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 lim1x

3x4x 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 ,thus2

1x

used. be will Rule sHopital'L' on Theorems limit said the evaluate To example. second the to applied be longerno can problems previous the in applied principle the Obviously,

00

0)0sin(

)0(2)0(2sin

2x2x 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 .1

0x

EXAMPLE:Evaluate the following limits.

00

00sin

0202sin

2x2x sin lim

0x

10 cos2

02cos2

122x2coslim

2xdxd

2x sindxd

lim2x

2x sin lim

:Rule s'Hopital'L gsinu By

0x0x0x

12x

2x sin lim 0x

.3y sin-y3y-y tan lim .2

0y

0

0 0000

0sin3-003-0tan

3y sin-y3y-y tan lim

0y

122

3131

0 3cos-130 sec

33y cos-113ysec lim

3y sin-ydxd

3y-y tandxd

lim3y sin-y3y-y tan lim

:LHR By

2

2

0y0y0y

13y sin-y3y-y tan lim

0y

.

2

4x x4

2x sin ln lim .3

4x42

2x2cos2x sin

1

limx4

dxd

2x sin lndxd

limx42x sin ln lim

:LHR By

4x2

4x

2

4x

.ateminerdetin still is This

00

082

2cot

448

42 cot 2

x482x 2cot lim

4x

0

00

2sinln

44

42 sinln

x42x sin ln lim 22

4x

01 ln:Note

ln 0 ln

.

32x2csc4 lim

)4(82x2csc2 lim

x48dxd

2x cot 2dxd

lim

:LHR peatRe

2

4x

2

4x

4x

811

81

42 csc

81x2csc

81 lim 2

2

2

4x

8

1x4

sin2x ln lim 2

4x

.

x

2

x ex lim .4

ee

x lim2

x

2

x

e

21e

2x lime

dxd

xdxd

limex lim

:LHR By

xxx

2

xx

2

x

02e2

1e12lim

edxd

2xdxd

lim

:LHR peatRe

xxx

x

0ex lim x

2

x

.3x tan ln3x cos ln lim .5

6x

- ln

0 ln

2 tan ln

2 cos ln

63 tan ln

63 cos ln

3x tan ln3x cos ln lim

6x

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

26x

6x

6x

x3cos1

3xcosx3sin

limx3secx3tan lim

3x3secx3tan

13x tan3 lim

2

2

2

6x

2

2

6x26

x

16

3sinx3sin lim2

2

6x

13x tan ln3x cos ln lim

6x

.

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 logarithmof properties the apply then function, the for y variable

a letting by evaluated be may forms ateindetermin Thesely.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

212x csc xlim

21

)0(cos21

2cos2x1 lim

2cos2x1 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 .3

1x

01

01

111

1 ln1

1x1

x ln1 lim

1x

0

01 ln 1-11 ln11

x ln 1-xx ln1xlim

1x1

x ln1 lim

:fraction simple a ot gminTransfor

1x1x

1x ln1x11x

1x11

limx ln 1-x

dxd

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

.

1xln1x1x1

1 limx ln x1-x

dxd

1-xdxd

limx ln x1-x

1-x lim

:LHR again Apply

1x1x1x

21

1 ln21

x ln21 lim

1x

21

1x1

x ln1 lim

1x

.

x2secx1

x 1 lim .4

220x

-

01

01

0sec01

01

x2secx1

x1 lim

220x

00

002cos1

xcos2x-1 lim

xx2cos

x1 lim

x2secx1

x1 lim

: function equivalent the to gminTransfor

20x220x220x

00

002sin

x22x2sinlim

xdxd

cos2x-1dxd

limx

cos2x-1 lim

:LHR Apply

0x2

0x20x

20cos2x2cos2lim

112x2coslim

xdxd

x2sindxd

limx

x2sinlim

:again LHR Apply

0x0x0x0x

2x2secx

1x1 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

.

1x1

1xxlim .6

111xlim 0

111

11x

1

1x

1x1

xy Let

1x

xlnxln1x

1x lny ln 1x1

00

111ln

1xxlnlimyln lim

:1x as sides both on itlim the Applying

1x1x

11

x1lim

1

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

xcosxsin2x2lim

x1

xcosxsin1

lim

x1

xsin1

xcosxsin

lim2

0x2

0x2

2

0x

0

00sin

02x2sin

x2lim22

0x

x2cosx2lim

2x2cosx4lim

x2sindxd

x2dxd

limx2sin

x2lim

:again LHR Apply

0x0x

2

0x

2

0x

010

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

x4sinxtanx2lim .1

0x

220y y1

ysin1lim.2

xsinx2lim.3 10x

ycosln

ylim.42

0y

x3x2lnlim.5

3

x

x2tan1

x1ln1lim.8 10x

x4

2

0xx1lim.9

x2

2

x ex3lim.10

2x2

2

0xxsin1lim.11

x2

x

0xx3elim.12

x2tanlnx2coslnlim.13

4x x

1

0xx2sinx2coslim.6

xcscxsinlim.15 1

0x

x

x2

0xe1lim.7

xlnxcoslim.14 1

0x

EXERCISES: Evaluate the following limits.