高数一极限存在准则二两个重要三小结及作业

第六节

极限存在准则两个重要极限一、极限存在准则二、两个重要极限三、小结及作业1一、极限存在准则2夹逼准则准则Ⅰ 如果数列xn

,

yn

及

zn

满足下列条件:yn

£

xn

£

zn

(n

=

1,2,3

)lim

yn

=

a, lim

zn

=

a,nfi

¥nfi

¥

nfi

¥那末数列xn

的极限存在,

且lim

xn

=

a.证

yn

fi

a,

zn

fi

a,"e

>0,

$N1

>0,

N

2

>0,

使得当

n

>

N时,

恒有当

n

>

N1时恒有

yn

-

a

<

e,当

n

>

N

2时恒有

zn

-

a

<

e,取

N

=

max{N1

,

N

2

},

上两式同时成立,即a

-

e<

yn

<

a

+

e,

a

-

e<

zn

<

a

+

e,a

-

e

<

yn

£

xn

£

zn

<

a

+

e,即xn

-a

<e

成立,3\

lim

xn

=

a.nfi

¥上述数列极限存在的准则可以推广到函数的极限0(2) lim

g(

x)

=

A, lim

h(

x)

=

A,4准则Ⅰ′

如果当

x

˛

U

(

x0

,

d

)

(或

x

>

M

)时,有(1)

g(

x)

£

f

(

x)

£

h(

x),xfi

x0(

xfi

¥

)xfi

x0(

xfi

¥

)那末

lim

f

(

x)存在,

且等于A.x

fi

x0(

x

fi

¥

)准则1

和准则2称为夹逼准则.注意:利用夹逼准则求极限关键是构造出yn与zn

,并且yn与zn的极限是容易求的.).111n2

+

nn2

+

2+n2

+

1+

+例1

求lim(nfi

¥解,11<<n2

+

1nn2

+

nn2

+

1n2

+

nn+

+

n111

+

nn2

+

n=

limnfi

¥又limnfi

¥=

1,11n1

+n2=

limnfi

¥limnfi

¥=

1,由夹逼定理得151n2

+

11)

=

1.+n2

+

nn2

+

2n2

+

1lim(nfi

¥+

+记住结果：(1)(2)(a

>

0)lim

n

n

=1nfi

¥lim

n

a

=1nfi

¥例2

lim

n

1+

2n

+

3n

+

4nnfi

¥<

4n

4解：

4

<

n

1+

2n

+

3n

+

4n而lim

4

n

4

=

4nfi

¥\

lim

n

1+

2n

+

3n

+

4n

=

4nfi

¥6xx1x2x3xn

xn+12.单调有界准则如果数列xn满足条件x1

£

x2

£

xn

£

xn+1

£

,单调增加x1

‡

x2

‡

xn

‡

xn+1

‡

,单调减少单调数列准则Ⅱ

单调有界数列必有极限.几何解释:AM78例3证明数列

xn

=式)的极限存在.3

+3

+

+3 (n重根证

显然

xn+1

>

xn

,

\

{x

}是单调递增的

;n又

x1

=3

<3,假定

xk

<

3,3

+

xkxk

+1

=<

3

+

3<

3,nfi

¥\

{xn

}是有界的;

\

lim

xn

存在.

xn+1

=

3

+

xn

,x2=

3

+

x

,n+1

nnlim

x2=

lim(3

+

x

),nfi

¥n+1nfi

¥A2

=

3

+

A,解得

A

=

1

+

13

,

A

=

1

-

132

2(舍去)2=

1

+

13

.n\

lim

xnfi¥9(n

=1,2,3,

)2

x=

1

(

x

+

a

)nnn+1例4

设xnfi

¥x1

>0,a

>0,求lim

xn

.nn解：x2

xn+1nnxa=

1

(

x

+

a

)

‡

x=

anxxn+1

=2nx

21

(1+

a

)£

1

(1+

a

)

=12

axn+1

£

xn即nnfi

¥\lim

xn存在，A

=

1

(

A

+

a

),设lim

x

=

A,

由nfi

¥

A

=

–

a

,2

A\

lim

xn

=

a

.nfi

¥10AC二、两个重要极限(1)xlim

sin

x

=

1x

fi

02设单位圆

O,

圆心角—

AOB

=

x, (0

<

x

<

p

)o

xBD作单位圆的切线，得DACO

.扇形OAB的圆心角为x

,DOAB的高为BD

,于是有sin

x

=

BD,

x

=

弧

AB,

tan

x

=

AC

,△AOB的面积＜圆扇形AOB的面积＜△AOC的面积即

1

sin

x

<

1

x

<

1

tanx,2

2

211\

sin

x

<

x<

tan

x,即cos

x

<sin

x

<1,x

lim

cos

x

=1,xfi

0x\

lim

sin

x

=

1.x

fi

0sin

x

cos

xx

1<\

1

<12例4x

2x

fi

0求

lim

1

-

cos

x

.解2x22

sin2

x原式=limxfi

022(

)2sin2

x2

xfi

0

x=

1

limf

(

x)说明：（1）更一般形式：

lim

sin

f

(

x)

=

1,f

(

x

)fi03=

1xsin

x3xfi

0如

lim（2）不要混淆：limsin

xx=

0.例3xlim

tan

x

=xfi

0x

cos

xxfi

¥lim

sin

x

1xfi

0=1

1

=1lim( 2

)21sin

x2

xfi

0

x=1212=2=

1

.x2例5

lim

arcsin

x

.xfi0解:令t

=arcsin

x,则x

=sin

t

,tt

fi

0

sin

t原式=limt13sin

t1=

limt

fi

0=

1(2)xx

fi

¥lim(1

+

1

)

x

=

en先证lim(1

+1

)n

存在：nfi

¥nn设

x

=

(1

+

1

)n1=

1

+

n

1

+

n(n

-1)1!

n

2!=

1

+1

+

1

(1

-

1

)

+

+

1

(1

-

1

)(1

-

2

)

(1

-

n

-1).2!

n

n!

n

n

nnn14n!n2

+

+n(n

-1)

(n

-

n

+1)

1).2121(1

-+)(1

-1+

1

(1

-n

+1n(n

+1)!

n

+1

n

+

2n

+

2)(1

-n!

n

+12!

n

+11n+1n

+1)

(1

-)

(1

-

n

-1)=

1

+1

+

1

(1

-

)

+

显然

xn+1

>

xn

,

\

{xn

}是单调递增的;nx

<

1

+1

+

1

+

+

11

12n-1<

1

+1

+

+

+21=

3

-2n-12!

n!<

3,\{xn

}是有界的;nfi

¥\lim

xn

存在.n15nfi

¥记为lim(1

+1

)n

=e(e

=

2.71828

)类似地,

xxxfi¥可证：lim

(1+

1

)

x

=

e.1xfi

0注：（1）等价形式：lim(1+

x)

x

=

e1（2）一般形式：lim

(1+

f

(

x))

f

(

x

)

=

ef

(

x

)fi

0x3例6

lim(1+xfi¥3xx

6)

3)2

x

=

lim(1+xfi¥316

x)

3

]6=

lim[(1+xfi

¥x=

e6xxfi

¥例7

求

lim(1-

2

)5

x

.解2-

x)

2

]-10-

x原式=lim[(1+xfi

¥-10=

e例8cot

x1-

x1+

xlim(

)xfi

01-x

2

x

cos

x)

2

x

1-x

sin

x1-

x2

x=

lim(1+xfi

0sin

x17)

2

x

]

1-x1-x

2cosx

x

1-

x2

x=

lim[(1+xfi

02=

e例9x

xlim(sin

1

+

cos

1

)x

.xfi

¥x

xx=

lim[(sin

1

+

cos

1

)2

]

2xfi

¥2xx=

lim

(1+

sin

2

)xfi

¥x18x21sin

2sin

2x

]

x=

lim[(1+

sin

2

)xfi¥=

e例101lim

(3x

+

9x

)xxfi

+¥(

)x1x11

x

3

x

+

1=

lim

9xfi

+¥x19

xfi

+¥

13x

1

3

3xx=

9 lim

1

+0=

9

e

=

9三、小结0asina1

lim

=

1;某过程20201lim

(1

+

a)a

=

e.某过程两个准则夹逼准则;

单调有界准则

.两个重要极限设a

为某过程中的无穷小,作业214(2,3,5)P562(1,2,4),习题1-61(3,4,5,6),思考题求极限122lim

(3x

+

9x

)xxfi

+¥思考题解答x1lim

(3x

+

9

)xfi

+¥(

)x111

xx

3

x

+

1

x

=

lim

9xfi

+¥x23

xfi

+¥

13x

1

3

3xx=

9 lim

1

+=

9

e0

=

9xfi

04、

limx

cot

3

x

=

.一、填空题:.xxfi

01、lim

sinw

x

=.xfi

0

sin

3

x2、lim

sin

2

x

=