微积分i2004上册课件

第6次课§1.5

极限存在准则两个重要极限简明扼要复习上一次课内容(3分钟左右)§1.5

极限存在准则两个重要极限一、极限存在准则1.夹逼准则(1)如果"

x

˛

N

(xˆ0

,d0

)(或

x

>

M

)，有g

(x

£

f

(x

£

h(x

,(2

lim

g

(xxfi

x0

xfi

x0(xfi

¥

)

(xfi

¥

)=

A,

lim

h(x

=

A,xfi

x0(xfi

¥

)那么

lim

f

(x

存在,

且等于A.分析(1

g

(x

£

f

(x

£

h(x

,

(2lim

g

(xxfi

x0xfi

x0=

A, lim

h(x

=

A,xfi

x0

lim

f

(x

=

A.xfi

x0lim

g

(x

=

A

xfi

x0\

"

e

>

0,

$d

>

0,"

x

˛

N

(

x0

,d

),

A

-e

<

g(

x)

<

A

+e.lim

h(x

=

A

\

"

e

>

0,

$d

>

0,"

x

˛

N

(

x0

,d

),

A

-

e

<

h(

x)

<

A

+e."

x

˛

N

(

x0

,d0

),

g

(x

£

f

(x

£

h(x

,\

"

e

>

0,

$d

>

0,"

x

˛

N

(

x0

,d

),

A

-e

<

f

(

x)

<

A

+

e.证xfi

x0

xfi

x0

lim

g

(x

=

A, lim

h(x

=

A.\

"

e

>

0,

$d1

>

0,

当0

<

x

-

x0

<

d1时,

有

g

(x

)-

A

<

e;

A

-

e

<

g

(x

)<

A

+

e$d2

>

0,

当0

<

x

-

x0

<

d2时,

有

h(x

)-

A

<

e;

A

-

e

<

h(x<

A

+

e

0

<

x

-

x0<

d0，有g

(x

<

f

(x

<

h(x\

取d

=

min{d1

,

d2

,

d0

},

当0

<

x

-

x0

<

d时,xfi

x0

f

(x

)-

A

<

e

lim

f

(x有

A

-

e

<

g

(x

<

f

(x

<

h(x

<

A

+

e=

A(1

yn

£

xn

£

zn

(n

=

1,

2,),(2 lim

yn

=

lim

zn

=

A,nfi

¥

nfi

¥

lim

xn

=

A.nfi

¥例1111).+n2

+

1n2

+

2+

+求lim(nfi

¥解11n2

+

1n2

+

n+

+nn2

+

n又limnfi

¥=

1,nlimnfi

¥n2

+

1=

1,由夹逼定理得111lim(nfi

¥+)

=

1.n2

+

1n2

+

2n2

+

n+

+,n2

+

nn<n2

+

1n<n2

+

n11=limnfi

¥1

+

n11=

limnfi

¥1

+

n2例2求lim

n

1

+2n

+3n

.nfi

¥n

3

£

n

1

+

2n

+

3n

£

n

3

·

3n=

3n

3解：

n

3n

£

n

1

+

2n

+

3n

£

n

3

·3n3

=

n

3n

£

n

1

+

2n

+

3n

£

n

3

·3n

=

3n

3lim 3

n

3

=

3nfi

¥\

lim

n

1

+

2n

+

3n

=

3.nfi

¥例3

证明:

limcos

x

=

1xfi

02解：

0

£

1

-

cos

x

=

2sin2

x

<2x

2且limx

fi

0xfi

0\

lim(1

-

cos

x

=

0

limcos

x

=

1xfi

0x2

x

2=2<

2

2

=

0sin

x

£

x2.

重要极限一(1)sin

xxlimxfi

0=

12设单位圆O,

圆心角—

AOB

=

x,

(0

<

x

<

p

)于是有sin

x

=

BD,

x

=

弧AB,

tan

x

=

AC

,过A点作单位圆的切线，得DACO.扇形OAB的圆心角为x,D

OAB的高为BD

,记住此公式x

cos

x

<

sin

x

<

1,2上式对于-p

<x

<0也成立.即当0

<

x

<

p

时,

有cos

x

<

sin

x

<

1;2

x

limcos

x

=

1,xfi

0lim1

=

1,xfi

0xxfi

0\

lim

sin

x

=

1.x

1sin

x

cos

x

1

<<

BD

<弧AB

<AC\

sin

x

<

x

<

tan

xxx

fi

0例4

(1)求lim

tan

x

.x

x

cos

xxfi

0xfi

0lim

tan

x

=

lim

sin

x

1解1limx

fi

0x

x

fi

0

cos

x=

lim

sin

x=

1.xx

fi

0(2)求lim

arcsin

x

.解

令t

=

arcsin

x,

x=sint,ttfi

0

sin

t=

1.原式=lim例5x2xfi

0求lim

1

-cos

x

.解2sin2

xx2原式=limxfi

0sin

2122xlimx

fi

02

=

x

2

2

1x

2

sin

2

=lim

2

x

fi

0

x

2

1212=12=

.x

3xfi

0例6求lim

tan

x

-sin

x

.3x解limx

fi

0xx

2x

fi

0x

fi

0lim

1

-

cos

x=

lim

tan

xx

2x

fi

0=

12lim

1

-

cosx

=

1

.前例6xx

2tan

x

-

sinx

=lim

tan

x1

-

cos

x

x

fi

0

x2sin

1sin

xxfi

0例7

求lim

x

.x2

sin

1xx

sin

1xxfi

0sin

x

xfi

0

sin

x解

lim

x

=

limxlim

x

sin

1xxfi

0

sin

xxfi

0=

lim=

0.xfi

x0sin

(x

-

x0x

-

x0例8 求

lim

.f

(x

)uxfi

x0ufi

0sin

f

(x

sin

ulim

=

lim

=

1.f

(x

)xfi

x0

lim

sin

f

(x=

1.一般地：lim

f

(

x)

=

0xfi

x0令u

=

f

(

x

),

x

fi

x0

u

fi

0,

故xfi

x0ufi

0解

令u

=

x

-

x0

,

当x

fi

x0时,

u

fi

0,故lim

sin

(x

-

x0

=

lim

sin

u

=

1.x

-

x0

ux

fi

0

sin

nx(m

、n

„0

).例9求lim

sin

mxsin

nxx

fi

0

sin

nx

mx

nxx

fi

0lim

sin

mx

=

lim

sin

mx

mx

nx解lim

mx

limnxmxx

fi

0x

fi

0nx

x

fi

0

sin

nx=

lim

sin

mxn=

1

m

1n=

m

.解 令

t

=

p

-

x

x

=

p

-

t

,x

fi

p

tan

5

xt

fi

0t

fi

0lim

sin

3

x

=

lim

sin

(3p

-

3

tsin

(p

-

3

t=

limtan

(5p

-

5

t

)tan

(p

-

5

t

)tan

5

tt

fi

0=

-

lim

sin

3

ttan

5

xx

fi

p例10求

lim

sin

3

x

.35sin

3ttan

5t5tt

fi

0=

-

lim

3t

5=

-

3

.tan

5

x

3

x5

x

tan

5

x

5x

fi

p

t

fi

0,x

fi

px

fi

plim

sin

3

x

=

lim

sin

3

x3

x

5

x

=

3x

fi

p3

1

+

x

2

-

1例10求lim

1

-

cos

2

x

.2x2xfi

0lim

1

-

cos

x

=

1xlimxfi

0n

1

+

x

-1

1=n322(2x)2xx21

-

cos

2x

4

x21

+

x

-

1解原式=limxfi

012

3=

1

4

=

6..xfi

1

cos

p

x1

-

x2例11

求lim2解

令1

-

x

=

u,

x

=

1

-

u;

当x

fi

1时,

u

fi

0,sin

u2pufi

0=

2lim

2

p2usin

p

u=

2

limufi

04=

.p2xfi

12p

u

2原式=

lim

(1

+

x)(1

-

x)

ucos

p

x

ufi

0

cos

p

(1

-

u)=2

lim单调增加有上界2.单调有界准则如果数列{xn

}满足条件

x1

£

x2£

xn

£

xn+1

£,x1

‡x2‡xn

‡xn+1

‡,单调减少有上界有上界,即$M

,"n,有xn

£

M有下界,即$m,"n,有xn

‡m准则Ⅱ

单调有界数列必有极限.nfi

¥证10证明有下界证明:lim

xn存在并求此极限值.a

xn2xn

2a

-

xn

=

-2

xn

又因

xn+1

-

xn

=

xn

+axn

2

n-1

n-1

x

=

1

x

+所以数列{xn

}有下界a

.2a

2

+

b2

‡

2ab

a

+

b

‡

ab()1

ax

+2

n

x

n

n

=1,

2,

3,,

a

>

0例12

x1

>

0,

且xn+1

=a

-

x2所以

xn+1

£

xn

,

数列{xn

}单调减少.xn-1n-1

a‡

x

=

a

,1n由已知x

>

0及递推公式知x

>

0,20

证明有单调减1

=

n

£0

.2

xnnfi

¥令lim

xn

=

A,

A

=

1

A

+

a

,2

A

nfi

¥故根据单调有界准则lim

xn存在.2naxnfi

¥

nfi

¥

n-1n-1

得

lim

x

=

lim

1

x

+

A

=

–

a

,nfi

¥所以

lim

xn

=

a

.axn

2

n-1n-1

对等式

x

=

1

x

+两边取极限因xn

>0,nfi

¥故lim

xn

=

A

‡

0.

A2

=

a例4

证明数列{xn

}:x1

=

3,

x2

=

3

+

3

,

x3

=nfi

¥极限lim

xn存在,并求此极限.解10

由数学归纳法证明数列的单调性.x1

=

3

<

3

+

3

=

x2

,假设xn-1

<

xn

,

由数列的递推关系式得xn

=

3

+

xn-1

<

3

+

xn

=

xn+1

,故数列{xn

}单调增加.

xn

<

xn

+1

,3

+

3

+

3

,,xn

=

3

+

xn-1xn

>

3,20证明数列有上界.nnxnnfi

¥由10

、20知lim

x

存在.nn

x2

<

3

+

xn

{x

}有上界.nxn

=

3

+

xn-1

<

3

+

x3n

x

<

3

+1

<

3

+1

=

3

+1(

x

>

3

)nfi

¥令lim

xn

=

A,nfi

¥

nfi

¥

lim

xn

=

lim 3

+

xn-1

A

=

3

+

A

A2

-

A

-

3

=

0

A

=

1

–

13

,2nfi

¥因为

xn

>

0,

且lim

xn存在,nfi

¥则lim

xn

=

A

‡

0,2n=

A

=

1

+

13

.所以

lim

xnfi

¥2nlim

xnfi

¥=

A

=

-1

+

5

.问:此种解法是否正确?为什么?nfi

¥解 由所给数列知一般项存在如下递推关系例5

设x1

=

1,

x2

=

1

-

1

,

x3

=

1

-

1

-

1

,求lim

xn

.xn

=

1

-

xn-1

,令

lim

xn

=

A

lim

xn

=

lim 1

-

xn-1nfi

¥

nfi

¥

nfi

¥2

A

=

-1

–

5

A

=

1-

A

A2

+

A

-1

=

0nfi

¥事实上,所给数列为:x1

=

1,

x2

=

0,

x3

=

1,

x4

=

0,.显然,limxn不存在.不正确注:通过递推公式两边取极限来求数列极限,必须要求数列的极限存在,否则其作法

是错误的.x1

=

1,

x2

=

1

-

1

,

x3

=

1

-1

-

1

,1x(2) lim(1

+xfi

¥)x

=

e1nnn1!

n

2!

n!n2

nnn

1

n(n-x

=(1+

)

=1++1) 1

++

n(n-1)(n-2)(n-n+1)

1=

1

+1

+

1

(1

-

1

)

++

1

(1

-

1

)(1

-

2)(1

-

n

-1).2!

n

n!

n

n

n1

n=

enfi

¥

n

首先证明：lim

1

+1n+1n+1x

=

1

+=

1

+

1

+

1

(1

-

1

)

+2!

n

+

11n

+

112

n

-

1+(1

-)(1

-n

+

1n

+

1)(1

-

)n!

n

+

1112)n+(1

-)(1

-(n

+

1)!

n

+

1n

+

1n

+

1)(1

->

xn

,

xn+1n\{x

}是增的;<

1

+

1

+

1

+

+

12!

n!122n-1<

1

+

1

+

1

+

+2

1

n1-

2

=1+

1-

1<3,\{xn

}是有界的;nfi

¥\limxn存在.nnfi

¥记为

lim(1

+

1

)n

=

e(e

=

2.718281828459045)n2!

n

n!

n

n

nx

=

1

+

1

+

1

(1

-

1

)

++

1

(1

-

1

)(1

-

2

)(1

-

n

-

1).

1

n-1=

3

-

2

"

x

>

1

,$n,

使n

£

x

<n

+1,n

nnnfi

¥nfi

¥nfi

¥lim(1

+

1

)而

lim(1

+

1

)n+1

=

lim(1

+

1

)n1

11)n+1nfi

¥nfi

¥lim(1

+nfi

¥)n

=

lim(1

+lim(1

+)-1

=

e,n

+

1

n

+

1n

+

1xxfi

+¥\

lim

(1

+

1

)x

=

e.由夹逼定理得1

x=

elim

1

+xfi

+¥

x

(1

+

1

<

1

£

1

n

+1

x

n11

+

1

<

1

+

1

£

1

+

1

n

+1

x

n1)n

<

(1

+

1

)n

(1

+)n

<

(1

+

1

)n

£

(1

+

1

)xx

xx n

+1n

+11n=

e,(1

+)n

<

(1

+

1

)n

£

(1

+

1

)x

£

(1

+

1

)x

£

(1

+

1

)n+1n

+1

x

x

n令t

=-x,x

txfi

-¥t

fi

+¥\

lim

(1

+

1

)x

=

lim

(1

-

1)-t1)

t=

lim

(1

+t

fi

+¥t

-

111=

lim

(1

+t

fi

+¥)t

-1

(1

+t

-

1t

-

1)

=

e.xxfi

¥\

lim(1

+

1

)x

=

e,1t令x

=1lim(1

+

x)

xxfi

01lim(1

+

x)

x

=

exfi01xxlim(1

+

)xfi

¥=

exxfi

-¥lim

(1

+

1

)x

，)

=

e.t1t=

lim(1

+t

fi

¥)

tt=

lim

(t

fi

+¥t

-

1例14xx

fi

¥求lim(1

-1

)x

.解1]-

x

-1-x原式

=

lim[(1

+

)xfi

¥.1e=

e-1

=例15xfi

¥2

+

x(1)求lim(3

+x

)2

x

.1)-4=

lim[(1

+xfi

¥)x+2

]2

(1

+x

+

2x

+

22=