Bài tập giải tích nhóm ngành 1 tập 2

P H A M N G O C T H A O (Chu bien)

B

A

l

T

A

P

(Giao trinh Toan , nhom nganh I )
TAP HAI
DAI HOC Q u 6 c GIA
HN

TRUNGTAM

thAngtin.THirVlfiN

510/33
V-GO
NHA XUAT

ban

DAI HOC QUOC GIA HA NOI

PHAM NGpC THAO (Chu bien)
LE MAU HAI - NGUy I n VAN KHUE
NGUy I n DINH s a n g - Bpl DAC TAC

BAI t a p GIAI TfCN
(Nhom ngdnK I)
t a p HAI

NHA XUAT b a n DAI HOC QUOC GIA HA NOI - 1998


L d l NOI DAU
Giao trinh bM tap nay dudc chiing toi soan thao tUcfng ting vcti cac
phan ly thuyet cua giao trinh giai tlch cua “Toan D ai cifdng" dung cho
nhom nganh I Dai hoc Quoc gia H a Ngi. N o la giao trinh chu yeu de
sinh vien luyen tap sau khi hoc ly thuyet b phan I cac tap 1 va 2 cua
“Toan D ai cUcfng” ve giai tich. Trong no gom cac loai bai tap sau: ly
thuyet gidi han, Topo va ham lien tuc trong R", phep tmh vi phan
trong R" va iing dung, tich phan 1 16p, tlch phan suy rong va tich phan
phu thupc tham so cung tich phan suy rong phu thuoc tham so'; chuoi
so' day va chuoi ham; tlch phan bpi, tlch phan dvtdng va mat.
Trong moi chifdng, sau phan tom tat cac van de ly thuyet, cac cong
thiic dupe dung c6 lien quan den chifdng do, chiing toi cho cac bai tap
cP ban cua chucfng do, nghia la cac bai tap ma sinh vien can lam de
n im dupe cac kien thiic va ky nang cd ban. N g a y trong do chung toi
cung sip xep bM tap theo cac loai: ren luyen ky nang ap dung thuan
tiiy cong thiic hoac ky nang tlnh toan, ren luyen tu duy suy luan va
kha nang sang tao, nhutng bM tap mang tlnh chat ly thuyet, cac bM
tap nang cao. Mot phan khong nho trong bai tap 6 moi chupng nham

bo sung va md rong kien thilc cho sinh vien do dieu kien thdi gian han
hep ma viec day ly thuyet chiia thiic hien dUdc. M ot so' bai tap kho
dupe chiing toi danh dau (*) danh cho cac sinh vien kha gioi c6 y muo'n
tim hieu sau them kien thiic va nang cao trinh dp cua minh.
Phan hu 6 ng din va tra Idi cho cac bai tap cung diWc chung toi lam
chi tiet va danh cong siic thoa dang. Cac bai tap d l dUdc cho dap so'
mot so bai tap kho dUdc giai ti mi. M ac du vay do sil han hep cua thdi
gian va tru 6 c nhu cau cap nhat cua sinh vien nen khong tranh khoi
nhiing thieu sot trong qua trinh bien soan. Chiing toi mong nhan dupe
nhiing y kien dong gop cua cac dong nghiep va doc gia de hoan thien
giao trinh hdn b cac Ian tai ban sau.
Nhan day chiing toi bay to su cam dn doi v 6 i cac thanh vien trong
hoi dong tham djnh, dac biet la G S TS N guyen V an M a u cung cac P G S
Nguyen Thiiy Thanh va Pham Chi Vinh da dong gop nhieu y kien cho
sU hoan thien cua ban thao.

Hd. ngi ngdy 18 thdng 2 ndm 1998
C a c tac g ia


G IA I H O A G H ir d N G D A N

C h iid n g I

G IC S l H A N

A. Gidl HAN CUA DAY
1. a) Xet

n+ 1

1

7n + 2

7

L a y no =

5

5
< — < B,
7(7n + 2)
7n

5
n >—
78

+ 1.

7c

b), c) tUdng til a).
2.b) Xet

X

2n

n

2
^
< ------- > 0

.
khi

n -> 00.

Tircfng til cho a) va c).
4. a) Gia sik k la so' tu nhien nho nhat v 6 i k > a. K h i do:
0<

a

n

n!

a ...a
1 .2 ...n

<

a

k-l

a

\ n-k+l

^ 0

( k - 1 )! vk/

khi n -> o).
b) Gia sii m la so' nguyen ducing v 6 i m >k. K h i do:
in
0 < —

a"

s

m"

n

n

a"

m/ n

-b ".

va

m

,1 b =

> 1,


nhimg:
n

n

0<

b"

n

[l+ (b -D
n

2 x1

<

->()

n(n - l)(b - I)

khi

n->co.

c) Ttf;
n

n

2

<

suy ra:

-> 0 khi n -> CO.
n -1

d) Bkng qui nap ta c6 ; n! >

U

Ttf do:
0 < 4 .<

^
/ \n
n
n
Ix /

n

■> 0 khi n ^ CO,

1

\ 11
5.T if

e - lim

n -^ o o

nen v 6 i day {nk},

-1- -

V

1

n J
-> + o o

1

+
e = lim
k->ooV

khi k

-> + o o

ta c6 :

1

Neu { pk), Pk>l, Pk -> +°o khi k-> +<» thi c6 day so nguyen ditdng
{Hk}; Hk < Pk < Hk + 1
6

nk->+oc.


Bdi vi:

1+

nk + 1.

<

1+

1 + ---

V

Pk^

Pk

< 1+
V

1+
V

—

—

nk + 1

SUV ra :

lim

k->oc

1+
P,.

Tritdng hdp da}’ {qj,} tien ra
<^k = -Qk

lim

k->oo

1+

-o o

thi ap dung ket qua tren clio day:

khi k ^ + 0 0 ta cung dUdc:

1

Qk
= e.

6 . a) Ro rang x„ <

Dong thdi:
1
1
^
1
1
1
X„ < 1 + ----+...4
— 1 + 1 ------ 1---------- h
1.2
(n - l)n
2
2

=

1

n-1

1

n

2 - 1n

suj’ ra: x„ <2 v 6 i V n >1.
Do do ton tai Iftii x „ .
U^cc
b) titdng W a).
7. b) x„ = lo g (n + l) -> + 00.
8 . a) Day tang vi x^+i - x„ =

n+l

10

10^

> 0 va bi chan tren vi:

10"

■■■■=Po +i'

7


l3 )“^ii±L
X II

1 +_ !_ >

n+l

l.V a y day t^ng.

TH:
+ ln

ln x „ = In
1 1

1

1

, '^ 4;

1

+...+ In
_

1

1+

2 'V

+...= 1

Vay x„ 9. a) Bang qui nap ta dudc x„ < Xn+i. M at khac x„ < Va +1 vdi moi
n. That vay, k h in g dinh diing vdi n = 1. Gia sii x„ 1 < Va +1. Khi do:
X „ < -y/a + V a + 1 < V a + 2-\/a + 1 - V a + 1.

Do do { x j CO gifli

Tii

han.

Dat b = lim x „ . Vay b>
n^co

Va .

x / = a + x„_, ditadenb® = a + b.

Do do

b^ - b - a = 0.

Tii: do:

b -

b )T a c 6;

X2 =

Ttfdo

X2 - X , - - ( x o - X i ) ,X 3 -X 2 = - ( X i - X 2 )

1+ Vl + 4a

Xi + Xn

X 2 + X]

. X3 = - ^ - —

1
2

1.

^

Bang qui nap:
k-2

Xk -X k -t = ( - 1 )

X2 - X 1
k-2

T\i do b^ng phep cong lien ti§p ta c6;
8

vdi k > 2

^


(-1)

X,, “ X| - (X2 - X i)

n-2

3.2

^ i^n-2 ^2

'^'^2 -'^]

3.2
Do do:

lim Xjj =,

2x2

11-2

2

n -> c c

c) Ro rang x„ > 0 vdi nioi n. Bang qui nap c h t o g minh duac
x„ < x„,, va x„ < 1 v6i moi n. Do do ton tai lim x,, - a.
n -> o c

1 x^_i

1

Q ua gidi han dang tliiic x„ = - + — ... ta difdc a = — + — .
2
2
2
2
'

Do do a = 1 .
d) Bang qui nap chtfng minh diWc daj- { x j giam va bi chan du 6 i
bcii 0. do do ton tai a = lim x „ , a > 0. Qua gidi han d in g thiic
n -> cc

Xn =

2 +x„_

ta CO a =

a

Tu’ do a = 0.

> ^/5

e) Ta CO Xn = 2

T\i do

x„ > x„,! va ton tai a = lim

n —>-oc

V X ,j_ i

+Xj^_

t a CO

v d i m o i n.

. Qua gidi han d in g thxi^

a = -n/5

1 0 . a)

< Mq

-t- q

M q

11+1

1— q

<8
9


LaV:
q

Ine
n>

M

In q

b)
sin(n + 1 )
•^11+])

-'n

I

211+1

1
, sin(n + p)
< ------------- + . . . +
ij+i
2 n+p

1

n+])

I
2*'

1
2n+p

< ---------------------------------------------------------------- < K

n+1

Lay n > log;;

c)

<

^ii+p

1

1

(n + l)(n + 2 )

(n + p)(n + p + 1 )

1

n+ 1

n+2

n+p

n+p

n+ 1

n

Lay n > —

c

11. D^t Z„ - 1 x 2 - Xi l + l x - x 2 l+...+lxn -X., J. Khi do v 6 i moi n

ZnDong thdi

= Z„ + I x „h - x j > Z„.

Do do ton tai lim Z„. Vay v 6 i s > 0 ton tai no sao cho n > no
n —>oc

p > 1 ta c6 :

10

^n+p

<8

Nhit vay

^n+p ^n+p-1 + . . . +

T a do:

^ii+p

< e.


Do do dav [ x j hoi tii do tieii chiian Caiichv.
1 - ( - 1 )'^

Xet x,i = -------------,n = L2 ,... R 6 rang lim x„ = 0 .
2n
n->x

Tuy nhien {x„( khong co

s i i t h a y doi bi chan. That v a y v 6 i A > 0

bat ky ta c6 :

,

1 1
= 1+ - +
3 5

+

x^ - x

ln(l + I) + ln

I

1+

V

3/

= ln

2n - 1

I

>

2 n -l

2 4

2‘‘

13

2n - I

> In J in + I > A

12. a) Xet:

1
(n + i r
n

l-a
a

>

1

+

^ 2 ii “

(n + 2 )

1

a

( 2 n )“

>

n
( 2 n)^

1
c/

vdi 0 < gq <

(X

b) X 2 „ - x „

I

n
n
+...+--------- > --------- >
ln(n + l)
ln( 2 n)
ln( 2 n)
2n
1

2

13. Gia stf {x,} la day dc3n dieii tang va lim x,. = C . Tii k < nj^
k->oo ^
nen x^

14. Dat lim X2 i, = a, lim x 2 ,1+1 = b. lim X3,, = c.V 6 i nu = 2k, day
D—>cc

n->cc

{X(3k} la day con cua {X;j,J. N o cung la day con cua {x.^J v 6 i ni^ = 3k. Vay
a = c. Tudng tu vdi n^ = 2k + 1, day { Xokfo} la day con cua {x^J va
{XiinHVay b = c. Do do a = b va lim x,i = a. Ket qua khong dung khi
n->x
thay 2 bdi so k > 2.

11

15. Trifd n g lidp

O v a liilii han thi l i m ------ -- - = 1. V6i a = :^
n—>cc x,j
a

tien tdi gidi han hQu han chang han nhif x„ = 2 ".

thi CO th§
Xn

Cung CO the lim
n->x

nhu day {x,J = {2

}.

Xj^

Tracing hdp a = 0 thi co th§ khong ton tai gidi lian nhu dol v 6 i
day
1
= — (q + ( - 0 ''X
n

16-a) Xet hai day con x,i = —
2^1

q ^ 1.

va x,, = --------- . Tu Uni x,j = 0
2 *’
n^oc

va lim x„ = 1 nen 0 va 1 la cac gidi han rieng. Hdn niJa cac day con
n-^oc

hoi tu cua day da cho phai thuoc vao mot trong hai daj’ naj- nen
khong con gidi han rieng nao khac.
b) R 6 rang moi so' hiJu ty r(0 < r < 1 ) la cac so hang cua day da
cho.
Gia svf a la so thilc, 0 < a < 1 . Vdi m tU nhien du Idn. xay ra;
(x + — ^ < 1
n+m

Vn = 1,2,3,...

Vdi moi so' tii nhien n, ton tai so* hiiu ty
a <

thuoc day da cho v d i:

< a + ---------.
n+m

Tu’ do lim r„ == a. Vay a la gidi han rieng. TiWng tu’ cho tntdng
n—>oc

hdp 0 < (X < 1 . Do d 6 gidi han cua day da cho la [0, 1 ].
c) Co hai gidi han rieng la 1 va 5^

12


17. a) V i cac phan tii cua { x j n im troiig hai day;
^ 2 ii-i = 2 +

_

va X 2 „ = -2 - —

j

,

x^„ < x^„

,

,

-{x^„ ,} ddn dieu

giam, {x^„ ,} ddn dieu tang nen:
lim x „ = lim X 2 „_, = 2
11—
>cc
n-^oc

I im x „ = lim Xj,, = -2
n->x

b)

lim x „ = + x ,

c)

lim x„ = 1-,

lim x ^ )= -c c .

lim

= 0.

18. Xet da}^;
ai + 1 , 02 + 1,

ap + 1,

1 1

+ 7 * ^ 2 + - ,...,a p

^

^

1

1

1

1

aj + ~ , a 2 + - , . . . , a p
n
n

2.

n

Ton g quat, xet:
^i> ill

^ 2 , a, + - , a 2 + | , a j, a, + j , 8 2 + j , a^ + | , a 4 ,...
-t

■+

19. a) Xet x„ = n(n = 1 ,2 ,...).
b) Gia su
} 1-

= a, a la hfiii han, con y„ = n. Khi do day x^,

3'a x.i, 3’-, ... la da}' phan ky va c6 gidi han rieng duy nhat la a.
c) Bai tap 18.
d) Xet day sau chi'fa moi so" hfiii ty.
I

O7A7I
n

n

2
)

2

’ ~ »’
2 .■> -■) .■>

2
1

1

n

»

2
n

?••*>

2

J ' 2 ’"

n - 1
n

J
n -

>

n

n
n -1 ’

n

n

n

n

n

n - r ' “ ’ 7 ” 7 ’T ” T

13


^ Do tinh tru mat cua tap so him ty treii

R

nen mpi so thiic la gidi

han rieng cua no.
21.

a) T a c6 nhan xet sau; neii { x n k }

limxn < limxn . Ttf dinh nghia gi6 i han dirdi ta c6:
+J’ n )=

k->=c

K hi do;
t o x„ + lim y„ < lira x,,^ + Um y„^ f) ^cC'
n—
k—
k“>cc
Um x „ , + mn y„, < lim x„
,n-»«:
'‘m i;->oo
N him g

+ lim y ,
k->»

+ > iik. )

N g o a ir a d a y |x„|^ | hoi tu nen day |yiik,„
Urn y„

j

tu nen
cung hoi tu

= lim yn.

Vay;

Iffl x„ + Um >■„ s

n->oo

n->cc

^

m->^oo

’

= lim (X n + y „ ).

n->oc
b) Chflng minh tiwng til a).
Dau dang thiic xay ra khi ton tai lim x„ va lim y„ • Cac day sau
®

n—>00

n->cc

cho ba't d in g thiic thilc sil:
n(ti+l)

X „ = (-1 )

14

2"

S in 2 l^ ,y „ = (-1 )

2

c o s '—

(n = l,2,...).


11( 11+ 1 )

Khi do: x„ + y „ = ( - 1 )

2

limx,, = - 1.

limy,, = - 1

limx„ = 1,

limy,) = 1

lim(x„ + y „ ) = -l,

lim(Xn + y n ) = 1

22. Cac phan a), b) diwc chting minh tUdng ti(.
N eu x„ = 0 (n = 1,2,...) hoac limx,, = 0 thi a) diWc chiing minh.
Gia sii limXn > 0. Khi do x„ > 0 bat dau ttf luc nao do trfl di.
Tif dinh nghia gi6 i han ditdi ta c6 ;
i ™ ( x , , . y „ ) = lm(x„|^ .y„^ ),limxn^ = lim x
*‘ni
11-^ sc

Ta c6 :
limxn.lim}',, < limx,,^ .lim>’,,^ = lim x„
m—>00
Vi

^ lu m -3 111

.limy.

•< lim x,,
---- —
m-»oc

k

la day hoi tu cua day hoi tu 'jxn.
l i m (x „ .y „ )= lim (x „
,y„
U-j^oo
‘‘nJ

NhUng da>’

|

‘m

i'nen

).

khac khong nen |y„^

cimg hoi tu. do do, lim yj,

= lim yj,
ni

lirnXn limyn ^ lini
I^j

. y„

.Iimv„

ni

|

.Do do:

in->cc

D1

=lim(x„.
.y„. ) = lim (x„y„).
» ...
^•m
m
I ■■
oj
in

Bay gicJ chiing minh: Hm (Xn,y„) < limx„.limy„
Truong hdp

limy,,

= 0

thi

limy,, = 0 (vi y„ > 0). Do do

lim(x „ y „ ) = 0 . Gia sik limy,, > 0. Khi do Urn

va diing
>’ 11

limyn

bat dang thtic viia chiing minh ta c6 :
15


Dan b to g xay ra khi ton tai lim x,, va liin y^. Dudi da> la vi du
ii-»x

n->cc
y

-

^

cho dau bat d in g thflc thuc sii:

)i( n+1)

1
x “ = 2 + ( - ! ) " , y" = 2 - ( - I ) " + y ( - l )

2

11(11 + 1 )

2 + (-l)
,
x „y „ = 3 + ----- V ^ . ( - D

Khido:

—

,
^

. va
.1

.

Unix„ =1, lirnx,, =3, limy,,
lim (x„y n ) =

7

, lim y „ - ^

lim (x„y„ )
1

23. Ttf gia thiet va ttf lim

Xi,

suy ra linxx,, = lim x,,. Do
limx.i

do {Xn) hoi tu.
24. Ta c6 ;
0 < x„,1<

Vay

Xn

X,

+...+

X

Xi

= nx,. Vay 0 < —

' bi chan va ton tai a - inf \

X

n

n

< x, (n - 1,2...)

r ■ V 6 i e < 0 nho tuy y,

ni

ton tai m sao cho a <

m

V i moi so nguyen n deii c6 bie'u dien n = qm + r, r - 0, 1, ..., m - 1 .
Do do:
Xn =Xgn5+i. X

n

n

X q n i+ r

qm + r

Xn

16

4 -X[.

<

a -f

qm + r

2J

X

111

m

qm
qm + r

X,
+

n


V i 0 < r < m-1 nen

bi chan v a ton tai N ( e ) sao cho:

0 < — < — khi
n
2
Khi d o a < ^ ^ < a + e khi
n

n > N(e).

n > N(8). Vay l i m ^ ^ = a.
n—>co n

25*. Tt( 4) suy ra c6 N = N ( e ) sao cho n > N ( e ):
x„ - a

Cung tiif 4) ton tai M > 0 sao cho Ix„ I< M, Ix„ - a I< 2M v6i moi
n. Tii 3) suy ra c6 n o =
Pnk <

no(

e

) >

N sao cho:

4NM

khi n > nn. Xet:
n

n

SPnkXk - a

n

n

2 Pnk^k “ Z P n k ^
k=l
k=I

k =l

2 P n k (X k
k=l

-a )

n

^ s Pnk X k - a
k=l

iiN

xn

= P n i X i - a + P n 2 X2 - a

- a + ? n N + l x n +1 - a +...+P nn Xn - a < N - ^ - 2 M +
4 NM
8

8

^(PnN+l + P n N + 2 + " + P n n ) < y + Y = S

Vay:

n>no.

n
hm t„ = Km SPnkXk = a.
n-> oo

n - > o o j^ _ j

26. D^t P,^ = — (k = 1,2,..., n). K hi do cac gia thiet cua dinh ly
n
,

n

Toephtz dudc thoa man v a tn = SPnk^k
k=l

V i du sau cho thay

dieu ngUdc lai khong diing: lay x„ = (-1)"

17


(bcii bai 28)

= lim ix
n->oo y

29. lini
n~>co

X|

X2

X

n—>'cc X jj

x„+, _ 3 + ( - l )
Xet x„ = ------ 771 — . K hi do Pn =
3+ (- l)"
n+1

Ta

c6 ;

X„

p2„ = 7 .
4

n+1

2 (3 + ( - 1 ) " )

n en khong tontj^ iim

P 2 n+i = 1

n->oc

^ii+i
X II

Tuy nhien:

1

lim V
n->^oo

n-><»

1

30*« Ta c6 :

v>n

n

^fa\

Vi

„n

n
j
n—
r = ^Xji,
V n!

_

x^=—
n!

.

\ n-1

X

lim — — = lim

n->^ooXn_j

n

1+

n~>co

= e
n-1

nen theo hki 29, lim —^ = = e .
n-^oo
!
31* Gia sii Urn ^ 5 - ^ ^ = a (a - hiiu han). Coi yo = Xo - 0 va
n-»“ Yn “ yn - 1
dat:
Pnk =

( k = 1,2....,11).

yn

Xn =

Xn

'

Cac gia thiet cua djnh ly Toepitz dUdc thoa man doi v 6 i P^k
n

Xn. Hdn nfla

= SPnkXk =
k=l

18

X
Yn

Do d6 :


X

lim

= lim t „ = lim X,i = lim

n —>cc

N eu

n->ocn-»cc

lim — —
n^coYj^ - Y n - 1

^n+] ^ Xn va lim

11^00

^n-1

a.

n^ooYn - Y n - l

= +00 thi dung lap luan vao day

Yn

vdi

X

= -hx .

31b*. b) D at x„ = (p +l)(lP +2 '’+...+n'^) lim

^

, y„ = (p+DnP. Khi do:

(p + l)(n + l)P ~ (n + l)P^’ +nP^' ^

" ^ “ Yn+l-Yn

(p + l)[(n + l)P - n P ]
+pn^

(P + i)

+

lim

n—>00

(p + 1 ) nP +p„P-> + P ( P - i > n P - 2 + ...+ l-n P
2

+

- n P + ‘ - ( p + l)nP - 5 ^ 5 _ i ) n P - '- . , , - i + nP+’
2
(P + 1) nP + p n P -' + P ( ^ n P - 2 + ...+ l-n P

2

P (P + 1 )

+0
VnV

= lim
n->oo
P (P + 0 + 0

vny

a) va c) diwc chiing minh tiwng til,
32*. Xet:
Xn+i - ^ n = - - ^ - l n ( n + I) + ln n = - ^ - I n ' i . i ' < 0
n+1
n+ 1
V
nJ

n+1

< In

1+

V

vi

nJ

M at khac:
19

V =1

-In n > ]n(l+ 1) +ln
V
n

2 3

l)

+]n

V

.

+...+ln

V

ru

-ln n =

n+ 1
1
3 4 n+1 1
>--- >0
=ln
In Z - . - . . . ----n

n+1
.2 3
n IV
Vay { x j ddn dieu giam va bi ch^n duo’i nen ton tai lim x,i = C.
n->cc
Dodo 1 + —+
2 3

— = C + ln n + e„,en ^ 0
n

----------1------------h...H------“ ^ 2n
n+1

V ay

n+2

Y n + l 2

Sj^

ln 2 + (S 2 n

V ^n yn

suy ra x„ > 0 , > 0 , (n - 1,2...).

~ V^nyn “ ^n+l

— V^n

»yn+1

N h a v4y: X„ < y„ < y i , y„ ^

n -> o o

Q ua gi6 i han cl d in g thfic:
Xn+Yn
Yn+l ;
T a dil0 c A = B.
35b*. D^t an

1+ 2^ +3^+...+n”
n

n

^

j

“

^ 11-

^ Xi . Do do ton tai A

B = lim Y n .

20

luXl

2n

n +2

35*. Ttt Xji =

”

—I n 2 l l + G 2 n

2n

= ln 2 .

lim

n->oo n + 1

va x n + 1

khin->oo.

lim Xn v i

n->co

T a c6;
n" ^
. n' + n^ + n 'V ...+ n '’
— < a „ < — ----------------------n
n”

( n " + '- n ) / ( n - l )

n

n

n“ - l

II

n

n

<

n -1

n
n -1

n

u-^oc

B. GI6 I HAN HAM SO, HAM SO LIEN TUC
36.
B & X ^ 2 uen c6 the' xem 1 < x < 3. K h i do I 3x^
-2 -1 0
3x^ - 12l = 3lx - 2l lx+2l < 1 5 lx -2 l< s .

f e

\

L a y 5 = min — ,1 .
V15 y
37. Coi -2 < X < 0, v a y 1 < x + 3 < 3,
2x + I

I
+ —
X + 3
2

5(x + 1 )
2(x + 3)

5
< — X + 1| < e.

2

L a y 5 = min — ,1
V5

38. Gia sii f bi chan tren Ian can V(xo) cua Xq eR. Bang cach thu
nho V (xo) di den trong Ian can V(xq) khong c6 so hflu ty nao. Dieu do
m a i l thuan v 6 i tinh tru mat cua tap so hflu ty.
39. Cho Xji = - —
2n7t

thi f(x,J - 2 n 7i ->+oo khi x ->+c».

Do do f khong bi chan. Vdi x„ = — — thi f ( x ’ ) = 0.
'
'TT
II
+ n?i

40. D S : ~ m n (n - m )
41. D S : 1

42. DS:

n(n +1)

21


43. DS: —
n
n(n -l)

44. ---------- a
. 2

n-2

45. DS: -

46. DS:

^

4 7 .D a t y = m + x - l

DS:

n

48. Them ya b6t 1 vao tii so, tach thanh hieii h ai gidi han
DS:

a

B

m

n

49. DS:

7
36

50. N han lien hdp va r iit x ra ngoai. DS : ~

51. N h an lien hdp v a rut

2
ra ngoai. DS: -

52. DS = 0
53. Bcii lim — - = 1 vk lim
X—>oc

54. DS:
55. DS: e
56 DS: 1
57. DS; Ve
58. DS: e
59. DS: 1

_|_ I

X -> 0 0 X + I

\ = 1, DS = 1


60. Rut e

va

ra ngoai. DS: —
2

61. Tach 21nx = In x +lnx. Chuyen

d m iu so len mu. DS: - —

62. DS: e'
64*. a) 0<

1

1

X. Jcos —
V
X

<

X

1

COS —
x^o V X

n e n lim x .,

= 0

b) G ia sii sinx > 0. D at a^^ - s i n ...........sin sinx (n da^ii sin) thi
0 < a„ = sinan.i ^ a,,.; < 1

Do do { a j ddn dieu giam va bi chan nen ton tai 1 = lim a „ . Qua
n^oo
gicfi han d in g thiic a^-^ =sina,, ta c6 1 = sinl. V a y 1 = 0.
c)

D at X = 3'

vi y = [y] + r, 0 < r < 1 nen:

lini X

x^O

X

y-> «

y

P
d) N eu X = - ,p, q la cac so' nguyen nguyen to cung nhau thi:

n!

X=

n! - e Z . Do do sin^(n!7ix) = 0.
q

V a y lim sgn [sin ^ ( n ! tix )] = 0.
n—>co
Tru’dng hdp x la s6' v6 ty th i n!x g Z va do do sin^(n!7tx) > 0. V ay
lim sgn[sin^(n!7tx)] = 1.
n-^cc
68 * . Lap lai cach ch iin g m inh cua bai tap 25.
69*. a) Gia sii

lim (f (x + 1) - f(x )) = 1.
X-^+00

D at P n .(x ) = ^ ^ , Pn, (X) =

I

, vdi 0 <

Xo <

x <

Xo+1, Xo>a.
23

u ,( X ) =
'

Un
X

(x) = f(x + n) - f(x + n - 1)

(n = 2,1,...).

+ 1
n
z -

f(x + n)
■■

■

'

k=l

Do do;
lira tn = lim

^ - - = lira [f(x + n) - f(x + n - 1 )] = 1

n—>00

X + n

n—

Do 1 khong phu thupc vao x nen:
f(x)
Um
= lim [ f (x + I) - f(x )] - 1
um ^
X->+CO
x
->+co

X

X—>+<«'

b)
.1

In f(x )

_

In f(x)

—

i i m -----------

lim [f(x )] ^ = lim e
X^+oo

X-> + oo

lim [ l n f ( x + l ) - l T i f ( x ) ]

=

f(x + 1)
= lim — ;-----x-^+=
f(x)

c)Tt( gia thiet v 6 i E>0 tuy y, ton tai

Xo>0

sao cho vdi x>x„:

f(x + l) - fix) > 2 E.
Do do

f(x+n) - f(x) > 2nE va;
f(x + n ) ^ f(x ) + 2 nE
x+n

x+n

Tif f(x) >C>0 khi x„ < x < Xo +1 va do bat dang thiic tren nen:

f ( X + n)
t:, i- ^
—^
------- > E vdi n > n o .
x+n

Do do v 6 i t = x+n, x„ < x < x„ +1, n>n„ thi
f(t)
>E.
t
Vay

f(

lim ------= h-oo .
x - > +o o

X

70*. G ia sft 1 hu:u han. TH gia thiet va dung khai trien Newton t
c6
lim

f(x + n ) - f ( x + n - l )

n ^ ® (x + n )”'+ ‘ - ( x + n - l ) “ ^^

24

_

1 ...

m+1’


Dat:

o

T.
(x +

(X + k )’" " ' - ( X ^ k - 1)-' +'

(

x

+ n) ' - '

(k = 2..1..,n),
()< X o < x < x „ - l,Xo >a,

f ( x + 1)
,„ 4 i • ■ •

u ,(x )=

(x + 1)" '" '

f ( x - n ) - f ( x + 11 - I)
“ "'■ ''^ ^ x + n )'" " ’ - ( x + i i - 1 ) " ^ '

,
ta ctifdc

=

f(x + n)
t „ = -----in+l
(x + n)

D iin g bai tap 68 ta clUdc:
lim tn
- 114-1 = hm ii,i(x)
''II = Im i ---------^
.1
'•Ml''**-/ =

n-»=c

n^cc

x-»co (x + n)

,
ni + 1

Do gidi han khong phu thuoc x nen:
f(x )

lim
11^ 4-0: ^ 111+1

1
ni + 1

Trudng hdp 1=+ co dvWc lap la i nhif bai tap 31.
76. Ta c6;
n '^ -1
-!
lim
n-»cc n 2 x + 1

,

X= 0

f(x ) = 0
lim
U —>00

0 < X < +x

1 -n

~2x

1+ n

-2x

=1

,

- oc < X < 0

Va}^ f(x ) gian doan tai x=0.
77*. N eu X(, la v6 ty thi f(Xo)=0. Cho
la cac so'nguyen difcJng sao cho —
Qi

b> 0.

Ton tai hQu han qi,...,qu

(i = l,2,...,s) . Do do trong Ian

can cua x,, chi c6 hiiu han so" hi5u ty — v6i man so*

(i=l,2,...,s) sao

q

cho I f(x ) I >e vdi x = — . L a y 6 > 0 du nho sao cho (Xo -

Xo + 5) khong

q

25


clilia cac so' hiiu ty -^ v d i q=q; (i = l,s ). K h i do v6i xe(x,, - cS, x,, + 6) t

q

CO I f(x)-f(xn) l<^:. V a y f lien tuc tai x=Xo. Trudng hdpxo = — . f (x o ) = ~
q
q

Co the coi q>0. K h i do vdi moi 6 > 0 va Go< — thi hion c6 diem x v6 t
q
xe(x,i - cS, X,, + 5) ma I f(x)-f(Xo) I = —
q

Do do f gian doan tai diei

hiJu ty.
78 . L a y Xq g R bat ky. N eu fc(Xo) = -C thi nghia la f(X(,)<-C. Do

lien tuc tai X(, nen c6 (x,, -

x„ + 5 ) sao cho f(x)< -C ; V ay f, I

5.

S)

-C va do do hen tuc tai x,„ L y liia n tifdng tu cho f, (Xo) = C. Tn fdn

hdp con la i thi f, (x )= f(x ) nen no hen tuc tai Xi,.
79^. Chi can xet cho m (x) = in f {f(£) : a < ! N eu x„ =a thi m ( X o )= m (a ) =f(a). Do f lien tuc phai tai a nen v6i

>

CO 6 >0 sao cho a < x < a +6 thi
f(a )- r, < f(x )< f(a )+ k
khi do vdi a
ta c6:

m (a)- s < m ix ) < ni(a) + s .
V ay

I m (x)-m (a) | <

h

va m lien tuc phai tai a.

Gia sii asao cho Xo-5 < X < Xo +

thi:

f(xn)- e < f(x ) < f(Xo) + c .
Vdi Xo
b

> m(Xo)-c -

M a t khac vdi t<x„ thi f(t) > m(Xo) > m(Xo) -c
N h it vay v6i moi a
x

< K q + S thi f(t) > m(xo) -

Do do m (x) > m(Xo)- e. R5 ra n g m (x) < m (x j Xom (x)-m (X o)lHdn n^ta tii dinh nghia m(Xo) ta thay vdi e>0 c6 a < Xj < X(, sao c l
f(Xi) < m(xo)+ e. K h i do vdi x, < x < Xq ta c6:
m(Xo)-6 < m (x) < f(Xi) < ni(Xo) +e .
26