数分课堂课件第三章

§3.7L’Hospital法则一、洛必达法则1

（

0

型）0设f

,g在区间(x0

,x0

+d

)有定义,g(x)„0,满足lim0(i)

f

(

x)

=

0,xfi

x

+0lim

g(

x)

=

0;xfi

x

+(ii

)f

,g在区间(x0

,x0

+d

)可导,且g

(x)„0;0(有限或无穷大).(iii

)lim

f

(

x)

=

a,g¢(

x)x

fi

x

+00g(

x)

g¢(

x)x

fi

x

+x

fi

x

+则有

lim

f

(

x)

=

lim

f

(

x)

=

a.0(以x

fi

x

+为例)000g(

x)g¢(

x)x

fi

x

+x

fi

x

+g¢(x)lim

f

(

x)

=

lim

f

(x)

=

lim

f

(

x)

=

a.xfi

x

+证明：补充定义:f

(x0

)=g(x0

)=0,"x

˛

(x0

,x0

+d

),f

,g在[x0

,x]连续.由Cauchy中值定理:

$x

˛

(

x0

,

x),

使得f

(

x)

=

f

(

x)

-

f

(

x0

)

=

f

(x

)

.g

(

x)

g

(

x)

-

g

(

x0

)

g

¢(x

)0

0而当x

fi

x

+时,x

fi

x

+,因此说明:00x

,x

fi

¥

,x

fi–¥

也成立.x

fi

x

-

,

x

fi例1.2

x

2x2lim

1

-

cos

x

=

lim

sin

x

=

1xfi

0

xfi

0x

-

sin

xx

xlimxfi

0+3yy

-

sinyx

=

y=

limyfi

0+yfi

0+6=

lim

1

-

cos

y

=

113

y2112=

1=

lim-x-

x21

+

x2xxfi

+¥

1

+

x2=

limxfi

+¥lim

2xfi

+¥p

-

arctan

x注意:

① 各种方法综合使用(提出常用因子,等价代换,变量替换)②

可多次连续使用例2.2x(e

x

-

1)2

e

x2e2

x

-

e

x

-

3

x

-

1

2e2

x

-

e

x

-

3

x

-

1=

limxfi

0limxfi

02==

limxfi

0=

limxfi

08e2

x

-

e

x

722

x4e2

x

-

e

x

-

3二、洛必达法则2

（

¥

型）¥设f

,g在(x0

,x0

+d

)内满足:0(i

)

lim

g(

x

)

=

¥

,x

+x

fi(ii

)f

,g在(x0

,x0

+d

)可导,且g

(x)„0,0(iii

)lim

f

(x)=l

(有限或无穷).g¢(

x)x

fi

x

+0(以x

fi

x

+为例)xfi

x

+xfi

x

+0

g(

x)

0

g¢(

x)则

lim

f

(

x)

=

lim

f

(

x)

=

l证明：xfi

x

+设

lim

f

(

x)

=

l

有限0

g¢(

x)"e

>0,$0

<d1

<d

,当x

˛

(x0

,x0

+d1

)时:g¢(

x)f

(

x)

-

l

<

e,(

x0

,

x0

+

d1

),对(x,c)g¢(x)g(

x)

-

g(c)l

-

e

<

f

(

x)

-

f

(c)

=

f

(x)

<

l

+

e.g¢(

x)l

-

e

<

f

(

x)

<

l

+

e.必存在x

˛

(x,c),使得<

l

+

eg(

x)

-

g(c)g(

x)l

-

e

<

f

(

x)

-

f

(c)

=

f

(

x)

-

f

(c)g(

x)

-

g(c)

g(

x)

g(

x)xfi

x

+g(

x)0有: lim

sup

f

(

x)

£

l

+

e由e

任意性,

£

lxfi

x+g(

x)0同理: lim

inf

f

(

x)

‡

l

-

e由e

任意性,

‡

lxfi

x

+\

lim

f

(

x)

=

l0

g(

x)0固定c

,令x

fi

x+,有说明:(1)并未要求:0lim

f

(

x)

=

¥xfi

x

+x

,

xfi

–¥

,

x

fi

¥00(2)

可推广到xfi

x-,x

fi0(3)

注意lim

f

(

x)

要存在或为无穷大!

(否则要用其它方法

)g¢(

x)x

fi

x

+x

1lim

x

+sin

x

=lim

1

+cos

x

不存在,不能用洛必达!xfi

¥xfi

¥xlim

1

+

sin

x

=

1xfi

¥x

fi

+¥11=

0=

lim(1)

lim

ln

x

=

limxxfi

+¥

a

xaxfi

+¥

a

xa

-1例3.x

fi+¥

时,

ln

x

<<

xa

(a

>

0)

<<

e

x

<<

x

x

.eexxxaxa=

limxfi

+¥(2)

limxfi

+¥a

xa

-1设m

-1

<a

£

mxex

fi

+¥=

lima

(a

-

1)(a

-

m

+

1)

xa

-mx

fi

+¥e

x

xm

-a=

lim

a

(a

-

1)(a

-

m

+

1)

=

0.=

0=

limx

(1-ln

x

)(3)

limxfi

+¥=

lim

exfi

+¥e

xx

x

xfi

+¥

e

x

ln

xe

x0

¥

,¥-¥

,00

,1¥

,¥

0

型例4解求

lim

x-2ex

.xfi

+¥(

0

¥

)exxfi

+¥

2

x原式=lim2ex=

lim

=

+¥

.xfi

+¥关键:将其它类型未定式化为洛必达法则可解决0(

)

.¥¥¥

,¥的类型(0

),1.

0

¥

型步骤:

0

¥

10或

0

¥

0

1

.三、其它不定型例5解1sin

x

x-

1

).求lim(xfi

0(

¥

-

¥

)¥

-

¥

1

-

1

0

-

0

.0

0

0

0x

sin

xxfi

0原式=lim

x

-sin

x1

-

cos

x=

limxfi

0

sin

x

+

x

cos

x=

0.2.

¥-¥

型步骤:步骤:3.

00

,1¥

,¥

0

型¥0

ln

¥ln10

ln

01¥¥

0

00

取对数fi

0

¥

.例6解求lim

xx

.xfi

0+(

00

)lim

x

ln

x=

ex

fi

0+原式=lim

ex

ln

xxfi

0+1x2lim

x

x

fi

0+

-

1=

e=

e0=

1.1xlim

ln

xx

fi

0+=

e例7解1求lim

x1-x

.xfi

1(

1¥

)ln

x1原式=lim

e1-xxfi

1lim

ln

x=

ex

fi

11-

x1lim

x=

ex

fi

1-1=

e-1

.例8解1xfi

0+求lim

(cot

x)ln

x

.11(

¥

0

)ln(cot

x

)=

e

ln

x

(cot

x

)

ln

x1xfi

0+

ln

x

lim1

1cot

x

sin2

x1-ln(cot

x)

=

limxfi

0+cos

x

sin

x-

x=

lim+xfi

0=

-1,x\原式=e-1

.再次强调:0

¥(1)

仅用于0

,¥

,g¢(

x)(2)

lim

f

¢(x)不存在,如何处理?x

fi

x0及时化简，多次使用.四、其它用法以Heine定理为媒介,计算数列极限.例9.nlimn100nfi

¥

100ln100

(ln100)解：lim100=

0=

limxxfi

+¥

100xxxfi

¥

100

xfi

+¥

100100

x99

100!

1x100=

=

lim100=

0nn100\原式=limnfi

¥例10.xx

fi

0设

f

˛

C

2

,

且lim

f

(

x)

=

0

,

f

¢(0)

=

4,1xfi

0

x求

lim1

+

f

(

x)

x

(1¥

)xf

(

x)

=

0

,xxfi

0

xfi

0xfi

0解：由

lim

f

(

x)

=

0

,

知

lim

f

(

x)

=

lim

x\

f

(0)

=

0,xlim

f

(

x)

=

f

¢(0)

=

0.x

fi

0)11xf

(

x

)xln

(1+=

lim

e

xxfi

0xfi

0

\

lim1

+

f

(

x)

xx2xxxfi

0limxfi

0ln[1

+

f

(

x)]\原式=e22=

lim

f

(

x)

=

2.xfi

02

xf

(

x)=

lim

f

(

x)

=

limxfi

0xfi

+¥axfi

+¥求证

lim

f

(

x)

=

b

.证明：a

-1aa

xa

xa

-1

f

(

x)

+

xa

f

(

x)xf

(

x)

xa=

limxfi

+¥limxfi

+¥a

a=

lim

a

f

(

x)

+

x

f

(

x)

=

bxfi

+¥例11.

f

在(a,+¥

)可导,若

lim

[a

f

(

x)

+

x

f

(

x)]

=

b

,(a

>

0)1例12.设lim时,(1

+ax2

)2

-1与cos

x

-1是等价无穷小,求axfi

0解：sin

x2ax1