cos2x-1 lim

:LHR Apply

0x2

0x20x==−−==

→→→

( ) ( )

( )( ) ( ) ( )

20cos2x2cos2lim1

12x2coslim

xdxd

x2sindxd

limx

x2sinlim

:again LHR Apply

0x0x0x0x=====

→→→→

2x2secx

1

x

1 lim

220x=

−∴

→

.

[ ] x

0xx2 lim .5

→

[ ] ( )[ ] 00x

0x00 2x2 lim ==

→

[ ] xx2y Let =

[ ]

[ ]x1

x2 ln2x ln xy ln

x2 lny ln x

==

=

( )∞∞−=

∞===

→→

0ln

01

02ln

x1

x2ln limyln lim

:sides both on itlim the Apply

0x0x

[ ] ( )( ) 0xlim

x1

2x2

1

x1

dxd

x2lndxd

lim

x1

x2lnlim

:LHR Apply

0x

2

0x0x=−=−=

=→→→

( )( ) 12x lim therefore then

2xy since

1y limey lim

:sides bothof function inverse the Take

x

0x

x

0x

0

0x

==

=→=

→

→→

0yln lim

0

x1

x2ln limyln lim

0x

0x0x

=

==

→

→→

.

( ) 1x

1

1xxlim .6 −

→ +

( ) ( ) ( ) ( ) ∞−−

→===

+111xlim 0

1

11

1

1x

1

1x

( ) 1x

1

xy Let −=

( ) ( )1x

xlnxln

1x1

x lny ln 1x

1

−=

−== −

00

111ln

1xxln

limyln lim

:1x as sides both on itlim the Applying

1x1x=

−=

−=

→

++ →→

+

( )

( )

( )11

x1

lim1

1x1

lim1x

dxd

xlndxd

lim1x

xlnlim

:member right the on LHR Apply

1x1x1x1x===

−=

− ++++ →→→→

.

( )

( ) 72.2exlim

xy but eylim

:sides bothof function inverse the take ,1ylnlim1x

xlnlim

,Thus

1x1

1x

1x1

1

1x

1x1x

==∴

==

==−

−→

−→

→→

+

+

++

( ) x

0xxcotlim .7

+→

( ) ( ) 00x

0x0cotxcotlim ∞==

+→

( )

( )

x1

xcot lnxcot ln xxcot lny ln

xcoty Let

x

x

===

=

( ) ( )∞∞=

∞∞=

=

+

++

→

→→

ln

01

0 cot lnlim

x1

xcotlnlimylnlim

:sides both on limit the pplyA

0x

0x0x

( ) ( )( )

2

2

0x0x0x

x1

1xcscxcot

1

lim

x1

dxd

xcotlndxd

lim

x1

xcotlnlim

:member right on LHR Apply

−

−=

=+++ →→→

xcosxsin2x2

lim

x

1xcosxsin

1

lim

x

1xsin

1xcosxsin

lim2

0x2

0x2

2

0x ⋅⋅==

=+++ →→→

( )( ) 0

00sin

02x2sin

x2lim

22

0x===

+→

( )

( ) ( ) ( ) x2cosx2

lim2x2cos

x4lim

x2sindxd

x2dxd

limx2sin

x2lim

:again LHR Apply

0x0x

2

0x

2

0x ++++ →→→→===

( )( ) 0

10

0cos02 ===

( ) ( ) 1x cotlim then x coty Since

1eylim

sides bothof function inverse the take ,0yln lim

x1

xcotlnlim

,Hence

x

0x

x

0

0x

0x0x

=∴=

==

==

+

+

++

→

→

→→

x4sin

xtanx2lim .1

0x

+→

−

→ 220y y

1

ysin

1lim.2

xsin

x2lim.3

10x −→

→ ycosln

ylim.4

2

0y

( )x3

x2lnlim.5

3

x +∞→

( )

−

+ −→ x2tan

1

x1ln

1lim.8

10x

( ) x

42

0xx1lim.9 +

→

+∞→ x2

2

x e

x3lim.10

( ) 2x

22

0xxsin1lim.11 +

→

( ) x

2x

0xx3elim.12 +

→

x2tanln

x2coslnlim.13

4x

π→( ) x

1

0xx2sinx2coslim.6 −

→

( )( )xcscxsinlim.15 1

0x

−

→

x

x

2

0xe1lim.7

+

+→

( ) xlnxcoslim.14 1

0x

−

→ +

EXERCISES: Evaluate the following limits.