limits (l'hospital's rule)

7/30/2019 Limits (L'Hospital's Rule)

1/15

Limits Indeterminate Forms and LHospitals Rule

I. Indeterminate Form of the Type

0

0

We have previously studied limits with the indeterminate form0

0as shown in the

following examples:

Example 1: ( ) 42222

)2)(2(

2

4limlimlim

22

2

2

=+=+=

+=

x

x

xx

x

x

xxx

Example 2: === xx

x

x

xx

x

x

xxx 2sin

1

3cos

1

1

3sin

2sin

3cos3sin

2sin

3tanlimlimlim

000

2

3)1)(1)(1(

2

3

2sin

2

3cos

1

3

3sin

2

3limlimlim

02003

==

x

x

xx

x

xxx

[Note: We use the given limit 1sin

lim0

=

.]

Example 3:12

1

83

1)8(

28

3 2

3

0lim ===

+

fh

h

h. [Note: We use the definition

of the derivativeh

afhafaf

h

)()()( lim

0

+=

where 3)( xxf =

and a = 8.]

Example 4: ( ) ( ) 23

3sin

3

3

21cos

lim3

===

fx

x

x

. [Note: We use the

definition of the derivativeax

afxfaf

ax

=

)()()( lim where

xxf cos)( = and 3=a .]

1


2/15

However, there is a general, systematic method for determining limits with the

indeterminate form0

0. Suppose thatfandgare differentiable functions atx = a

and that)(

)(lim

xg

xf

axis an indeterminate form of the type

0

0; that is, 0)(lim =

xf

ax

and 0)(lim =

xgax

. Sincefandgare differentiable functions atx = a, thenfandg

are continuous atx = a; that is, )()( lim xfafax

= = 0 and )()( lim xgagax

= = 0.

Furthermore, sincefandgare differentiable functions atx = a, then = )(af

ax

afxf

ax

)()(lim and

ax

agxgag

ax

=

)()()( lim . Thus, if 0)( ag , then

)(

)(

)(

)(

)()(

)()(

)()(

)()(

)(

)(limlimlimlim

xg

xf

ag

af

ax

agxgax

afxf

agxg

afxf

xg

xf

axaxaxax

=

=

=

=

iff and

g are continuous atx = a. This illustrates a special case of the technique known as LHospitals Rule.

LHospitals Rule for Form0

0

Suppose thatfandgare differentiable functions on an open interval

containingx = a, except possibly atx = a, and that 0)(lim =

xfax

and

0)(lim = xgax . If )()(

lim xg

xf

ax

has a finite limit, or if this limit is + or

, then)(

)(

)(

)(limlim

xg

xf

xg

xf

axax

=

. Moreover, this statement is also true

in the case of a limit as ,,, + xaxax or as .+x

In the following examples, we will use the following three-step process:

Step 1. Check that the limit of)(

)(

xg

xfis an indeterminate form of type

00 . If it

is not, then LHospitals Rule cannot be used.

Step 2. Differentiatefandgseparately. [Note: Do not differentiate)(

)(

xg

xf

using the quotient rule!]

2


3/15

Step 3. Find the limit of)(

)(

xg

xf

. If this limit is finite, + , or , then it is

equal to the limit of)(

)(

xg

xf. If the limit is an indeterminate form of type

00 , then simplify )(

)(

xg

xf

algebraically and apply LHospitals Rule again.

Example 1: 4)2(21

2

2

4limlim

2

2

2

===

x

x

x

xx

Example 2:2

3

)1(2

)1(3

2cos2

3sec3

2sin

3tan 2

00limlim === x

x

x

x

xx

Example 3:121

)8(3

1

)8(3

11

)1()8(

3

1

283

23

20

32

0

3

0limlimlim ==

+=

+=+

h

h

hh

hhh

Example 4: ( )2

3

3sin

1

sin

3

21cos

limlim33

==

=

x

x

x

xx

Example 5:2

1

22

11limlimlim

002

0

==

=

x

x

x

x

x

x

e

x

e

x

xe[Use LHospitals Rule

twice.]

Example 6: ( ) ( )( ) ( ) 010

1cos

2

11cos

2

1sin

1

limlimlim2

32

===

=

+++x

x

xx

x

x

x

xxx

, or

( )0

1

)0(2

cos

2

sin1sin

1

limlimlim0

2

0

2

====++ + y

y

y

y

x

x

yyx

where xy 1= .

Example 7: ( ) 0)0(0ln1

ln limlim 00 === xxxx

xx [This limit is not an indeterminate

form of the type0

0, so LHospitals Rule cannot be used.]

3


4/15

II. Indeterminate Form of the Type

We have previously studied limits with the indeterminate form

as shown in the

following examples:

Example 1: =

+

+=

+

+

++222

2

222

2

2

2

132

753

132

753limlim

xx

x

x

x

xx

x

x

x

xx

xx

xx

2

3

002

003

132

753

limlim

2

2=

+

+=

+

+

++ xxxx

xx

Example 2: =+

1

13

2lim x

x

x =

+

22

2

22

1

13

lim

xx

x

xx

x

x 0

1

0

01

00

11

13

2

2

lim ==+

=+

x

xx

x

Example 3: =+

12

43

2

3

limx

x

x

=

+

33

2

33

3

12

43

lim

xx

x

xxx

x =

+

=+

0

3

00

03

12

43

3

3

lim

xx

x

x

limit does not exist.

Example 4: =++

1

14 2

limx

x

x

=+

+

x

xx

x

x 1

14 2

lim

x

x

x

x

x 1

14

2

2

lim +

+

xx =2(

becausex < 0 and thus 2xx = ) = =+

+

x

x

x

x

x 1

14

2

2

lim

4


5/15

21

4

11

14

2

2

lim =

=+

+

x

x

x

.

However, we could use another version ofLHospitals Rule.

LHospitals Rule for Form

Suppose thatfandgare differentiable functions on an open interval

containingx = a, except possibly atx = a, and that =

)(lim xfax

and

= )(lim xgax . If )( )(lim xg xfax has a finite limit, or if this limit is + or

, then)(

)(

)(

)(limlim

xg

xf

xg

xf

axax

=

. Moreover, this statement is also true

in the case of a limit as ,,, + xaxax or as .+x

Example 1:2

3

4

6

34

56

132

753limlimlim 2

2

==+

=++

+++ xxx x

x

xx

xx

Example 2: 0)0(2

31

2

3

2

3

1

13limlimlim 2 ====+

xxx

x

xxx

Example 3: ===+

4

18

4

9

12

43limlimlim

2

2

3 x

x

x

x

x

xxx

Example 4: +

=+=++

144

1142

8

114

2

22

limlimlimxxx

x

xx

xxx

LHospitals

Rule does not help in this situation. We would find the limit as we

did previously.

5


6/15

Example 5:( )( ) 24

4

22

3

3

2

2

3

2

33

22

13

12

1

3

1

2

)1ln(

)1ln(limlimlimlim

xx

xx

xx

xx

x

x

x

x

x

x

xxxx ++

=++

=

+

+=++

++++=

3

2

72

48

72

48

636

24

612

28

limlimlim 2

2

3

3

===+=++

+++ x

x

x

x

xx

x

xxx

Example 6: 02

0

222

1

1

ln 22

0

3

03

02

0limlimlimlim =

=

=

=

=++++

x

x

x

x

x

x

x

xxxx

Example 7: 02

)0(arctan1arctan

limlimlim =

=

=

+++

x

xx

x

xxx

[This limit is

not an indeterminate form of the type

, so LHospitals Rule

cannot be used.]

III. Indeterminate Form of the Type 0

Indeterminate forms of the type 0 can sometimes be evaluated by rewriting theproduct as a quotient, and then applying LHospitals Rule for the indeterminate

forms of type

0

0or

.

Example 1: 0)(1

1

1

lnln limlimlimlimlim

0

2

02

000

==

=

==+++++

xx

x

x

x

x

xxx

xxxxx

Example 2: =

=

==++++ x

xx

xx

x

x

xxx

xxxx

tansin

cotcsc

1

csc

lnln)(sin limlimlimlim

0000

0)0)(1(tan

sin

limlim 00 ==

++ xx

x

xx

Example 3: ( )(

1sin

1

1sin1sin limlimlim

0

===+++ y

y

x

xx

xyxx

[Let xy 1= .]

6


7/15

IV. Indeterminate Form of the Type

A limit problem that leads to one of the expressions

)()( ++ , )()( , )()( ++ , )()( ++

is called an indeterminate form of type . Such limits are indeterminatebecause the two terms exert conflicting influences on the expression; one pushes

it in the positive direction and the other pushes it in the negative direction. However,

limits problems that lead to one the expressions

)()( +++ , )()( + , )()( + , )()( +

are not indeterminate, since the two terms work together (the first two produce alimit of + and the last two produce a limit of ). Indeterminate forms of thetype can sometimes be evaluated by combining the terms and manipulating

the result to produce an indeterminate form of type00 or

.

Example 1: =+

=

=

+++ xxx

x

xx

xx

xx xxx sincos

1cos

sin

sin

sin

11limlimlim

000

02

0

coscossin

sinlim

0

==++

+ xxxx

x

x

Example 2: ( )[ ] =

=

2

0

2

0

cos1lnln)cos1ln( limlim

x

xxx

xx

=

=

2

1ln

2

sinln

cos1ln limlim

02

0 x

x

x

x

xx

V. Indeterminate Forms of the Types 1,,0 00

Limits of the form [ ])(

)(limxg

ax

xf

[ ]

)()(lim

xg

x

xfor frequently give rise to

indeterminate forms of the types 1,,0 00 . These indeterminate forms can

sometimes be evaluated as follows:

(1) [ ] )()( xgxfy =

(2) [ ] [ ])(ln)()(lnln )( xfxgxfy xg ==

(3) [ ] [ ]{ })(ln)(ln limlim xfxgyaxax

=

7


8/15

The limit on the righthand side of the equation will usually be an

indeterminate limit of the type 0 . Evaluate this limit using the

technique previously described. Assume that [ ]{ })(ln)(lim xfxgax

=L.

(4) Finally, [ ] L

axaxax

eyLyLy ==

=

limlimlim lnln .

Example 1: Findx

x

xlim0+

.

This is an indeterminate form of the type 00 . Let === xx xyxy lnln

xx ln . ( ) =====

+++++ x

x

x

x

xxxy

xxxxxlimlimlimlimlim

02

0000 1

1

1

lnlnln 0.

Thus, 10

0lim ==

+exx

x.

Example 2: Findxx

x

e2

)1(lim

++ .

This is an indeterminate form of the type 0 . Let +=

xxey2

)1(

x

eey

x

xx)1ln(2

)1(lnln2 +

=

+=

.

x

ey

x

xx

)1ln(2ln limlim

+=

++=

22

1

2

1

12

limlimlim =

=+

=

+

+++x

x

xx

x

x

x

x

x e

e

e

ee

e

. Thus,xx

x

e2

)1(lim

++ =

2e .

Example 3: Find ( ) xx

x1

0

coslim+

.

This is an indeterminate form of the type 1 . Let = xxy1

)(cos

x

xxy x

)ln(cos)(coslnln

1

=

= . ==

++ x

xy

xx

)ln(cosln limlim

00

( ) 0tanlim0

=+x

x. Thus, ( ) x

x

x 1

0

coslim+

= 10 =e .

8


9/15

Practice Sheet for LHospitals Rule

(1) =

)2cos(1

3

0lim

x

xxex

x

(2) =+ 3)(ln

limx

x

x

(3) =

)]ln()cos1[ln( 2

0lim xxx

(4) =

+

x

x x

32

1lim

(5) =

+ 12

cos

11

cos

lim

x

x

x

(6) =

x

xx

x

2

0

11lim

(7) =

2

1

0

)(coslim xx

x

(8) =+

++

1683

275

234

234

1lim

xxx

xxxx

x

(9) =

x

x

x

5819lim

0

(10)( ) =++ 15ln

2ln3

3

limxx

x

(11) ( ) =+

+

xx

x

e2

1lim

9


10/15

(12) =

30

)2sin(2)4sin(lim

x

xx

x

(13) =

xx

x

1

sin

1lim

0

(14) =

+1

1

lim xx

ex

(15) =+

xxx

ln3

0lim

(16) =

++

x

x

x

x

15

12ln

lim0

(17) =

+

+

21lim

x

x x

e

(18) = x

x

x

)3arctan(sinlim

0

(19)( )

=+

3

0

2

0

sin

limx

dttx

x

(20) ( ) =+

xx

x

xe1

2

0lim

(21) =+ x

x

x

arctanlim

(22) = )tan2arcsin())3(arctan(sin

lim0 x

x

x

(23) = 20

)ln(coslim

x

x

x

10


11/15

(24) =

+

+

x

x x2

11lim

(25) =+

x

x

x1

)(lnlim

(26) ( =++

xex

x

)1ln(lim

Solution Key for LHospitals Rule

(1) 2

3

4

6

2cos4

339

2sin2

13

)2cos(1

333

0

33

0

3

0 limlimlim ==++

=+

=

x

eexe

x

exe

x

xxexxx

x

xx

x

x

x

(2) ( ) ( )=====

+++++ x

x

xxx

x

xxx

x

xxxxx ln61ln6

1

)(ln31)(ln3

1

)(lnlimlimlimlimlim 223

( )+==

++ 616

1limlim

x

xxx

(3) =

=

=

2

0

2

0

2

0

cos1ln

cos1ln)]ln()cos1[ln( limlimlim

x

x

x

xxx

xxx

2ln2

1ln

2

sinln lim

0

=

=

x

x

x

(4) Let =x

y1

( ) yy

x

x

yx

3

0

3

212

1 limlim =

++. Now, let ( ) == zyz y ln21

3

( )( )

61

21

6

)21ln(3ln

21ln321ln limlimlim

000

3

=

=

=

=+++

y

y

yz

y

yy

yyy

y. Thus,

==

=

+++

6

000limlimlim 6ln6ln ezzzyyy

( ) yy

x

x

yx

3

0

3

2121 limlim =

++=

6

0lim

=

+ez

y.

11


12/15

(5) Let =x

y1 =

=

=

++ + )2sin(2

)sin(

1)2cos(

1)cos(

12

cos

11

cos

limlimlim00 y

y

y

y

x

x

yyx

41

cos41

cossin4sin limlim

00== ++ yyy

yyy

.

(6) =+

+

=

2

22

0

2

0 11

111111limlim

xx

xx

x

xx

x

xx

xx

( ) ( )( ) ( ) 2

1

11

1

1111

11

202

2

02

2

0limlimlim =

+

=

+

=

+

xx

x

xxx

xx

xxx

xx

xxx.

(7) Let ( ) ===== 200211

)ln(cosln)ln(cos)ln(cosln)(cos limlim22

xxy

xxxyxy

xx

xx

2

1

cos2

1sin

cos2

sin

2

cos

sin

limlimlim000

=

=

=

xx

x

xx

x

x

x

x

xxx

. Thus, ( ) = 2

1lnlim

0

yx

==

21

00limlim

2

1ln eyy

xx

21

0

1

0limlim

2

)(cos

== eyx

x

x

x

.

(8) =+

+=

+

+=

+

++

124836

24260

122412

122120

1683

2752

2

123

23

1234

234

1

limlimlimxx

xx

xxx

xxx

xxx

xxxx

xxx

0

20limit does not exist.

(9) ( )=

+

=++

=

xx

x

x

x

x

x

x

x

xxx 5819

)581(81

5819

581958195819limlimlim

000

( ) 185

5819

5

5819

5limlim

00

=+

=+ xxx

x

xx.

(10)( )( )

13015

315

)2(15

)15(3

15

15

2

3

15ln

2ln3

3

3

3

3

2

3

2

3

3

limlimlimlim =+

=+

=

+=+

++++ x

x

x

x

x

x

x

x

x

x

xxxx

12


13/15

(11) Let ( ) ( ) =+=+=+=+

y

x

eeyey

x

x

xxxx ln)1ln(2

1lnln1 lim22

2

2

1

2

1

1

2

)1ln(2limlimlimlim =

=

+

=+

=+

++++x

x

xx

x

x

x

x

x

x

x e

e

e

ee

e

x

e . Thus, =+

yx

lnlim

==

++

2limlim 2ln2 eyyxx

( ) 22

limlim 1

+

+==+ eye

x

xx

x.

(12) =+

=

=

x

xx

x

xx

x

xx

xxx 6

)2sin(8)4sin(16

3

)2cos(4)4cos(4)2sin(2)4sin(limlimlim

02

03

0

86

48

6

)2cos(16)4cos(64lim

0

=

=+

xx

x

.

(13) =+

=

=

xxx

x

xx

xx

xx xxx sincos

cos1

sin

sin1

sin

1limlimlim

000

02

0

coscossin

sinlim

0

==++ xxxx

x

x

.

(14)(

.11

1

1

11 0

1

2

2

11

1

limlimlimlim ===

=

=

++++ee

x

xe

x

eex x

x

x

x

x

x

x

x

(15) .0)3(3

31

1ln

ln 3

0

34

034

031

0

3

0limlimlimlimlim ==

=

==

+++++ x

x

x

x

x

x

xxx

xxxxx

(16)3

)15)(12(

3

1

)15(

)12(5)15(2

12

15

15

12ln

limlimlim0

2

00

=++

=

+

++

++

=

++

xx

x

xx

x

x

x

x

x

xxx

.

(17) Let =x

y 1 ( ) yy

x

x

eyx

e2

1

0

2

11 limlim +=

+

++. Next, let ( ) =+= zeyz y ln1 2

1

( )

22

1

2

)1ln(ln

2

)1ln(1ln limlimlim

000

21 eey

e

y

eyz

y

eyey

yyy

y =+

=+

=+

=++++

. Thus,

13


14/15

==

=

+++

2

000limlimlim

2ln

2ln

e

yyy

eze

ze

z ( ) yy

x

x

eyx

e2

1

0

2

11 limlim +=

+

++=

2

0

lime

y

ez=+

. ( ) ( ) ( ) =+=+=+=

y

x

xexeyxey

x

x

xxxx lnln

lnln lim0

21

21

2

(18)3

1

3sin1

3cos3

)3arctan(sin 2

00limlim =

+=

x

x

x

x

xx

.

(19)( )

3

1

3

)cos(

6

)cos(2

3

)sin(sin 2

0

2

02

2

03

0

2

0limlimlimlim ====

++++

x

x

xx

x

x

x

dtt

xxx

x

x

.

(20) Let ( ) ( ) ( ) =+=+=+=

yx

xexeyxeyx

x

xxxx lnlnlnln lim0

2

1212

( )3

1

12

ln 2

2

0

2

0limlim =+

+

=+

xe

e

x

xe x

x

x

x

x

. Thus =

=

3ln3ln limlim

00

yyxx

=

3

0lim eyx

( ) 30

12

0limlim eyxex

xx

x

==+

.

(21)0

2arctan

lim =+=+

x

x

x

.

(22)2

3

121

3

tan41

sec2

3sin1

3cos3

)tan2arcsin(

))3(arctan(sin

2

2

2

00limlim ==

+=

x

x

x

x

x

x

xx.

(23)

2

1

cos2

1sin

2

cos

sin

)ln(coslimlimlim

002

0

=

=

= xx

x

x

x

x

x

x

xxx

.

(24) Let =x

y1

.2

11

2

11

1

0limlim

y

y

x

x

yx

+=

+

++Let =

+= zyz

y

ln2

11

1

14


15/15

.

2

1

1

211

21

2

11ln

ln2

11ln

2

11ln limlimlim

000

1

=+

=

+

=

+

=

+

+++

y

y

y

zy

y

yyyy

y

Thus, ==

=

+++

21

000limlimlim

2

1ln

2

1ln ezzz

yyy

=

+

+

x

x x2

11lim

.2

11 2

1

0

1

0limlim ezyy

y

y

==

+

++

(25) Let =====++ x

xy

x

xxyxy

xx

xx)ln(ln

ln)ln(ln

)ln(lnln)(ln limlim11

0

ln

1

1

ln1

limlim ==++ xx

xx

xx

. Thus, ==

=

+++yyy

xxxlimlimlim 0ln0ln

=10e 1)(ln limlim1

==++

yxx

x

x.

(26) ( ) ( )( ) =

+=+=+

+++x

x

x

xx

x

x

x e

eeexe

1lnln1ln)1ln( limlimlim

01lnln1

ln limlim ==

=

+

++x

x

xx

x

x e

e

e

e.

15

limits (l'hospital's rule)

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