1

羅必達法則(L’Hospital’s Rule)

1.不定式 (Indeterminate Forms)

2.羅必達定理 (L’Hopital’s Rule)

3. 例題 page 659-663

2

Indeterminate Forms

1. The Indeterminate Forms of Type

2. The Indeterminate Forms of Type

3. The Indeterminate Forms and

4. The Indeterminate Forms , and

EX:

0 00 01

0

0

,,3

cos1limlim 2

0x

xx e

x

xx

x

x

x

x

xx

xx

x

x

)3sin(,

ln,)1( limlimlim

0

cos

0

3

The Indeterminate Forms of Type 0/0

Take for example

When &

Divide both numerator and denominator by x-1

2

1

1 0limx

x

2

1

1

1limx

x

x

1

1lim 1

1x

x

x

2

1

12

1limx

x

x

1x 1

1 0limx

x

2

1

1 0

1 0limx

x

x

4

The Indeterminate Forms of Type 0/0

22

21

1 1

1

2

221 0 1

1 0 1

1

111 211

1 11 11 1

1( 1) 21 21 2

11 ( 1) 1 11

limlim lim

lim

lim |lim

lim

x

x x

x

x x x

x x x

x

xxx xx

x xxx x

xx xx x

xx xx

5

The Indeterminate Forms of Type 0/0

Replace by

Replace by

Replace x −1 by

if

, ,

exist and , then the weak form of L’Hopital’s Rule

limx

f x

( ) ( ), ( ) 0f x f u f u

1x ux

)('

)('

)(

)(lim xg

xf

xg

xf

ux

0)( ug

2 1x

limx

g x

( ) ( )lim ,limx u x u

f x f u g x g u

x u x u

( ) ( ), ( ) 0g x g u g u

6

L’Hospital’s Rule Let f and g be functions and let a be a real numb

er such that

Let f and g have derivative that exist at each point in some open interval containing a

If , then

If does exist because becomes large without bound for values of x near a, then also does not exist

'

lim'x a

f xL

g x

lim 0,lim 0x a x a

f x g x

lim

( )x u

f x

g u

limx a

f xL

g x

'

lim'x a

f x

g x

'

'

f x

g x

7

EX1 L’Hospital’s Rule

FindCheck the conditions of L’Hospital’s Rule

If then f’(x)=2xIf f(x)=x-1 then f’(x)=1By L’Hospital’s Rule, this result is the desired limit:

2

1

1 0limx

x

2

1

1

1limx

x

x

22

1 1 1

1 '1 22

1 1 ' 1lim lim limx x x

xx x

x x

1

1 0limx

x

2

1

1 0

1 0limx

x

x

2 1f x x

8

EX2 L’Hospital’s Rule

FindCheck the conditions of L’Hospital’s Rule

If then f’(x)=If f(x)= then f’(x)=2(x-1)

Because does not existThen does not exist

1

ln 0limx

x

21

ln

1lim

x

x

x

21

ln

1lim

x

x

x

2

1

1 0limx

x

2

1

ln 0

01lim

x

x

x

lnf x x 1x

21x

1

1 2 1limx

x

x

9

Using L’Hospital’s Rule

1. Be sure that leads to the indeterminate form 0/0.

2. Take the derivates of f and g seperately.

3. Find the limit of ; this limit, if it exists, equals the limit of f(x)/g(x).

4. If necessary, apply L’Hospital’s rule more than once.

limx a

f x g x

lim ' 'x a

f x g x

10

EX3 L’Hospital’s Rule

FindCheck the conditions of L’Hospital’s Rule

If then f’(x)=If f(x)= then f’(x)=

3

1

0limx

x

3

0 1lim xx

x

e

3

1

01lim x

x

x

e

1

1 0lim x

x

e

3

1

0

1 0lim xx

x

e

3f x x 23x

1xe 22

0

00

lim33 00

lim 1lim xx x

xx

xx

e e

xe

11

EX4-1 L’Hospital’s Rule

Find

If then f’(x)=If f(x)= then f’(x)=2x

0

1 0lim x

x

e x

20

limx

x

e x x

x

2

0

0limx

x

2

0

1 0

0limx

x

e x

x

1xf x e x 1xe 2x

0

1 0

2 0limx

x

e

x

12

EX4-2 L’Hospital’s Rule

If then f’(x)=If f(x)= then f’(x)=2

2 00

1 1 0lim

2 0limx x

xx

e x e

x x

1xf x e xe

2x

2 0 00

1 1 1lim lim

2 2 2limx x x

x xx

e x e e

x x

0

1

2 2limx

x

e

13

EX5 L’Hospital’s Rule

Find

(by substitution)

2

1

1 0limx

x

2

1

1lim

x

x

x

1

1limx

x

1

1 0 0

1 0limx

x

e x

x

12

2

11 1 2

1 21lim lim

x x

x x

x x

22

1

1

1

11 0

01

limlim

limx

x

x

xx

x x

14

Proof of L’Hospital’s Rule-1

We can prove the theorem for special case f, g, f’, g’ are continuous on some open interval containing a, and g’(a)=0. With these assumptions the fact that

and

means that both

f(a)=0 and g(a)=0

lim 0x a

g x

lim 0x a

f x

15

Proof of L’Hospital’s Rule-2

Thus,

Multiplying the numerator and denominator by 1/(x-a) gives

( )lim lim

( ) ( ) ( )x a x a

f x f x f a

g x g x g a

( )

lim lim( ) ( )( )x a x a

f x f af x x a

g x g ag xx a

16

Proof of L’Hospital’s Rule-3

By the property of limits, this becomes,

the limit of numerator is f’(a)

the limit of denominator is g’(a) and

( )

limlim

( ) ( )( ) lim

x a

x a

x a

f x f af x x a

g x g ag xx a

lim '' 'lim

'( ) lim '( ) '( )x a

x ax a

f xf a f x

g a g x g x

( ) 0g a

17

Proof of L’Hospital’s Rule-4

Thus,

' ''

lim lim lim( ) '( ) ''( )x a x a x a

f x f x f x

g x g x g x

18

Example: Find (0/0).sin

lim30 x

xxx

6

1

6

coslim

6

sinlim

3

1coslim

sinlim

0

02030

xx

x

x

x

x

xx

x

xxx

3

0 0lim(sin 1) 0, lim 0x x

x x

19

Example: Find (0/0)xx

xx 3

cos1lim

20

032

sinlim

3

cos1lim

020

x

x

xx

xxx

2

0 0lim(1 cos ) 0, lim 3 0x x

x x x

20

Example: Find (0/0)20

1)1ln(lim

x

xex

x

0lim,0)1)1ln((lim 2

00

xxe

x

x

x

12

)1(1

lim

21

1

lim1)1ln(

lim

2

0

020

xe

xx

e

x

xe

x

x

x

x

x

x

21

The Indeterminate Forms of Type

If and

Then

lim ( )x u

f x

)('

)('

)(

)(limlim xg

xf

xg

xf

uxux

lim ( )x u

g x

22

xx

xxx 53

2lim

2

2

56

21lim

53

2lim

2

2

x

x

xx

xxxx

3

1

6

2lim

x

Example (∞/∞) Find

2 2

0 0lim( 2 ) , lim 3 5x x

x x x x

23

Example: Find , where p>0 。x

p

x e

x

lim

lim ,p

xx

x

xelim

1

lim limp p

x xx x

x px

e e

)1(lim p

xpx Nk01 kp

x

p

xx

p

x e

px

e

x 1

limlim

0)1()1(

lim

x

kp

x e

xkppp

24

Example: Find (∞/∞)x

xx tan1

seclim

)2/(

x

xx

x

xxx 2)2/()2/( sec

tanseclim

tan1

seclim

1sinlim)2/(

xx

25

x

xx 2

lnlim

Example: Find (∞/∞)

x

x

x

xxax /1

/1lim

lnlim

01

lim xx

26

Example: Find (a>0) (∞/∞)ax x

xlnlim

xx

lnlim

a

xxlim

1

/1lim

lnlim

axax ax

x

x

x0

1lim

ax ax

27

Example: Find (∞/∞)xx

x

2

)(lnlim

2

2)(lnlim xx

x

x2lim

xxxx

xxx

2)2(ln

1)ln2(

lim2

)(lnlim

2

xx x

x

2)2(ln

)ln2(lim

]2)2(ln2[)2(ln

/2lim

xxx x

x

0

28

Example: Find (∞/∞)x

xx cot

lnlim

0

xx

lnlim

xx

cotlim0

x

x

x

xxx 200 csc

/1lim

cot

lnlim

x

xx

2

0

sinlim

0 0

sinlim sin limx x

xx

x

010

29

The Indeterminate Forms andTo evaluate

Rewrite

Or

Then apply L’Hospital’s Rule

0

1 0

f xf x g x

g x

0

lim 0n z

f x g x

1

g xf x g x

f x

30

The Indeterminate Forms and To evaluate

F(x)-g(x) must rewrite as a single term. When the trigonometric functions are involved, switching to all sines and cosins may help.

0

limn z

f x g x

31

Example: Find

xxx

lnlim 2/1

0

0lim 2/1

0

x

x

xx

lnlim0

0

1/ 21/ 20 0

lnlim ln lim x x

xx x

x

02lim

21

/1lim

02/30

x

x

xxx

32

Example: Findx

xx

1sinlim

0

x

xx

xxx 1

1sin

lim1

sinlim

t

tt

sinlim

0 1

33

Example: Find )sinln(tanlim)(

2

xxx

xx

tanlim)(

2

0sinlnlim)(

2

xx

0

2 2( ) ( )

ln sin 0lim (tan ln sin ) lim

cot 0x x

xx x

x

x

xx

x 2)( csc

cossin

1

lim2

0)sincos(lim

)(2

xxx

34

Example: Find (∞−∞))ln

1

1(lim

1 xx

xx

1

lim1 x

xx

xx ln

1lim

1

00

1 1

1 ln 1lim( ) lim

1 ln ( 1) lnx x

x x x x

x x x x

xxx

xxxx /1)1(ln

1/1lnlim

1

1ln

lnlim

1

xxx

xxx

2

1

2ln

1lnlim

1

x

xx

35

Example: Find (∞−∞))1

1

ln

1(lim

1

xxx

1 1

1 1 1 ln 0lim( ) lim

ln 1 ( 1) ln 0x x

x x

x x x x

xx

x

xx 1

ln

/11lim

1

1ln

1lim

1

xxx

xx

2

1

ln2

1lim

1

xx

36

)tan(seclim)(

2

xxx

Example: Find (∞−∞)

xx

seclim)(

2

xx

tanlim)(

2

2 2

00

( ) ( )

1 sinlim (sec tan ) lim

cosx x

xx x

x

0sin

coslim

)(2

x

xx

37

Example: Find )]1ln(2[lnlim

xxx

)1ln(lim xx

xx

2lnlim

2lim[ln 2 ln( 1)] lim ln

1x x

xx x

x

)1

2limln(

x

xx 2ln)

1

2limln(

x

38

Example: Find 0x

xx

1tanlim

00

1tan1

lim tan lim 1x x

xxx

x

t

tt

tanlim

0 1

1

seclim

2

0

tt

39

Example: Find 0xxx

tan)2

(lim2

]2

)2

tan[()2

(limtan)2

(lim22

xxxxxx

)2

tan()(lim0

tt

t

)cot)((lim0

ttt

t

ttt sin

coslim

0

t

ttt sin

coslim

0 t

t

ttt

coslimsin

lim00

tt tt

coslimcos

1lim

00 111

40

The Indeterminate Forms , and

In these cases

1. Let

2.

3. If exists and equal L,

then

g xy f x

00 0 1

ln lny g x f x lim ln

x ag x f x

limg x L

x af x e

41

Example: Find

x

xx cot

0)1(lim

1)1(lim0

xx

cotlim

0x

)1ln(cot

0

cot

0lim)1(lim xx

x

x

xex

]tan

)1ln(limexp[

0 x

xx

1]sec)1(`

1limexp[

20

xxx

cot

0lim (1 ) 1x

xx

42

Example: Find n

nn

lim

let ny nn

ny

lnln

01

1

limln

limlnlim

xx

xy

nxn

and

Then

1lim0)limln(lnlim

yyynnn

1lim

n

nn

43

Example: Findx

x x)

11(lim

1)1

1(lim xx

xxlim 1

)1

1ln(lim)

11(lim x

x

x

x

xe

x

)]1

1ln(limexp[x

xx

1 lim ln(1 ) 0

xx

x

44

00

1ln(1 )1

lim ln(1 ) lim 1x x

xxx

x

t

tt

)1ln(lim

0

1

11

1

lim0

tt

Then ex

x

x

)1exp()

11(lim

45

Example: 。xxx cot

0)4sin1(lim

1)1(lim0

xx

cotlim

0x

1

)4sin1ln(cot

0

cot

0lim)1(lim xx

x

x

xex

]tan

)4sin1ln(limexp[

0 x

xx

420

]sec)4sin1(`

4cos4limexp[ e

xx

xx

46

Example: Find

x

xx

0lim 00

xx

x

x

xex ln

00limlim

)lnlimexp(

0xx

x

x

xxx

xx 1ln

limlnlim00

2

0 1

1

lim

x

xx

0)(lim0

xx

0 0 lim exp( lim ln )x

x xx x x

1)0exp(

47

Example: Find x

x x)

1(lim

20 0

2

1ln

020lim)

1(lim x

x

x

x

xe

x )

1lnlimexp(

20 xx

x

20

1lnlim

xx

xReplace the result of

48

x

xx

xxx 1

1ln

lim1

lnlim2

020

2

32

0 1

2

lim

x

xx

x

02lim0

xx

1)0exp()1

lnlimexp()1

(lim2020

x

xx x

x

xThen

49

Example:

)ln(/1

0

/1

0lim)(lim xex

x

xx

x

x

exe

1

))ln(

exp(lim0 x

xex

x

xx

xxe /1

0)(lim

))ln(

limexp(0 x

xex

x

50

1

1

lim)ln(

lim00

xee

x

xe x

x

x

x

x

1and

Then ))ln(

limexp()(lim0

/1

0 x

xexe

x

x

xx

x

e )1exp(

Example: xx

xxe /1

0)(lim