Giới hạn dãy số, giới hạn hàm số và hàm số liên tục – Diệp Tuân

Tài liệu gồm 156 trang, được biên soạn bởi thầy giáo Diệp Tuân, phân dạng và hướng dẫn giải các bài tập chuyên đề giới hạn dãy số, giới hạn hàm số và hàm số liên tục (Đại số và Giải tích 11 chương 4).

󰉪󰉗󰉭 󰉼󰉴-Bài 1. 󰉵󰉗 󰉯
1
󰉵󰉚󰉪 Tel: 0935.660.880
4
GI󰉇I H󰈩N
A. 󰈸
I. DÃY S󰉁 CÓ GI󰉇I H󰈩N
0.
1. 󰉬 Ta nói r󰉟ng dãy s󰉯
n
u
có gi󰉵i h󰉗n 0 n󰉦u v󰉵i m󰉭i s󰉯 󰉼󰉴󰉮 bao nhiêu tùy ý cho
󰉼󰉵c, m󰉭i s󰉯 h󰉗ng c󰉻a dãy s󰉯, k󰉨 t󰉾 s󰉯 h󰉗󰉷 󰉧u có giá tr󰉬 tuy󰉪󰉯i nh󰉮 󰉴󰉯
󰉼󰉴
B󰉟ng cách s󰉿 d󰉺ng các kí hi󰉪u toán h󰉭c, 󰉬nh ngha trên có th󰉨 vi󰉦t nh󰉼 sau:
00
lim 0 0, :
nn
u n n n u

.
hi󰉪u:
ho󰉢c
lim 0
n
u
ho󰉢c
0
n
u
󰉺 1. 󰉽h 󰉯
1
45
n
n
u
n
󰉵󰉗
L󰉶i gi󰉘i
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2. Nh󰉝n xét
lim 0 lim 0.
nn
uu
N󰉦u
n
u
0
n
u
,
*
n
thì
lim lim0 0
n
u 
.
Cho hai dãy s󰉯
n
u
n
v
. N󰉦u
lim 0
nn
n
uv
v

thì
lim 0
n
u
.
󰉳󰉝 󰉭󰉨󰉽󰉵󰉗󰉟
0
󰉟󰉬(
󰉵󰉗󰉣
).
3. Các dãy s󰉯 có gi󰉵i h󰉗n
0
󰉼󰉶ng g󰉢p.
1
lim 0
n
1
lim 0
n
0
lim 0
C
n
v󰉵i
C
là h󰉟ng s󰉯
1
lim 0
k
n
n

k
1
lim 0
k
n
2, kk
lim 0
n
q
1.q
4. Ví d󰉺 minh h󰉭a.
󰉺. 󰉽󰉟󰉯󰉵󰉯󰉗󰉱󰉵󰉗
a).
1
32
n
n
u
n
. b).
3
sin2
2
n
nn
u
n
. c).
1 cos
n
n
n
u
n
. d.)
2
3sin 4cos
21
n
nn
u
n
.
L󰉶i gi󰉘i
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 GI󰉇I H󰈩N C󰉍A DÃY S󰉁
󰉪󰉗󰉭 󰉼󰉴-Bài 1. 󰉵󰉗󰉯
2
󰉵󰉚󰉪 Tel: 0935.660.880
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II. DÃY S󰉁 CÓ GI󰉇I H󰈩N H󰉒U H󰈩N.
1. 󰉬 Ta nói dãy s󰉯
n
u
có gi󰉵i h󰉗n là s󰉯 th󰊁c
L
n󰉦u
lim 0
n
uL
.
󰉦t
lim
n
n
uL

, vi󰉦t t󰉞t là
lim
n
uL
ho󰉢c
lim
n
uL
.
Nh󰉝n xét:
󰉨 ch󰉽ng minh dãy s󰉯
n
u
có gi󰉵i h󰉗n là s󰉯 th󰊁c
L
ta chuy󰉨n v󰉧 vi󰉪󰉽ng minh
lim 0
n
uL
.
lim
nn
u a u a
nh󰉮 bao nhiêu cng 󰉼󰉹c v󰉵i
n
󰉻 l󰉵n.
Không ph󰉘i m󰉭i dãy s󰉯 󰉧u có gi󰉵i h󰉗n h󰊀u h󰉗n.
󰉺󰉽󰉟
a).
3
3
lim 1
1
n
n




. b).
2
2
3 2 1
lim
22
nn
nn




.
L󰉶i gi󰉘i
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󰉺󰉽󰉟
a).
3.3 sin3
lim 3
3
n
n
n



. b).
2
1
lim
2
n n n
.
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 1. 󰉵󰉗 󰉯
3
󰉵󰉚󰉪 Tel: 0935.660.880
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2. M󰉳t s󰉯 󰉬nh lý.
󰉬nh 1.
( tìm gi󰉵i h󰉗n c󰉻a hàm tr󰉬 tuy󰉪󰉯i ho󰉢󰉽c)
Gi󰉘 s󰉿
lim
n
uL

3
3
lim
n
uL
.
N󰉦u
0
n
u
v󰉵i m󰉭i
n
thì
0L
lim
n
uL
.
󰉬nh 2. Gi󰉘 s󰉿
lim
n
uL
,
lim
n
vM
C
là m󰉳t h󰉟ng s󰉯
lim
nn
u v L M
.
lim . .
nn
u v L M
lim
n
Cu CL
.
lim
n
n
u
L
vM



v󰉵i
0M
.
limcc
(c là h󰉟ng s󰉯).
Nh󰉝n xét.
Cho ba dãy s󰉯
,
nn
uv
n
w
. N󰉦u
,
n n n
u v w n
lim lim ,
nn
u w a a
thì
lim
n
va
(g󰉭i 󰉬nh lí k󰉣p).
i󰉧u ki󰉪n 󰉨 m󰉳t dãy s󰉯 tng ho󰉢c dãy s󰉯 gi󰉘m có gi󰉵i h󰉗n h󰊀u h󰉗n:
M󰉳t dãy s󰉯 tng và b󰉬 ch󰉢n trên thì có gi󰉵i h󰉗n h󰊀u h󰉗n.
M󰉳t dãy s󰉯 gi󰉘m và b󰉬 ch󰉢n d󰉼󰉵i thì có gi󰉵i h󰉗n h󰊀u h󰉗n.
3. T󰉱ng c󰉻a c󰉙p s󰉯 nhâni vô h󰉗n
Cho c󰉙p s󰉯 nhân
n
u
có công b󰉳i
q
và th󰉮a
1q
.
Khi ó t󰉱ng
1 2 3 n
S u u u u
󰉼󰉹c g󰉭i là t󰉱ng vô h󰉗n c󰉻a c󰉙p s󰉯 nhân và
1
1
1
lim lim
11
n
n
uq
u
SS
qq

.
V󰉝y c󰉙p s󰉯 nhân
n
u
có công b󰉳i
q
th󰉮a mãn
1q
thì
1
12
...
1
u
S u u
q
.
󰉺󰉱
a).
2
1 1 1
... ...
3 3 3
n
S
b).
1 1 1
1 1 .
2 4 2
n
n
S
c).
16 8 4 2 ...S
L󰉶i gi󰉘i
󰉪󰉗󰉭 󰉼󰉴-Bài 1. 󰉵󰉗󰉯
4
󰉵󰉚󰉪 Tel: 0935.660.880
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󰉺. 󰉨󰉩󰉯󰉝󰉗󰉚󰉼󰉵󰉗󰉯
a).
0,353535...A
. b).
5,231231...B
.
L󰉶i gi󰉘i
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III. DÃY S󰉁 CÓ GI󰉇I H󰈩N VÔ C󰉓C
1. Dãy s󰉯 có gi󰉵i h󰉗n

󰉬 Ta nói r󰉟ng dãy s󰉯
n
u
có gi󰉵i h󰉗n là

n󰉦u v󰉵i m󰉲i s󰉯 󰉼󰉴󰉼󰉵c, m󰉭i
s󰉯 h󰉗ng c󰉻a dãy s󰉯, k󰉨 t󰉾 s󰉯 h󰉗󰉷 󰉧u l󰉵󰉴󰉯 󰉼󰉴 
󰉦t
lim
n
u
ho󰉢c
lim
n
u 
.
T󰉾 󰉬󰉦t qu󰉘
3
lim ; lim ; limn n n
.
2. Dãy s󰉯 có gi󰉵i h󰉗n

󰉬 Ta nói r󰉟ng dãy s󰉯
n
u
gi󰉵i h󰉗n là

n󰉦u v󰉵i m󰉲i s󰉯 󰉼󰉵c, m󰉭i s󰉯
h󰉗ng c󰉻a dãy s󰉯, k󰉨 t󰉾 s󰉯 h󰉗󰉷 󰉧u nh󰉮 󰉴󰉯 âm 
󰉦t
lim
n
u
ho󰉢c
lim
n
u 
.
Nh󰉝n xét.
N󰉦u
lim
n
u 
thì
lim
n
u 
.
󰉪󰉗󰉭 󰉼󰉴-Bài 1. 󰉵󰉗 󰉯
5
󰉵󰉚󰉪 Tel: 0935.660.880
c dãy s󰉯 có gi󰉵i h󰉗n

ho󰉢c

󰉼󰉹c g󰉭i chung là các dãy s󰉯 có gi󰉵i h󰉗n vô c󰊁c hay d󰉚n
󰉦n vô c󰊁c.
N󰉦u
lim
n
u 
thì
1
lim 0
n
u
.
3. Các quy t󰉞c m gi󰉵i h󰉗n vô c󰊁c
Quy t󰉞c nhân:
lim
n
u
lim
n
v
lim .
nn
uv
lim
n
u
lim 0
n
vL
lim .
nn
uv




















Quy t󰉞c chia
lim 0
n
uL
có d󰉙u
lim 0, 0
nn
vv
có d󰉙u
lim
n
n
u
v




4. Các ví d󰉺 minh h󰉭a.
󰉺. Tìm gi󰉵i h󰉗n c󰉻a dãy
n
u
bi󰉦t:
a).
32
3 2 2
n
u n n
b).
43
23
n
u n n n
c).
43
2
4 2 1
1
n
nn
u
n

d).
1
4 2. 3
n
n
n
u
e).
2
2
cos 4
10
n
n
u n n

f).
4 2 2
3
4 1 2
32
n
n n n
u
n n n

.
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 1. 󰉵󰉗󰉯
6
󰉵󰉚󰉪 Tel: 0935.660.880
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B. PN D󰈩NG VÀ BÀI T󰈯P MINH H󰈿A.
󰉗󰉽󰉯󰉵󰉗0.
1. 󰉼󰉴
Cách 1: Áp d󰉺󰉬
Cách 2: S󰉿 d󰉺󰉬nh lí sau:
N󰉦u
k
là s󰉯 th󰊁󰉼󰉴
1
lim 0
k
n
.
V󰉵i hai dãy s󰉯
n
u
n
v
, n󰉦u
nn
uv
v󰉵i m󰉭i
n
lim 0
n
v
thì
lim 0
n
u
.
N󰉦u
1q
thì
lim 0
n
q
.
2. 󰉝󰉭
󰉝 󰉽󰉯
n
u
󰉵󰉗
a).
cos4
3
n
n
u
n
b).
3
1 cos
23
n
n
u
n
c).
11
1
1
23
n
n
nn
u


L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 1. 󰉵󰉗 󰉯
7
󰉵󰉚󰉪 Tel: 0935.660.880
󰉝 󰉽 󰉟󰉯󰉵󰉯󰉗󰉱󰉵󰉗
a).
33
21
n
u n n
. b).
3 sin 2 4
2 4.5
nn
n
nn
n
u
.
c).
2
sin2
n
nn
nn
u
. d).
cos
5
n
n
n
u
n n n
.
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 1. 󰉵󰉗󰉯
8
󰉵󰉚󰉪 Tel: 0935.660.880
󰉝 󰉯
n
u
󰉵
3
n
n
n
u
.
a). 󰉽󰉟
1
2
3
n
n
u
u
󰉵󰉭
*
n
.
b). 󰉟󰉼󰉴󰉗 󰉽󰉟
2
0
3
n
n
u




󰉵󰉭
*n
.
c). Dãy
n
u
󰉵󰉗
L󰉶i gi󰉘i
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3. 󰉮󰉞󰉪.
Câu 1. 󰉦󰉘󰉻󰉵󰉗
sin5
lim 2
3
n
n



󰉟
A.
2.
B. 3. C. 0. D.
5
.
3
󰉶󰉘
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󰉪󰉗󰉭 󰉼󰉴-Bài 1. 󰉵󰉗 󰉯
9
󰉵󰉚󰉪 Tel: 0935.660.880
Câu 2. 󰉯󰊁󰉡
k
󰉨
1
2 cos
1
lim .
22
k
nn
n
n
A.
0.
B. 1. C.
4.
D. 󰉯
󰉶󰉘
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Câu 3. 󰉦󰉘󰉻󰉵󰉗
3sin 4cos
lim
1
nn
n
󰉟
A.
1
. B.
0
. C.
2
. D.
3
.
󰉶󰉘
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Câu 4. 󰉦󰉘󰉻󰉵󰉗
2
cos2
lim 5
1
nn
n



󰉟
A.
4
. B.
1
.
4
C.
5
. D.
4.
󰉶󰉘
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Câu 5. 󰉦󰉘󰉻󰉵󰉗
23
lim sin 2
5
n
nn



là:
A.
.
B.
2.
C.
0
. D.
.
󰉶󰉘
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Câu 6. 󰉬󰉻󰉵󰉗
1
lim 4
1
n
n




󰉟
A.
1
. B.
3
. C.
4
. D.
2
.
󰉪󰉗󰉭 󰉼󰉴-Bài 1. 󰉵󰉗󰉯
10
󰉵󰉚󰉪 Tel: 0935.660.880
󰉶󰉘
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Câu 7. 󰉯
n
u
n
v
2
1
1
n
n
u
n
2
1
.
2
n
v
n

lim
nn
uv
󰉬
󰉟
A.
3
. B.
0
. C.
2
. D.
1
.
󰉶󰉘
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󰉗󰉬󰉽󰉯
n
u
󰉵󰉗󰊀󰉗
L
.
1. 󰉼󰉴
Ta bi󰉦󰉱i
lim
n
uL
v󰉧 d󰉗ng tìm lim có gi󰉵i h󰉗n b󰉟ng
0
t󰉽c là ch󰉽ng minh
lim 0
n
uL
.
K󰉦t lu󰉝n
lim
n
uL
.
2. 󰉝󰉭
󰉝 󰉽
a).
2 3 1
lim
4 5 2
n
n
b).
4.3 5.2 2
lim
6.3 3.2 3
nn
nn
c).
2
lim 2 1n n n
.
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 1. 󰉵󰉗 󰉯
11
󰉵󰉚󰉪 Tel: 0935.660.880
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󰉗m gi󰉵i h󰉗n c󰉻a dãy
n
u
có gi󰉵i h󰉗n h󰊀u h󰉗󰉟󰉞󰉬
Bài toán 1. Dãy
n
u
là m󰉳t phân th󰉽c h󰊀u t󰉫 d󰉗ng
n
Pn
u
Qn
(v󰉵i
,P n Q n
là hai a th󰉽c).
1. Ph󰉼󰉴ng pháp:
Chia c󰉘 t󰉿 m󰉜u cho
k
n
v󰉵i
k
n
ly th󰉾a s󰉯 m l󰉵n nh󰉙t c󰉻a
Pn
Qn
(ho󰉢c rút
k
n
là ly th󰉾a có s󰉯 m l󰉵n nh󰉙t c󰉻a
Pn
Qn
ra làm nhân t󰉿) sau ó áp d󰉺ng các 󰉬nh
lý v󰉧 gi󰉵i h󰉗n.
N󰉦u
k
là s󰉯 th󰊁󰉼󰉴
1
lim 0
k
n
.
N󰉦u
1q
thì
lim 0
n
q
.
2. 󰉝󰉭
󰉝. Tìm gi󰉵i h󰉗n c󰉻a dãy
n
u
bi󰉦t:
a).
2
2
2 3 1
53
n
nn
u
n

b).
32
43
2 3 4
4
n
nn
u
n n n

c).
42
2
23
2 1 1 3 2 1
n
n n n
u
n n n

d).
22
11
2 2 3
n
u
n n n


e).
2
3
32
2 1 3 4
4 2 2
n
nn
u
nn


f).
2
21
23
n
nn
u
nn

L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 1. 󰉵󰉗󰉯
12
󰉵󰉚󰉪 Tel: 0935.660.880
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󰉪󰉗󰉭 󰉼󰉴-Bài 1. 󰉵󰉗 󰉯
13
󰉵󰉚󰉪 Tel: 0935.660.880
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󰉝. 󰉵󰉗
a).
2
2
42
lim
21
nn
nn

. b).
2
22
31
lim 2 1
2 3 1
n
n n n n




.
L󰉶i gi󰉘i
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4. 󰉮󰉞󰉪.
Câu 8. 󰉬󰉻󰉵󰉗
2
3
lim
4 2 1nn

là:
A.
3
.
4
B.
.
C.
0
. D.
1.
󰉶󰉘
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Câu 9. 󰉬󰉻󰉵󰉗
2
3
2
lim
31
nn
nn

󰉟
A.
2
. B.
1
. C.
2
.
3
D.
0
.
󰉶󰉘
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󰉪󰉗󰉭 󰉼󰉴-Bài 1. 󰉵󰉗󰉯
14
󰉵󰉚󰉪 Tel: 0935.660.880
Câu 10. 󰉬󰉻󰉵󰉗
3
4
3 2 1
lim
4 2 1
nn
nn


là:
A.
.
B. 0. C.
2
.
7
D.
3
.
4
󰉶󰉘
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Câu 11. 󰉬󰉻󰉵󰉗
󰉟
A.
3
.
2
B. 2. C. 1. D. 0.
󰉶󰉘
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Câu 12. 󰉯
n
u
n
v
1
1
n
u
n
2
.
2
n
v
n

lim
n
n
v
u
󰉬󰉟
A. 1. B. 2. C. 0. D. 3.
󰉶󰉘
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Câu 13. 󰉯
n
u
󰉵
4
53
n
an
u
n

a
󰉯󰊁󰉨󰉯
n
u
󰉵󰉗
󰉟
2
󰉬󰉻
a
là:
A.
10.a
B.
8.a
C.
6.a
D.
4.a
󰉶󰉘
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Câu 14. 󰉯
n
u
󰉵
2
53
n
nb
u
n

b
󰉯󰊁󰉨󰉯
n
u
󰉵󰉗
󰊀󰉗󰉬󰉻
b
là:
A.
b
󰉳󰉯󰊁 B.
2.b
C. 󰉰󰉗
.b
D.
5.b
󰉶󰉘
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󰉪󰉗󰉭 󰉼󰉴-Bài 1. 󰉵󰉗 󰉯
15
󰉵󰉚󰉪 Tel: 0935.660.880
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Câu 15. 󰉵󰉗
2
2
5
lim .
21
nn
L
n

A.
3
.
2
L
B.
1
.
2
L
C.
2.L
D.
1.L
󰉶󰉘
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Câu 16. 󰉯
n
u
󰉵
2
2
42
.
5
n
nn
u
an

󰉨󰉯󰉵󰉗󰉟
2
󰉬󰉻
a
là:
A.
4.a 
B.
4.a
C.
3.a
D.
2.a
󰉶󰉘
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Câu 17. 󰉵󰉗
23
3
3
lim .
2 5 2
nn
L
nn

A.
3
.
2
L 
B.
1
.
5
L
C.
1
.
2
L
D.
0.L
󰉶󰉘
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Câu 18. 󰉙󰉘󰉬󰉻󰉯
a
󰉨
24
4
53
lim 0.
1 2 1
n an
L
a n n

A.
0; 1.a a
B.
0 1.a
C.
0; 1.aa
D.
0 1.a
󰉶󰉘
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Câu 19. 󰉵󰉗
32
4
2 3 1
lim .
2 1 7
n n n
L
nn


󰉪󰉗󰉭 󰉼󰉴-Bài 1. 󰉵󰉗󰉯
16
󰉵󰉚󰉪 Tel: 0935.660.880
A.
3
.
2
L 
B.
1.L
C.
3.L
D.
.L 
󰉶󰉘
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Câu 20. 󰉵󰉗
23
42
2 2 1 4 5
lim .
3 1 3 7
n n n n
L
n n n
A.
0.L
B.
1.L
C.
8
.
3
L
D.
.L
󰉶󰉘
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Câu 21. 󰉵󰉗
3
3
1
lim .
8
n
L
n
A.
1
.
2
L
B.
1.L
C.
1
.
8
L
D.
.L
󰉶󰉘
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Bài Toán 2. Dãy
n
u
là m󰉳t phân th󰉽c d󰉗ng
n
Pn
u
Qn
(v󰉵i
,P n Q n
các bi󰉨u th󰉽c ch󰉽a cn
c󰉻a
n
).
1. 󰉼󰉴
󰉼󰉵c 1. Chia c󰉘 t󰉿 m󰉜u cho
k
n
v󰉵i
k
n
ly th󰉾a s󰉯 m l󰉵n nh󰉙t c󰉻a
Pn
Qn
(ho󰉢c rút
k
n
ly th󰉾a s󰉯 m l󰉵n nh󰉙t c󰉻a
Pn
Qn
ra làm nhân t󰉿) sau ó áp d󰉺ng
các 󰉬nh lý v󰉧 gi󰉵i h󰉗n.
󰉪󰉗󰉭 󰉼󰉴-Bài 1. 󰉵󰉗 󰉯
17
󰉵󰉚󰉪 Tel: 0935.660.880
󰉼󰉵c 2. 󰉦t qu󰉘 sau
N󰉦u
lim lim
n
Pn
uC
Qn
gi󰉵i h󰉗n c󰉻a dãy s󰉯
.C
N󰉦u
0
lim lim
n
Pn
u
Qn


thì ta nói gi󰉵i h󰉗󰉗󰉬nh.
󰉨 tính ti󰉦p gi󰉵i h󰉗n ta ph󰉘i kh󰉿 d󰉗󰉬nh b󰉟ng các k󰊄 thu󰉝t sau:
󰉯i v󰉵󰉽c: 󰉼󰉹ng liên h󰉹p c󰉻a b󰉝c hai và b󰉝󰇛󰇜
2
a b a b
ab
ab
a b a b


2
a b a b
ab
ab
a b a b


2
2
3 3 3
3
3
22
22
3 3 3 3
.
..
a b a a b b
ab
ab
a a b b a a b b



.
2
2
3 3 3
3
3
22
22
3 3 3 3
.
..
a b a a b b
ab
ab
a a b b a a b b



2
2
3 3 3
3
3
22
22
3 3 3 3
.
..
a b a a b b
ab
ab
a a b b a a b b



2
2
3 3 3
3
3
22
22
3 3 3 3
.
..
a b a a b b
ab
ab
a a b b a a b b



22
3 3 3 3 3 3
33
2 2 2 2
3 3 3 3 3 3 3 3
.
..
a b a a b b
ab
ab
a a b b a a b b



.
22
3 3 3 3 3 3
33
2 2 2 2
3 3 3 3 3 3 3 3
.
..
a b a a b b
ab
ab
a a b b a a b b



󰉯i v󰉵i 󰉽c: 󰉢t nhân t󰉿 chung b󰉟󰉼󰉴th󰉽c thành
nhân t󰉿.
Th󰉾a s󰉯 chung.
H󰉟󰉠ng th󰉽󰉵.
.
Nhóm, tách và thêm b󰉵t h󰉗ng t󰉿.
Nh󰉵:
N󰉦u
2
ax bx c
có hai nghi󰉪m
12
,xx
thì
2
12
ax bx c a x x x x
N󰉦u b󰉝c ba
32
ax bx cx d

0
xx
r󰉰󰉽󰉨 󰉼󰉧
32
0
ax bx cx d x x f x
, v󰉵i
fx
là hàm b󰉝c hai.
2. 󰉝󰉭
󰉝. Tìm gi󰉵i h󰉗n c󰉻a dãy
n
u
bi󰉦t:
󰉪󰉗󰉭 󰉼󰉴-Bài 1. 󰉵󰉗󰉯
18
󰉵󰉚󰉪 Tel: 0935.660.880
a).
2
2
41
93
n
n n n
u
nn
b).
2 1 3
45
n
nn
u
n
c).
3
2 3 2
24
4
4 1 8 2 3
16 4 1
n
n n n
u
n n n
d).
3
23
4
4
3
16 1
n
n n n n
u
n
L󰉶i gi󰉘i
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󰉝. Tìm gi󰉵i h󰉗n c󰉻a dãy
n
u
bi󰉦t:
a).
2
35
n
u n n n
b).
2
9 3 4 3 2
n
u n n n
c).
3
32
3
n
u n n n
d).
3
32
8 4 2 2 3
n
u n n n
e).
3
2 3 2
4 3 7 8 5 1
n
u n n n n
f).
3
4 2 6
11
n
u n n n
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 1. 󰉵󰉗 󰉯
19
󰉵󰉚󰉪 Tel: 0935.660.880
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󰉪󰉗󰉭 󰉼󰉴-Bài 1. 󰉵󰉗󰉯
20
󰉵󰉚󰉪 Tel: 0935.660.880
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Bài 󰉝. 󰉵󰉗
a).
2
2
lim
4 3 2
n n n
n n n


b).
2
3
23
24
4
n
n n n
u
n n n


c).
22
lim 2 9 2n n n n n
d).
3
2 2 3 2
lim 2 2 8 3n n n n n n
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 1. 󰉵󰉗 󰉯
21
󰉵󰉚󰉪 Tel: 0935.660.880
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󰉪󰉗󰉭 󰉼󰉴-Bài 1. 󰉵󰉗󰉯
22
󰉵󰉚󰉪 Tel: 0935.660.880
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󰉝. 󰉵󰉗
a).
2
9 2 3
lim
43
n n n
n

. b).
5
4
5
4
3 4 2
lim
23
nn
nn

. c).
2
lim 4 2 2n n n
.
d).
3
3
lim 2 1 .n n n
e).
3
26
42
1
lim
1
nn
nn


f).
3
2 2 3
lim 2 3n n n n
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 1. 󰉵󰉗 󰉯
23
󰉵󰉚󰉪 Tel: 0935.660.880
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5. 󰉮󰉞󰉪.
Câu 22. 󰉦󰉘󰉻󰉵󰉗
2
91
lim
42
nn
n

󰉟
A.
2
.
3
B.
3
.
4
C. 0. D. 3.
󰉶󰉘
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Câu 23. 󰉦󰉘󰉻󰉵󰉗
2
4
21
lim
32
nn
n
󰉟
A.
2
.
3
B.
1
.
2
C.
3
.
3
D.
1
.
2
󰉶󰉘
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󰉪󰉗󰉭 󰉼󰉴-Bài 1. 󰉵󰉗󰉯
24
󰉵󰉚󰉪 Tel: 0935.660.880
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Câu 24. 󰉦󰉘󰉻󰉵󰉗
23
lim
25
n
n
là:
A.
5
.
2
B.
5
.
7
C.
.
D.
1.
󰉶󰉘
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Câu 25. 󰉦󰉘󰉻󰉵󰉗
14
lim
1
n
nn


󰉟
A. 1. B. 0. C.
1.
D.
1
.
2
󰉶󰉘
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Câu 26. 󰉦󰉟
2
2
1
lim sin .
4
2
nn
ab
nn



Tính
33
.S a b
A.
1.S
B.
8.S
C.
0.S
D.
1.S 
󰉶󰉘
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Câu 27. 󰉦󰉘󰉻󰉵󰉗
42
10
lim
1nn
là:
A.
.
B. 10. C. 0. D.
.
󰉶󰉘
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Câu 28. 󰉦󰉘󰉻󰉵󰉗
42
22
lim 1
1
n
n
nn

là:
A.
.
B. 1. C. 0. D.
.
󰉪󰉗󰉭 󰉼󰉴-Bài 1. 󰉵󰉗 󰉯
25
󰉵󰉚󰉪 Tel: 0935.660.880
󰉶󰉘
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Câu 29. 󰉦󰉟
3
32
2
57
lim 3
32
an n
bc
nn



󰉵
,,abc
󰉯󰉬󰉻󰉨󰉽
3
.
ac
P
b
A.
3.P
B.
1
.
3
P
C.
2.P
D.
1
.
2
P
󰉶󰉘
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Câu 30. 󰉦󰉘󰉻󰉵󰉗
5
52
lim 200 3 2nn
là:
A.
.
B. 1. C. 0. D.
.
󰉶󰉘
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Câu 31. 󰉬󰉻󰉵󰉗
lim 5 1nn
󰉟
A.
0.
B.
1.
C.
3.
D.
5.
󰉶󰉘
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Câu 32. 󰉬󰉻󰉵󰉗
2
lim 1n n n
là:
A.
1
.
2
B.
0.
C.
1.
D.
.
󰉶󰉘
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󰉪󰉗󰉭 󰉼󰉴-Bài 1. 󰉵󰉗󰉯
26
󰉵󰉚󰉪 Tel: 0935.660.880
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Câu 33. 󰉬󰉻󰉵󰉗
22
lim 1 3 2nn
là:
A.
2.
B.
0.
C.
.
D.
.
󰉶󰉘
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Câu 34. 󰉬󰉻󰉵󰉗
22
lim 2 2n n n n
là:
A.
1.
B.
2.
C.
4.
D.
.
󰉶󰉘
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Câu 35. 󰉬󰉻
a
󰉨
2 2 2
lim 2 1 0.n a n n a n
A.
0.
B. 2. C.
1.
D. 3.
󰉶󰉘
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Câu 36. 󰉬󰉻󰉵󰉗
22
lim 2 1 2 3 2n n n n
là:
A.
0.
B.
2
.
2
C.
.
D.
.
󰉶󰉘
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󰉪󰉗󰉭 󰉼󰉴-Bài 1. 󰉵󰉗 󰉯
27
󰉵󰉚󰉪 Tel: 0935.660.880
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Câu 37. 󰉬󰉻󰉵󰉗
22
lim 2 1 2n n n n
là:
A.
1.
B.
1 2.
C.
.
D.
.
󰉶󰉘
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Câu 38. 󰉬󰉻
a
󰉮
22
lim 8 0n n n a
.
A.
0.
B. 2. C. 1. D. 󰉯
󰉶󰉘
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Câu 39. 󰉬󰉻󰉵󰉗
2
lim 2 3n n n
là:
A.
1.
B.
0.
C.
1.
D.
.
󰉶󰉘
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󰉪󰉗󰉭 󰉼󰉴-Bài 1. 󰉵󰉗󰉯
28
󰉵󰉚󰉪 Tel: 0935.660.880
Câu 40. 󰉯
n
u
󰉵
22
51
n
u n an n

a
󰉯󰊁
a
󰉨
A.
3.
B.
2.
C.
2.
D.
3.
󰉶󰉘
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Câu 41. 󰉬󰉻󰉵󰉗
33
33
lim 1 2nn
󰉟
A.
3.
B.
2.
C.
0.
D.
1.
󰉶󰉘
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Câu 42. 󰉬󰉻󰉵󰉗
3
23
lim n n n
là:
A.
1
.
3
B.
.
C.
0.
D.
1.
󰉶󰉘
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Câu 43. 󰉬󰉻󰉵󰉗
3
32
lim 2n n n
󰉟
A.
1
.
3
B.
2
.
3
C.
0.
D.
1.
󰉶󰉘
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󰉪󰉗󰉭 󰉼󰉴-Bài 1. 󰉵󰉗 󰉯
29
󰉵󰉚󰉪 Tel: 0935.660.880
Câu 44. 󰉬󰉻󰉵󰉗
lim 1 1n n n


là:
A.
1.
B.
.
C.
0.
D.
1.
󰉶󰉘
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Câu 45. 󰉬󰉻󰉵󰉗
lim 1n n n



󰉟
A.
0.
B.
1
.
2
C.
1
.
3
D.
1
.
4
󰉶󰉘
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Câu 46. 󰉬󰉻󰉵󰉗
22
lim 1 3n n n



󰉟
A.
1.
B.
2.
C.
4.
D.
.
󰉶󰉘
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Câu 47. 󰉬󰉻󰉵󰉗
22
lim 1 6n n n n n



là:
A.
7 1.
B.
3.
C.
7
.
2
D.
.
󰉶󰉘
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󰉪󰉗󰉭 󰉼󰉴-Bài 1. 󰉵󰉗󰉯
30
󰉵󰉚󰉪 Tel: 0935.660.880
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Câu 48. 󰉬󰉻󰉵󰉗
2
1
lim
24nn
là:
A.
1.
B.
0.
C.
.
D.
.
󰉶󰉘
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Câu 49. 󰉬󰉻󰉵󰉗
2
92
lim
32
n n n
n
là:
A.
1.
B.
0.
C.
3.
D.
.
󰉶󰉘
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Câu 50. 󰉬󰉻󰉵󰉗
3
3
1
lim
1nn
là:
A.
2.
B.
0.
C.
.
D.
.
󰉶󰉘
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương IV-Bài 1. Giới Hạn y s
31
Lớp Toán Thầy–Diệp Tuân Tel: 0935.660.880
Bài Toán 3. Dãy
n
u
là mt phân thc hu t dng
n
Pn
u
Qn
( trong đó
,P n Q n
là các biu
thc chứa m
,,
n n n
abc
,…
1. Phương pháp.
Chia c t và mu cho
n
a
vi
a
là cơ số ln nht .
Ri áp dng kết qu ca gii hn
lim 0
n
q
1.q
2. i tp minh họa.
Bài tập 11. Tìm gii hn ca dãy
n
u
biết:
a).
24
43
nn
n
nn
u
b).
3.2 5
5.4 6.5
nn
n
nn
u
c).
21
13
46
5 2.6
nn
n
nn
u


d).
2
2
21
32
n
n
n
u
e).
1
3 4.5
2.4 3.5
n
n
n
nn
u

f).
2
1 2 1
2 3 4.5
2 3 5
n n n
n
n n n
u


g).
11
2 3 4
lim
2 3 4
n n n
n n n


h).
2
2
1 2 2 ... 2
lim
1 3 3 ... 3
n
n
Li gii
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Trung Tâm Luyện Thi Đại Học Amsterdam Cơng IV. Giới Hạn Dãy Số
32
Lớp Toán Thầy Diệp Tuân Tel: 0935.660.880
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3. u hỏi trắc nghiệm.
Câu 51. Kết quả của giới hạn
2
25
lim
3 2.5
n
nn
bằng:
A.
25
.
2
B.
5
.
2
C.
1.
D.
5
.
2
Lời giải.
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Câu 52. Kết quả của giới hạn
1
1
3 2.5
lim
25
nn
nn
bằng:
A.
15.
B.
10.
C.
10.
D.
15.
Lời giải.
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương IV-Bài 1. Giới Hạn y s
33
Lớp Toán Thầy–Diệp Tuân Tel: 0935.660.880
Câu 53. Kết quả của giới hạn
1
3 4.2 3
lim
3.2 4
nn
nn

là:
A.
0.
B.
1.
C.
.
D.
.
Lời giải.
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Câu 54. Kết quả của giới hạn
31
lim
2 2.3 1
n
nn

bằng:
A.
1.
B.
1
.
2
C.
1
.
2
D.
3
.
2
Lời giải.
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Câu 55. Biết rằng
1
2
1
2
5 2 1
2 3 5
lim
1
5.2 5 3
n
n
n
n
na
c
nb







với
,, .abc
Tính gtrcủa biểu
thức
2 2 2
.S a b c
A.
26.S
B.
30.S
C.
21.S
D.
31.S
Lời giải.
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Câu 56. Kết quả của giới hạn
2
22
32
lim
3 3 2
n n n
n n n


là:
A.
1.
B.
1
.
3
C.
.
D.
1
.
4
Lời giải.
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Trung Tâm Luyện Thi Đại Học Amsterdam Cơng IV. Giới Hạn Dãy Số
34
Lớp Toán Thầy Diệp Tuân Tel: 0935.660.880
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Câu 57. Kết quả của giới hạn
lim 3 5
n
n



là:
A.
3.
B.
5.
C.
.
D.
.
Lời giải.
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Câu 58. Kết quả của giới hạn
41
lim 3 .2 5.3
nn
là:
A.
2
.
3
B.
1.
C.
.
D.
1
.
3
Lời giải.
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Câu 59. Kết quả của giới hạn
1
3 4.2 3
lim
3.2 4
nn
n
n

là:
A.
0.
B.
1.
C.
.
D.
.
Lời giải.
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Câu 60. Kết quả của giới hạn
1
2
2 3 10
lim
32
n
n
nn


là:
A.
.
B.
2
.
3
C.
3
.
2
D.
.
Lời giải.
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Câu 61. Tìm tất cả giá trị nguyên của
a
thuộc
0;2018
để
1
4
.
42
l
4
im
23
1
04 1
nn
n n a
A.
2007.
B.
2008.
C.
2017.
D.
2016.
Lời giải.
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Câu 62. Kết quả của giới hạn
2
1
2
lim
3 1 3
n
n
nn
n




bằng:
A.
2
.
3
B.
1.
C.
1
.
3
D.
1
.
3
Lời giải.
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Câu 63. Kết quả của giới hạn
3 1 cos3
lim
1
n
nn
n





bằng:
A.
3
.
2
B.
3.
C.
5.
D.
1.
Lời giải.
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Câu 64. bao nhiêu giá trị nguyên của
a
thuộc
0;20
sao cho
2
2
11
lim 3
32
n
an
n

một số
nguyên.
A.
1.
B.
3.
C.
2.
D.
4.
Trung Tâm Luyện Thi Đại Học Amsterdam Cơng IV. Giới Hạn Dãy Số
36
Lớp Toán Thầy Diệp Tuân Tel: 0935.660.880
Lời giải.
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Câu 65. Kết quả của giới hạn
lim 2.3 2
n
n
là:
A.
0.
B.
2.
C.
3.
D.
.
Lời giải.
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Dạng 4. nh giới hạn mà y
n
u
cho ới dạng công thức truy hồi.
1. Phương pháp.
Đưa dãy số
n
u
v dng tng quát ri làm giống như ba dạng trên.
T dãy cho dưới dng truy hi ta công thc tuy hồi ta đưa về công thc tng quát.
Chng minh dãy s gii hn hu hn (có nghĩa chng minh dãy s tăng b chn trên
hoc dãy s gim và b chn dưới) sau đó da vào h thc truy hi để tìm gii hn.
2. i tp minh họa.
Bài tập 12. Tìm gii hn ca dãy
n
u
biết:
a).
1 1 1
1.2 2.3 ( 1)
n
u
nn
b).
2 2 2 2
1 2 3
( 1)( 2)
n
n
u
n n n

c).
1 1 1 1
1.4 4.7 7.10 (3 2)(3 1)
n
u
nn

d).
2
1 3 5 (2 1)
34
n
n
u
n
e).
2 2 2
1 1 1
1 1 ... 1
23
n
u
n
f).
3 3 3
43
12
lim
41
n
nn

g).
1 1 1
lim
1.2.3 2.3.4 ( 1)( 2)n n n




h).
2 2 2 2
4
2.1 3.2 ( 1) ( 1)
lim
n n n n
n



k).
1 1 1
lim
1.3 3.5 2 1 2 1nn





Li gii
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương IV-Bài 1. Giới Hạn y s
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Lớp Toán Thầy–Diệp Tuân Tel: 0935.660.880
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Trung Tâm Luyện Thi Đại Học Amsterdam Cơng IV. Giới Hạn Dãy Số
38
Lớp Toán Thầy Diệp Tuân Tel: 0935.660.880
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Bài tập 14. Cho dãy s
n
u
xác định bởi
01
21
1; 6
3 2 0,
n n n
uu
u u u n


. Tìm
lim
3.2
n
n
u
.
Li gii
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Bài tập 15. Cho dãy s
n
u
được xác định như sau:
12
21
1, 3
2 1, *
n n n
uu
u u u n


. Tính
2
lim
n
u
n
.
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39
Lớp Toán Thầy–Diệp Tuân Tel: 0935.660.880
Li gii
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4. u hỏi trắc nghiệm.
Câu 66. Giá trị của giới hạn
2
13
1 ...
2 2 2
lim
1
n
n
bằng:
A.
1
.
8
B.
1.
C.
1
.
2
D.
1
.
4
Lời giải.
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Câu 67. Giá trị của giới hạn
2 2 2
1 2 1
lim ...
n
n n n



bằng:
A.
0
. B.
1
.
3
C.
1
.
2
D.
1
.
Lời giải.
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Câu 68. Giá trị của giới hạn
2
1 3 5 2 1
lim
34
n
n



bằng:
A. 0. B.
1
.
3
C.
2
.
3
D.
1
.
Lời giải.
Trung Tâm Luyện Thi Đại Học Amsterdam Cơng IV. Giới Hạn Dãy Số
40
Lớp Toán Thầy Diệp Tuân Tel: 0935.660.880
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Câu 69. Giá trị của giới hạn
1 1 1
lim ...
1.2 2.3 1nn




là:
A.
1
.
2
B.
1.
C.
0.
D.
.
Lời giải.
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Câu 70. Giá trị của giới hạn
1 1 1
lim ...
1.3 3.5 2 1 2 1nn





bằng:
A.
1
.
2
B.
1
.
4
C.
1
. D.
2
.
Lời giải.
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Câu 71. Giá trị của giới hạn
1 1 1
lim ......
1.4 2.5 3nn




bằng:
A.
11
.
18
B.
2
. C.
1
. D.
3
.
2
Lời giải.
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương IV-Bài 1. Giới Hạn y s
41
Lớp Toán Thầy–Diệp Tuân Tel: 0935.660.880
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Câu 72. Giá trị của giới hạn
2 2 2
2
1 2 ...
lim
1
n
nn
bằng:
A. 4. B.
1
. C.
1
.
2
D.
1
.
3
Lời giải.
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Câu 73. Cho dãy số có giới hạn
n
u
xác định bởi
1
1
2
.
1
, 1
2
n
n
n
u
un
u

Tính
lim .
n
u
A.
B.
lim 0.
n
u
C.
1
lim .
2
n
u
D.
lim 1.
n
u
Lời giải.
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Câu 74. Cho dãy số có giới hạn
n
u
xác định bởi
1
1
2
.
1
, 1
2
n
n
u
u
un

Tính
lim .
n
u
A.
lim 1.
n
u
B.
lim 0.
n
u
C.
lim 2.
n
u
D.
lim .
n
u 
Lời giải.
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Dạng 5. nh giới hạn dựa vào định lý kẹp.
1. Phương pháp.
Trung Tâm Luyện Thi Đại Học Amsterdam Cơng IV. Giới Hạn Dãy Số
42
Lớp Toán Thầy Diệp Tuân Tel: 0935.660.880
Da vào định lí: Cho ba dãy s
,
nn
uv
n
w
.
Nếu
,
n n n
u v w n
lim lim ,
nn
u w a a
thì
lim
n
va
.
2. i tp minh họa.
Bài tập 16. Tìm gii hn ca dãy
n
u
biết:
a).
1 3 5 2 1
2 4 6 2
n
n
u
n
b).
2 2 2
1 1 1
12
n
u
n n n n
c).
*
1.3.5.7.... 2 1
,
2.4.6...2n
n
n
un

d).
1 2 3 ...
n
n
u
nn
e).
2 2 2
1 1 1
l
4 1 4 2 4
n
u
n n n n
Li gii
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương IV-Bài 1. Giới Hạn y s
43
Lớp Toán Thầy–Diệp Tuân Tel: 0935.660.880
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Trung Tâm Luyện Thi Đại Học Amsterdam Cơng IV. Giới Hạn Dãy Số
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Lớp Toán Thầy Diệp Tuân Tel: 0935.660.880
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Dạng 6. Giới hạn có kết qulà vô cực (

hoặc

)
1. Phương pháp.
Chia c t mu cho
k
n
vi
k
n
lũy tha s mũ ln nht ca
Pn
Qn
(hoc rút
k
n
là lũy tha s mũ ln nht ca
Pn
Qn
ra làm nhân t) sau đó áp dng các quy
tc nhân và quy tc chia).
Nếu gp dạng vô định thì ta kh nhé.
Nếu gp dng
0
L
, là bc ca t lớn hơn bậc ca mu thì ta tách cùng bậc đưa về tích.
Quy tc nhân:
lim
n
u
lim
n
v
lim .
nn
uv
lim
n
u
lim 0
n
vL
lim .
nn
uv




















Quy tc chia
lim 0
n
uL
có du
lim 0, 0
nn
vv
có du
lim
n
n
u
v




2. i tp minh họa.
Bài tập 17. Tìm các giới hạn sau
a).
5 4 3
32
2
lim
4 6 9
n n n n
nn

. b).
3
63
7 5 8
lim
12
n n n
n
c).
lim 2 3 1nn
d).
2
42
1 2 3
lim
1
nn
nn


e).
4
lim 1 1nn
.
f).
3
1
lim 2sin 2 3
3
nn




.
Li gii
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Lớp Toán Thầy–Diệp Tuân Tel: 0935.660.880
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Lớp Toán Thầy Diệp Tuân Tel: 0935.660.880
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Bài tập 18. Tìm các giới hạn sau
a).
1
lim 5 3
nn
. b).
1
1
36
lim
35
n
n
n
n


.
Li gii
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5. u hỏi trắc nghiệm.
Câu 75. Kết quả của giới hạn
5
52
lim 200 3 2nn
là:
A.
.
B.
1
. C.
0
. D.
.
Lời giải.
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Câu 76. Giá trị của giới hạn
lim 5 1nn
bằng:
A.
0.
B.
1.
C.
3.
D.
5.
Lời giải.
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Câu 77. Giá trị của giới hạn
22
lim 1 3 2nn
là:
A.
2.
B.
0.
C.
.
D.
.
Lời giải.
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Câu 78. Giá trị của giới hạn
2
1
lim
24nn
là:
A.
1.
B.
0.
C.
.
D.
.
Lời giải.
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Lớp Toán Thầy Diệp Tuân Tel: 0935.660.880
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Câu 79. Kết quả của giới hạn
3
2
2
lim
13
nn
n
là:
A.
1
.
3
B.
.
C.
.
D.
2
.
3
Lời giải.
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Câu 80. Kết quả của giới hạn
3
2
23
lim
4 2 1
nn
nn

là:
A.
3
.
4
B.
.
C. 0 D.
5
.
7
Lời giải.
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Câu 81. Kết quả của giới hạn
4
3
lim
45
nn
n
là:
A.
0.
B.
.
C.
.
D.
3
.
4
Lời giải.
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Câu 82. Trong các giới hạn sau đây, giới hạn nào bằng 0?
A.
B.
2
3
23
lim .
24
n
n

C.
3
2
23
lim .
21
nn
n

D.
24
42
23
lim .
2
nn
nn

Lời giải.
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Lớp Toán Thầy–Diệp Tuân Tel: 0935.660.880
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Câu 83. Dãy số nào sau đây có giới hạn bằng
1
3
?
A.
2
2
2
.
35
n
nn
u
n
B.
43
32
21
.
3 2 1
n
nn
u
nn

C.
23
32
3
.
91
n
nn
u
nn

D.
2
3
25
.
3 4 2
n
nn
u
nn

Lời giải.
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Câu 84. Dãy số nào sau đây có giới hạn là
?
A.
B.
2
3
2
.
55
n
n
u
nn
C.
2
2
2
.
55
n
nn
u
nn
D.
2
12
.
55
n
nn
Lời giải.
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Câu 85. Dãy số nào sau đây có giới hạn là
?
A.
2
12
.
55
n
nn
B.
3
3
21
.
2
n
nn
u
nn


C.
24
23
23
.
2
n
nn
u
nn
D.
2
2
.
51
n
nn
u
n
Lời giải.
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Câu 86. Tính giới hạn
2
lim 3 5 3 .L n n
A.
3.L
B.
.L 
C.
5.L
D.
.L 
Lời giải.
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Câu 87. Có bao nhiêu giá trị nguyên của tham số
a
thuộc khoảng
10;10
để
23
lim 5 3 2L n a n 
.
A. 19. B. 3. C. 5. D. 10.
Lời giải.
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Câu 88. Tính giới hạn
42
lim 3 4 1 .n n n
A.
7.L
B.
.L 
C.
3.L
D.
.L 
Lời giải.
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Câu 89. Cho dãy số
n
u
với
2
2 2 ... 2 .
n
n
u
Mệnh đề nào sau đây đúng ?
A.
lim .
n
u 
B.
2
lim .
12
n
u
C.
lim .
n
u 
D. Không tồn tại
lim .
n
u
Lời giải.
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Câu 90. Tổng của một cấp số nhân lùi hạn bằng
2
, tổng của ba số hạng đầu tiên của cấp số
nhân bằng
9
4
. Số hạng đầu
1
u
của cấp số nhân đó là:
A.
1
3.u
B.
1
4.u
C.
1
9
.
2
u
D.
1
5.u
Lời giải.
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Lớp Toán Thầy–Diệp Tuân Tel: 0935.660.880
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Câu 91. Tính tổng
3
1 1 1
9 3 1
3 9 3
n
S
.
A.
27
.
2
S
B.
14.S
C.
16.S
D.
15.S
Lời giải.
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Câu 92. Tính tổng
1 1 1 1
21
2 4 8 2
n
S


.
A.
2 1.S 
B.
2.S
C.
2 2.S
D.
1
.
2
S
Lời giải.
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Câu 93. Tính tổng
2 4 2
1
3 9 3
n
n
S
.
A.
3.S
B.
4.S
C.
5.S
D.
6.S
Lời giải.
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Câu 94. Tổng của cấp số nhân vô hạn
1
1
1
1 1 1
, , ,..., ,...
2 6 18 2.3
n
n
bằng:
A.
3
.
4
B.
8
.
3
C.
2
.
3
D.
3
.
8
Lời giải.
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Lớp Toán Thầy Diệp Tuân Tel: 0935.660.880
Câu 95. Tính tổng
1 1 1 1 1 1
... ...
2 3 4 9 2 3
nn
S
.
A.
1.
B.
2
.
3
C.
3
.
4
D.
1
.
2
Lời giải.
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Câu 96. Giá trị của giới hạn
2
2
1 ...
lim 1, 1
1 ...
n
n
a a a
ab
b b b

bằng:
A.
0.
B.
1
.
1
b
a
C.
1
.
1
a
b
D. Không tồn tại.
Lời giải.
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Câu 97. Rút gọn
2 4 6 2
cos cos cos c s1 o
n
S x x x x 
với
cos 1.x 
A.
2
sin .Sx
B.
2
cos .Sx
C.
2
1
.
sin
S
x
D.
2
1
.
cos
S
x
Lời giải.
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Câu 98. Rút gọn
2 4 6 2
1 sin si 1n sin .sin
n
n
S x x x x 
với
sin 1.x 
A.
2
sin .Sx
B.
2
cos .Sx
C.
2
1
.
1 sin
S
x
D.
2
tan .Sx
Lời giải.
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương IV-Bài 1. Giới Hạn y s
53
Lớp Toán Thầy–Diệp Tuân Tel: 0935.660.880
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Câu 99. Thu gọn
23
1 tan tantanS
với
0.
4

A.
1
.
1 tan
S
B.
cos
.
2 sin
4
S



C.
tan
.
1 tan
S
D.
2
tan .S
Lời giải.
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Câu 100. Cho
,mn
là các số thực thuộc
1;1
và các biểu thức:
23
1M m m m
23
1N n n n
2 2 3 3
1A mn m n m n
Khẳng định nào dưới đây đúng?
A.
.
1
MN
A
MN

B.
.
1
MN
A
MN

C.
1 1 1
.A
M N MN
D.
1 1 1
.A
M N MN
Lời giải.
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Câu 101. Số thập phân hạn tuần hoàn
0,5111
được biểu diễn bởi phân số tối giản
a
b
. Tính
tổng
.T a b
A.
17.
B.
68.
C.
133.
D.
137.
Lời giải.
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Trung Tâm Luyện Thi Đại Học Amsterdam Cơng IV. Giới Hạn Dãy Số
54
Lớp Toán Thầy Diệp Tuân Tel: 0935.660.880
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Câu 102. Số thập phân hạn tuần hoàn
0,353535...A
được biểu diễn bởi phân số tối giản
a
b
.
Tính
.T ab
A.
3456.
B.
3465.
C.
3645.
D.
3546.
Lời giải.
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Câu 103. Số thập phân hạn tuần hoàn
5,231231...B
được biểu diễn bởi phân số tối giản
a
b
.
Tính
.T a b
A.
1409.
B.
1490.
C.
1049.
D.
1940.
Lời giải.
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Câu 104. Số thập phân hạn tuần hoàn
0,17232323
được biểu diễn bởi phân số tối giản
a
b
.
Khẳng định nào dưới đây đúng?
A.
15
2.ab
B.
14
2.ab
C.
13
2.ab
D.
12
2.ab
Lời giải.
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󰉪󰉗󰉭
󰉼󰉴-Bài 2. 󰉵󰉗 󰉯
55
󰉵󰉚 󰉪 Tel: 0935.660.880
A. 󰈸
I. Gi󰉵i h󰉗n c󰉻a hàm s󰉯 t󰉗i m󰉳󰉨m
0
xx
.
1. Gi󰉵i h󰉗n h󰊀u h󰉗n: Cho
0
;,x a b f x
là hàm s󰉯 󰉬nh trên t󰉝p h󰉹p:
0
;\D a b x
, n󰉦u
v󰉵i m󰉭i dãy s󰉯
0
;\
n
x a b x
sao cho
0
lim
n
xx
󰉧u có:
lim
n
f x L

s󰉯
fx
có gi󰉵i h󰉗n là
L
khi d󰉚󰉦n
0
x
󰉼󰉹c kí hi󰉪u:
0
lim
xx
f x L
.
Chú ý:
0
0
lim
xx
xx
0
lim
xx
CC
, (
C
: h󰉟ng s󰉯).
󰉺󰉵󰉗
a).
2
1
lim 3 2 1
x
xx


. b).
3
2
2
31
lim
3
x
x x x
x

. c).
2
2
2
3 10
lim .
3 5 2
x
xx
xx


d).
32
3
2
3 9 2
lim
6
x
x x x
xx

. e).
2
2
4 1 3
lim
4
x
x
x

. f).
2
22
lim .
73
x
x
x


h).
3
0
1 4 1
lim
x
x
x

. g).
3
2
3
2
1
3 2 4 2
lim
32
x
x x x
xx

. k).
3
2
1
1
lim
32
x
x
x


l).
0
9 16 7
lim
x
xx
x
. m).
3
32
0
1 3 1 1
lim
x
x x x
xx
. n).
3
22
0
3 1 2 1
lim
1 cos
x
xx
x
L󰉶i gi󰉘i
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 GI󰉇I H󰈩N C󰉍A HÀM S󰉁
󰉪󰉗󰉭 󰉼󰉴-Bài 2. 󰉵󰉗 󰉯
56
󰉵󰉚 󰉪 Tel: 0935.660.880
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2. Gi󰉵i h󰉗n vô c󰊁c:
Cho
0
;,x a b f x
là hàm s󰉯 󰉬nh trên t󰉝p h󰉹p:
N󰉦u v󰉵i m󰉭i dãy s󰉯
n
x
v󰉵i
0
;\
n
x a b x
sao cho
0
lim
n
xx
󰉧u có:
lim
n
fx 
thì
ta nói
0
lim
xx
fx
.
N󰉦u v󰉵i m󰉭i dãy s󰉯
n
x
v󰉵i
0
;\
n
x a b x
sao cho
0
lim
n
xx
󰉧u có:
lim
n
fx 
thì ta
nói
0
lim
xx
fx
.
II. Gi󰉵i h󰉗n c󰉻a m s󰉯 t󰉗i vô c󰊁c
x 
1. 󰉬
Cho
fx
là hàm s󰉯 󰉬nh trên
;a 
, n󰉦u v󰉵i m󰉭i dãy s󰉯
n
x
v󰉵i
n
x
;a 
lim
n
x 
󰉧u có:
lim
n
f x L
thì ta nói
lim
x
f x L

󰉪󰉗󰉭
󰉼󰉴-Bài 2. 󰉵󰉗 󰉯
57
󰉵󰉚 󰉪 Tel: 0935.660.880
Các gi󰉵i h󰉗n
lim ; lim ; lim ; lim ; lim
x x x x x
f x f x f x L f x f x
    
   
󰉼󰉹c
󰉬󰉼󰉴󰊁.
2. Các gi󰉵i h󰉗󰉢c bi󰉪t:
lim
k
x
x


2
lim
21
k
x
khi k n
x
khi k n



lim
x
CC

, (
C
là h󰉟ng s󰉯)
lim 0
k
x
C
x

󰉺󰉵󰉗
a).
3
3
31
lim
31
x
xx
xx



. b).
2
lim 1
x
x x x

. c).
2
lim 3 1
x
x x x

d).
3
23
31
lim
2 6 6
x
xx
xx



. e).
32
1
lim
4
x
x
x
xx

. f).
2
31
lim
9 2 1
x
x
xx


g).
5 3 1
lim
1
x
xx
x


. h).
2
2
23
lim
4 1 2
x
x x x
xx


. k).
20 30
50
2 3 3 2
lim
21
x
xx
x


l).
2
lim 2 1 4 4 3
x
x x x

. m).
3
23
lim 1 1
x
xx

. n).
3
32
lim 3 1 2
x
xx

L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 2. 󰉵󰉗 󰉯
58
󰉵󰉚 󰉪 Tel: 0935.660.880
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3. M󰉳t s󰉯 󰉬nh lí v󰉧 gi󰉵i h󰉗n
󰉬nh 1: N󰉦u
0
lim
xx
x
f x L

0
lim
xx
x
g x M

thì:
0
lim
xx
x
f x g x L M



0
lim
xx
x
f x g x L M



0
lim . .
xx
x
f x g x L M



0
lim . .
xx
x
C f x C L



, (C là h󰉟ng s󰉯);
0
0
lim . .
kk
xx
C x C x
,(C là h󰉟ng s󰉯,
k
0
lim 0
xx
x
fx
L
khi M
g x M


󰉬nh 2: Gi󰉘 s󰉿
0
lim
xx
x
f x L

0
lim
xx
x
f x L

;
0
3
3
lim
xx
x
f x L

;
N󰉦u
0 0 0
0, ; \f x x x x x

0
lim
xx
f x L
thì
0L
0
lim
xx
f x L
󰉺󰉵󰉗
a).
2
3
3
lim
6
x
x
xx

. b).
4
3
2
2
2 3 2
lim
2
x
xx
xx



. c).
2
2
2
32
lim
4
x
xx
x

d).
2
2 2 5
lim
2
x
xx
x


. e).
3
23
31
lim
2
x
xx
xx


.
󰉪󰉗󰉭
󰉼󰉴-Bài 2. 󰉵󰉗 󰉯
59
󰉵󰉚 󰉪 Tel: 0935.660.880
L󰉶i gi󰉘i
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󰉬nh 3: Cho 3 hàm s󰉯
,,f x g x h x
󰉬nh trên t󰉝p:
D
0 0 0
; \ , 0x x x
N󰉦u
,f x g x h x x D
00
lim lim
x x x x
g x h x L


thì:
0
lim
xx
h x L
.
T󰉾 󰉽󰉼󰉹c:
0 0 0
sin
sin
lim 0 lim 1 lim 0
x x x x x x
ux
x
khi u x
x u x
.
󰉺󰉵󰉗
a).
0
sin5
lim
x
x
x
. b).
0
tan2
lim
3
x
x
x
. c).
0
1 cos
lim .
sin
x
x
x
d).
2
0
1 cos
lim
x
x
x
. e).
3
0
sin5 .sin3 .sin
lim
45
x
x x x
x
. f).
0
sin7 sin5
lim .
sin
x
xx
x
g).
0
1 cos5
lim
1 cos3
x
x
x
. h).
2
0
1 cos 2
lim
.sin
x
x
xx
. k).
0
.sin
lim
1 cos
x
x ax
ax
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 2. 󰉵󰉗 󰉯
60
󰉵󰉚 󰉪 Tel: 0935.660.880
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󰉪󰉗󰉭
󰉼󰉴-Bài 2. 󰉵󰉗 󰉯
61
󰉵󰉚 󰉪 Tel: 0935.660.880
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3. Các quy t󰉞cm gi󰉵i h󰉗n vô c󰊁c:
N󰉦u
0
lim
xx
fx

thì
0
1
lim 0
xx
fx
.
N󰉦u
0
lim 0
xx
f x L

0
lim
xx
gx

thì
0
0
0
'
'
lim
r
l
m t ái d
im .
li âu
xx
xx
xx
cùng dh âk ui L g x
f x g x
khi L g x


N󰉦u
0
lim 0
xx
f x L

0
lim 0
xx
gx
thì
0
0
0
'
'
lim
t
l
l rái
i
d
m
im âu
xx
xx
xx
cùng dh âk ui L g x
fx
gx
khi L g x


󰉺. 󰉵󰉗
a).
3
2
31
lim
31
x
xx
xx



. b).
2
2
7
lim 1 2 3
1
x
xx
x
x










. c).
2 1 3
lim
42
x
x x x
xx


L󰉶i gi󰉘i
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B. PN D󰈩NG VÀ BÀI T󰈯P MINH H󰈿A.
󰉗󰉵󰉗󰉻󰉯󰉟󰉬
1. 󰉼󰉴
a). 󰉨 tìm
0
lim
xx
fx
󰉼
Xét dãy s󰉯
n
x
b󰉙t k thu󰉳c t󰉝󰉬nh D v󰉵i
0n
xx
0
lim
n
xx
Tìm
lim
n
fx
:
N󰉦u ta có
lim
n
f x L
thì
0
lim
xx
f x L
.
N󰉦u ta có
lim
n
fx 
thì
0
lim
xx
fx
.
󰉪󰉗󰉭 󰉼󰉴-Bài 2. 󰉵󰉗 󰉯
62
󰉵󰉚 󰉪 Tel: 0935.660.880
b). 󰉨 tìm
lim
x
fx

ho󰉢c
lim
x
fx

󰉼 :
Xét dãy s󰉯
n
x
b󰉙t k thu󰉳c t󰉝󰉬nh mà
lim
n
x 
.
Tìm
lim
n
fx
:
N󰉦u ta có
lim
n
f x L
thì
lim
x
f x L

.
N󰉦u ta có
lim
n
fx 
thì
lim
x
fx


.
󰉼󰉴󰊁 khi tính
lim
x
fx

.
c). 󰉨 ch󰉽ng minh hàm s󰉯
fx
không có gi󰉵i h󰉗n khi
0
xx
󰉼󰉶󰉼 :
Ch󰉭n hai dãy s󰉯
n
u
n
v
ng thu󰉳c t󰉝󰉬nh c󰉻a hàm s󰉯 sao cho
00
,
nn
u x v x
0
lim lim
nn
u v x
.
Ch󰉽ng minh
lim lim
nn
f u f v
ho󰉢c m󰉳t trong hai gi󰉵i h󰉗n này không t󰉰n t󰉗i.
󰉬󰉯 không có gi󰉵i h󰉗n khi
0
xx
.
󰉯i v󰉵󰉼󰉶ng h󰉹p
00
, , ,x x x x x x


󰉼󰉴󰊁.
2. 󰉝󰉭
Bài 󰉝1. 󰉬󰉵󰉗
a).
2
1
34
lim
1
x
xx
x


. b).
2
2
1
4
lim
1
x
x
x
.
c).
2
lim 9 2
x
xx


. d).
2
2
lim 2 sin
4
x
x
x
x
.
L󰉶i gi󰉘i
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󰉪󰉗󰉭
󰉼󰉴-Bài 2. 󰉵󰉗 󰉯
63
󰉵󰉚 󰉪 Tel: 0935.660.880
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󰉝 󰉬󰉵󰉗󰉻󰉯
a).
0
1
lim .sin
x
x
x



. b).
2
2
31
lim
1
x
xx
x

. c).
2
2
1
23
lim
21
x
xx
xx


.
L󰉶i gi󰉘i
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Bài 󰉝. 󰉽󰉟
1
3
lim sin
1
x
x

󰉰󰉗
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 2. 󰉵󰉗 󰉯
64
󰉵󰉚 󰉪 Tel: 0935.660.880
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󰉗m g󰉵󰉗󰉻󰉯󰉗󰉳󰉨
0
xx
󰉟󰉞󰉬:
0
lim
xx
fx
1. 󰉼󰉴 chung.
Thay
0
xx
vào
fx
n󰉦u k󰉦t qu󰉘 b󰉟ng m󰉳t s󰉯
C
thì ta k󰉦t lu󰉝n gi󰉵i h󰉗󰉟ng
C
, t󰉽c là:
00
00
lim lim
x x x x
f x f x f x C

Thay
0
xx
vào
fx
n󰉦u k󰉦t qu󰉘 b󰉟ng
0
0
thì ta k󰉦t lu󰉝n gi󰉵i h󰉗n 󰉬nh, 󰉨 kh󰉿 nó ta xét 3
bài toán sau.
Bài toán 1. 󰉯
Px
fx
Qx

,P x Q x
󰉽󰉦
x
.
N󰉦u
0
lim 0
xx
Px
;
0
lim 0
xx
Qx
thì
0
lim
xx
Px
Qx
󰉼󰉹c g󰉭i là có d󰉗󰉬nh
0
0
.
Ta bi󰉦󰉱i v󰉧 d󰉗ng
01
01
.
.
m
n
x x P x
x x Q x
r󰉰i gi󰉘󰉼󰉵c các th󰉾a s󰉯d󰉗ng
0
k
xx
; max ,k m n
Px
Qx
Các cách bi󰉦󰉱i :
󰉽c thành nhân t󰉿󰉭n bi󰉨u th󰉽c làm c󰉘 t󰉿 và m󰉜u b󰉟ng 0.
Phân 󰉽c thành nhân t󰉿 󰉼󰉴
S󰉿 d󰉺ng b󰉘y h󰉟󰉠ng th󰉽󰉵.
N󰉦u tam th󰉽c b󰉝c hai thì s󰉿 d󰉺ng
2
12
,0ax bx c a x x x x a
v󰉵i
12
,xx
nghi󰉪m c󰉻󰉼󰉴
2
0ax bx c
.
S󰉿 d󰉺󰉼󰉴
󰉽c
4 3 2
P x ax bx cx dx e
cho
0
()xx
󰉴󰉰 Hoocner:
2. 󰉝󰉭
󰉝 󰉵󰉗
a).
3
2
2
8
lim
11 18
x
x
xx


b).
32
32
3
2 5 2 3
lim
4 13 4 3
x
x x x
L
x x x
c).
32
32
1
2 5 4 1
lim
1
x
x x x
x x x

d).
3
2
1 12
lim
28
x
xx




e).
3
42
1
1
lim
43
x
x
xx

f).
22
2
11
lim
3 2 5 6
x
x x x x



L󰉶i gi󰉘i
󰉪󰉗󰉭
󰉼󰉴-Bài 2. 󰉵󰉗 󰉯
65
󰉵󰉚 󰉪 Tel: 0935.660.880
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󰉪󰉗󰉭 󰉼󰉴-Bài 2. 󰉵󰉗 󰉯
66
󰉵󰉚 󰉪 Tel: 0935.660.880
Bài toán 2. 󰉯
Px
fx
Qx

,P x Q x
󰉨 󰉽󰉽󰉽
x
.
1. 󰉼󰉴
󰉼󰉵c 1. N󰉦u bi󰉨u th󰉽󰉼󰉵i d󰉙u gi󰉵i h󰉗n có ch󰉽a d󰉙󰉨u th󰉽c liên h󰉹p c󰉻a
bi󰉨u th󰉽c ch󰉽󰉦n v󰉧 0, r󰉰󰉼󰉴󰊁 󰉼d󰉗ng trên ta s󰉥 kh󰉿 󰉼󰉹c d󰉗󰉬nh.
22
22
22
ab
ab
ab
a b a b a b
ab
ab
ab


33
22
ab
ab
a ab b


33
22
ab
ab
a ab b


.
2
2
3 3 3
3
3
22
22
3 3 3 3
.
..
a b a a b b
ab
ab
a a b b a a b b



.
2
2
3 3 3
3
3
22
22
3 3 3 3
.
..
a b a a b b
ab
ab
a a b b a a b b



2
2
3 3 3
3
3
22
22
3 3 3 3
.
..
a b a a b b
ab
ab
a a b b a a b b



2
2
3 3 3
3
3
22
22
3 3 3 3
.
..
a b a a b b
ab
ab
a a b b a a b b



22
3 3 3 3 3 3
33
2 2 2 2
3 3 3 3 3 3 3 3
.
..
a b a a b b
ab
ab
a a b b a a b b



.
22
3 3 3 3 3 3
33
2 2 2 2
3 3 3 3 3 3 3 3
.
..
a b a a b b
ab
ab
a a b b a a b b



󰉼󰉵c 2: 󰉽c thành nhân t󰉿󰉭n h󰉗ng t󰉿 chung c󰉻a c󰉘 t󰉿 m󰉜u.
2. 󰉝󰉭
Bài 󰉝. 󰉵󰉗
󰇛󰉿󰉝󰇜:
a).
1
32
lim
1
x
x
x

b).
2
7
23
lim
49
x
x
x

c).
22
2
3
2 6 2 6
lim
43
x
x x x x
xx

d).
2
22
lim
73
x
x
x


e).
2
4
1
21
lim
x
x x x
xx

f).
2
22
lim
13
x
xx
xx

.
g).
1
4 5 3 6
lim
32
x
xx
x

h).
3
1 3 5
lim
2 3 6
x
xx
xx
k).
2
2
0
11
lim
93
x
x
x


l).
2
1
3 2 4 2
lim
1
x
x x x
x
m).
2
2
0
42
lim
93
x
x
x


n).
2
0
1 2 1
lim
x
x x x
x
L󰉶i gi󰉘i
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󰉪󰉗󰉭
󰉼󰉴-Bài 2. 󰉵󰉗 󰉯
67
󰉵󰉚 󰉪 Tel: 0935.660.880
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󰉪󰉗󰉭 󰉼󰉴-Bài 2. 󰉵󰉗 󰉯
68
󰉵󰉚 󰉪 Tel: 0935.660.880
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Bài 󰉝. 󰉵󰉗
󰇛󰉿󰉝󰇜:
a).
3
2
42
lim
2
x
x
x
b).
3
3
2
1
10 2 1
lim
32
x
xx
xx


c).
3
3
2
3
27
lim
1 4 28
x
x
xx
d).
3
3
1
1
lim
21
x
x
x

e).
33
1
21
lim
1
x
xx
x

f).
4
1
4 3 1
lim
1
x
x
x

.
g).
3
1
32
lim
1
x
xx
x

h).
3
2
3 58
lim
2
x
xx
x

k).
3
3
1
1
lim
4 4 2
x
x
x

.
L󰉶i gi󰉘i
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󰉪󰉗󰉭
󰉼󰉴-Bài 2. 󰉵󰉗 󰉯
69
󰉵󰉚 󰉪 Tel: 0935.660.880
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󰉪󰉗󰉭 󰉼󰉴-Bài 2. 󰉵󰉗 󰉯
70
󰉵󰉚 󰉪 Tel: 0935.660.880
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Bài toán 3: 󰉵󰉯󰉗󰉢󰉳󰉨󰉽󰉞󰉨󰉿󰉼󰉹󰉗󰉬
󰇛󰉿 󰉝hai và 󰉝ba)
1. 󰉳󰉯󰉗󰉼󰉶󰉢
0
0
lim
kk
xx
f x g x c
xx

ho󰉢c
0
0
lim
km
xx
f x g x c
xx

ho󰉢c
0
0
lim
km
n
xx
f x g x c
xx

.

*
,,k m n N
min( , )n k m
.
2. 󰉼󰉴:
Bài toán 1.
0
( ) ( )
lim
nm
xx
f x g x
x
Ta tách v󰉧 t󰉱ng ho󰉢c hi󰉪u c󰉻a hai
lim
: vi󰉪c tách h󰉟ng s󰉯
C
(bi󰉦n s󰉯󰇜󰉼󰉹󰉼󰉘
s󰉿 ta tính gi󰉵i h󰉗n:
0
( ) ( )
lim
nm
xx
f x g x
x
.
Ta ti󰉦n hành tách ra:
0 0 0
12
( ) ( ) ( ) ( )
lim lim lim
n m n m
x x x x x x
f x c c g x f x c c g x
II
x x x
.
Tìm
C
b󰉟ng cách: th󰉦
0
xx
vào:
00
00
( ) 0 ( )
( ) 0 ( )
nn
mm
f x c c f x
c
f x c c f x






N󰉦u gi󰉵i h󰉗n có d󰉗ng t󰉱ng quát :
0
1
( ) ( )
lim
....
nm
nn
xx
f x g x
ax bx c
.
Ta ti󰉦n hành tách ra:
0
1 2 1 2
1
( ) ( ... ) ( ... ) ( )
lim
....
n n n n
nm
nn
xx
f x x x x x g x
ax bx c
S󰉿 d󰉺󰉼󰉴󰉰ng nh󰉙t th󰉽c.
3. 󰉝󰉭
󰉪󰉗󰉭
󰉼󰉴-Bài 2. 󰉵󰉗 󰉯
71
󰉵󰉚 󰉪 Tel: 0935.660.880
Bài 󰉝. 󰉵󰉗 :
a).
1
2 2 5 4 5
lim
1
x
xx
x
b).
3
2
3 2 5 6
lim
2
x
xx
x
c).
3
2
2
2
2 4 11 7
lim
4
x
x x x
x
d).
3
2
3 2 3 2
lim
2
x
xx
x
e).
22
1
5 1 3 1 5 2 1
lim
1
x
x x x x
x
f).
3
32
2
1
57
lim
1
x
xx
x
g).
3
0
2 1 8
lim
x
xx
x
h).
3
0
3 8 5 4
lim
x
xx
x
k).
2
3
2
1
3 2 4 2
lim
32
x
x x x
xx

l).
3
2
2
8 11 7
lim
2 5 2
x
xx
xx

L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 2. 󰉵󰉗 󰉯
72
󰉵󰉚 󰉪 Tel: 0935.660.880
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󰉪󰉗󰉭
󰉼󰉴-Bài 2. 󰉵󰉗 󰉯
73
󰉵󰉚 󰉪 Tel: 0935.660.880
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󰉪󰉗󰉭 󰉼󰉴-Bài 2. 󰉵󰉗 󰉯
74
󰉵󰉚 󰉪 Tel: 0935.660.880
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󰉝 󰉵󰉗
a).
3
2
0
1 4 1 6
lim
x
xx
x
b).
3
22
2
2
2 6 5 3 9 7
lim
2
x
x x x x
x
c).
3
3 2 2
3
0
8 6 9 9 27 27
lim
x
x x x x x
x
d).
3
2
32
1
6 2 2
lim
1
x
xx
x x x

e).
2 2 2
2
1
3 2 4 19 3 46
lim
1
x
x x x x
x
f).
3
3
2
2
3 4 24 2 8 2 3
lim
4
x
x x x
x
L󰉶i gi󰉘i
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󰉪󰉗󰉭
󰉼󰉴-Bài 2. 󰉵󰉗 󰉯
75
󰉵󰉚 󰉪 Tel: 0935.660.880
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󰉪󰉗󰉭 󰉼󰉴-Bài 2. 󰉵󰉗 󰉯
76
󰉵󰉚 󰉪 Tel: 0935.660.880
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󰉪󰉗󰉭
󰉼󰉴-Bài 2. 󰉵󰉗 󰉯
77
󰉵󰉚 󰉪 Tel: 0935.660.880
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󰉝 󰉵󰉗
a) .
3
4
1
1
lim
1
x
x
x
b).
3
2
4
2
0
1 1 2
lim
x
xx
xx
.
c) .
4
3
1
21
lim
21
x
x
x



d) .
3
4
2
6 7 2
lim
2
x
xx
x
L󰉶i gi󰉘i
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Bài toán 2. Tính
0
0
( ) 1 ( )
lim
n
xx
P x ax Q x
xx

b󰉟ng cách thêm b󰉵t
Px
r󰉰󰉢t nhân t󰉿 chung.
󰉪󰉗󰉭 󰉼󰉴-Bài 2. 󰉵󰉗 󰉯
78
󰉵󰉚 󰉪 Tel: 0935.660.880
Tách
00
00
( ) 1 ( ) ( ) ( )
( ) 1 ( )
lim lim
n
n
x x x x
P x ax P x Q x P x
P x ax Q x
x x x x



00
12
00
( ) P( )
( ) 1 ( )
lim lim
n
x x x x
Q x x
P x ax P x
II
x x x x



Tính hai
1
I
2
.I
󰉝 󰉵󰉗
a).
0
1 4 . 1 6 1
lim
x
xx
x
b).
3
0
4 . 1 2 2
lim
x
xx
x
c).
3
1
3 1 2 2
lim
1
x
xx
x
d).
3
2
0
4 8 3 4
lim
x
xx
xx
e).
3
4
0
1.2 1 2.3 1 3.4 1 1
lim
x
xxx
x
f).
L󰉶i gi󰉘i
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79
Lớp Toán Thầy Diệp Tuân Tel: 0935.660.880
4. u hỏi trc nghiệm
Mức độ 1. Nhn biết
Câu 1. Giá trị của giới hạn
2
2
lim 3 7 11
x
xx

là:
A.
37.
B.
38.
C.
39.
D.
40.
Lời giải.
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Câu 2. Giá trị của giới hạn
2
3
lim 4
x
x
là:
A.
0.
B.
1.
C.
2.
D.
3.
Lời giải.
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Câu 3. Giá trị của giới hạn
2
0
1
lim sin
2
x
x
là:
A.
1
sin .
2
B.
.
C.
.
D.
0.
Lời giải.
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Câu 4. Giá trị của giới hạn
2
3
1
3
lim
2
x
x
x

là:
A.
1.
B.
2.
C.
2.
D.
3
.
2
Lời giải.
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Câu 5. Giá trị của giới hạn
3
4
1
lim
2 1 3
x
xx
xx

là:
A.
1.
B.
2.
C.
0.
D.
3
.
2
Lời giải.
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Câu 6. Giá trị của giới hạn
4
1
1
lim
3
x
x
xx


là:
A.
3
.
2
B.
2
.
3
C.
3
.
2
D.
2
.
3
Lời giải.
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Câu 7. Giá trị của giới hạn
2
1
31
lim
1
x
xx
x


là:
Trung Tâm Luyện Thi Đại Học Amsterdam Chương IV-Bài 2. Giới Hạn Hàm S
80
A.
3
.
2
B.
1
.
2
C.
1
.
2
D.
3
.
2
Lời giải.
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Câu 8. Giá trị của giới hạn
2
4
3
9
lim
2 1 3
x
xx
xx

là:
A.
1
.
5
B.
5.
C.
1
.
5
D.
5.
Lời giải.
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Câu 9. Giá trị của giới hạn
2
3
2
2
1
lim
2
x
xx
xx

là:
A.
1
.
4
B.
1
.
2
C.
1
.
3
D.
1
.
5
Lời giải.
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Câu 10. Giá trị của giới hạn
3
2
2
3 4 3 2
lim
1
x
xx
x
là:
A.
3
.
2
B.
2
.
3
C.
0.
D.
.
Lời giải.
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Mức độ 2. Thông Hiu
Câu 11. Giá trị của giới hạn
3
2
2
8
lim
4
x
x
x
là:
A.
0.
B.
.
C.
3.
D. Không xác định.
Lời giải.
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Câu 12. Giá trị của giới hạn
5
3
1
1
lim
1
x
x
x

là:
A.
3
.
5
B.
3
.
5
C.
5
.
3
D.
5
.
3
Lời giải.
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Câu 13. Giá trị của giới hạn
2
2
3
6
lim
3
x
xx
xx

là:
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A.
1
.
3
B.
2
.
3
C.
5
.
3
D.
3
.
5
Lời giải.
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Câu 14.(Sở Quảng Ninh Lần1)
2018
2 2018
2018
2
4
lim
2
x
x
x
A.
2019
2
B.
2018
2
C. 2 D.

.
Lời giải
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Câu 15. Biết rằng
3
2
3
2 6 3
lim 3 .
3
x
x
ab
x


Tính
22
.ab
A.
10.
B.
25.
C.
5.
D.
13.
Lời giải.
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Câu 16. (THPT Thái Tổ) Tính giới hạn
2
2
2
2 3 3
lim
4
x
xx
L
x


.
A.
2
7
L 
. B.
7
24
L 
. C.
9
31
L 
. D.
0L
.
Lời giải
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Câu 17. Giá trị của giới hạn
2 21 21
7
0
12
lim
x
xx
x

là:
A.
21
2
.
7
B.
21
2
.
9
C.
21
2
.
5
D.
21
12
.
7
Lời giải.
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Câu 18.(THPT Hoàng Hoa Thám) Tính
2
2
28
lim
2 5 1
x
xx
x



.
A.
3
. B.
1
2
. C.
6
. D.
8
.
Lời giải
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Câu 19.(Tạp Chí Tn Học)Giới hạn
3
1 5 1
lim
43
x
xx
xx

bằng
a
b
(Phân số tối giản). Giá trị thực của
ab
A.
1
. B.
1
9
. C.
1
. D.
9
8
.
Lời giải
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Câu 20. (Tạp Chí Tn Học)Giới hạn
1
2 7 2
lim
54
x
xx
xx

bằng
a
b
(Phân số tối giản).
Giá trị thc của
ab
A.
10
. B.
1
9
. C.
8
. D.
10
9
.
Lời giải
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Mức độ 3. Vn dng
Câu 21. Giá trị của giới hạn
3
3
1
1
lim
4 4 2
x
x
x

là:
A.
1.
B.
0.
C.
1.
D.
.
Lời giải.
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Câu 22. Giá trị của giới hạn
3
0
2 1 8
lim
x
xx
x
là:
A.
5
.
6
B.
13
.
12
C.
11
.
12
D.
13
.
12
Lời giải.
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Câu 23. Biết rằng
0, 5b a b
3
0
11
lim 2
x
ax bx
x
. Khẳng định nào dưới đây sai?
A.
1 3.a
B.
1.b
C.
22
10.ab
D.
0.ab
Lời giải.
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u 24. (THPT Thái Tổ) Tìm tất cả các giá trị của tham số
m
để
2B
với
32
1
lim 2 2 5 5
x
B x x m m
.
A.
0;3m
. B.
1
2
m
hoc
2m
. C.
1
2
2
m
. D.
23m
.
Lời giải
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Câu 25.(THPT Thái Tổ) Nếu
2
lim 5
x
fx
thì
2
lim 3 4
x
fx


bằng bao nhiêu?
A.
18
. B.
1
. C.
1
. D.
17
.
Lời giải
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Câu 26.(THPT Gia Lộc Hải Dương 2019) Cho hàm số
y f x
thỏa mãn:
2 1 3 5
2 2 1
xx
f
xx





1
2;
2
xx



. Tìm
lim ?
x
fx

A.
4
3
. B.
1
5
. C.
3
2
. D.
2
3
.
Lời giải
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Câu 27.(THPT Sơn y Nội 2019) Cho hàm số
y f x
đạo hàm tại điểm
0
2x
. Tính
2
22
lim
2
x
f x xf
x
.
A.
2 2 2ff
. B. 0. C.
2f
. D.
2 2 2ff
.
Lời giải
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Câu 28.(THPT Thái Tổ) Cho
1
1
lim 1
1
x
fx
x

. Tính
2
1
2
I lim
1
x
x x f x
x

A.
5I
. B.
4I 
. C.
4I
. D.
5I 
.
Lời giải
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Câu 29. (HSG Lớp 12 Bắc Giang) Cho
a
,
b
các số thực dương thỏa mãn
8ab
2
0
2 1 1
lim 5
x
x ax bx
x
. Trong các mệnh đề ới đây, mệnh đề nào đúng?
A.
2;4a
. B.
3;8a
. C.
3;5b
. D.
4;9b
.
Lời giải
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Câu 30. (Kênh truyền Hình GD Quốc Gia 2019) Cho
,mn
các s thực khác
0
. Nếu giới hạn
2
1
lim 3
1
x
x mx n
x

thì
.mn
bằng
A.
3
. B.
1
. C.
3
. D.
2
.
Lời giải
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Câu 31. (S GD Và Đào Tạo Vĩnh Phúc) Tính
2
3
1
21
lim
1
x
x a x a
x
.
A.
2
3
a
B.
2
3
a
C.
3
a
D.
3
a
Lời giải
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Câu 32. Biết rằng
4ab
3
1
lim
11
x
ab
xx




hữu hạn. Tính giới hạn
3
1
lim
11
x
ba
L
xx





.
A.
1.
B.
2.
C.
1
. D.
2.
Lời giải.
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Câu 33. Kết quả của giới hạn
0
1
lim 1
x
x
x






là:
A.
.
B.
1.
C.
0.
D.

.
Lời giải.
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Câu 34. Kết quả của giới hạn
2
2
0
1
lim sin
x
xx
x



là:
A.
0
. B.
1
. C.
.
D.
.
Lời giải.
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Câu 35. (THPT Chuyên Bc Giang) Cho biết
2
3
1
2
12
lim
4 3 1
x
ax bx
c
xx

với
,,abc
.
Tập nghiệm của phương trình
42
2 2 0ax bx c
trên có số phần tử là
A.
1
. B.
3
. C.
0
. D.
2
.
Lời giải
Trung Tâm Luyện Thi Đại Học Amsterdam Chương IV-i 2. Giới Hạn Hàm S
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Lớp Toán Thầy Diệp Tuân Tel: 0935.660.880
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Câu 36.(Chuyên KHTN 2019) Cho hàm số
y f x
xác định trên thỏa mãn
2
16
lim 12
2
x
fx
x
.
Tính giới hạn
3
2
2
5 16 4
lim
28
x
fx
xx


.
A.
5
24
. B.
1
5
. C.
5
12
. D.
1
4
.
Lời giải
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Câu 37. Tính
2018 2018
2 2 2018
1
1 2 2.3
lim
1 2017
x
x x x
xx

A.
2017
5.3
. B.
2017
3
. C.
2017
8.3
. D.
2017
2.3
.
Lời giải
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Dạng 3. m giới hạn của m số khi
x 
1. Phương pháp.
Để tìm
lim
x
fx

ta làm như sau:
Chia t mu cho
k
x
vi
k
x
lũy tha s mũ lớn nht ca t mu, (hoc rút
k
x
làm
nhân tử) sau đó áp dụng các định lý v gii hn hu hn hoc các quy tc v gii hn vô cc.
Nếu
lim
x
f x C

thì kết lun hàm s đó có giới hn là
.C
Nếu
lim lim .
xx
f x P x Q x
 
vi
lim , lim
xx
P x L Q x
 

thì ta da vào quy tc
dấu để kết lun kết qu ca gii hn là

hoc
.
Nếu
lim lim
xx
Px
fx
Qx
 
với đa thức
,P x Q x
có bc lần lượt mà
,mn
lim , lim 0
xx
P x L Q x
 

ta đặt mũ cao nht ca t mu ri tách ra cùng bậc đưa về
lim
ca tích.
Nếu gp dạng vô định
,0 ;
thì ta kh dạng vô định:
Dng
: Nếu
0
lim
xx
x
fx


;
0
lim
xx
x
gx


thì
0
lim
xx
x
fx
gx

đưc gi là có dạng vô định.
Dng
0
: Nếu
0
lim 0
xx
x
fx

;
0
lim
xx
x
gx


thì
0
lim .
xx
x
f x g x



đưc gi là có dng
vô định .
Dng
: Nếu
0
lim
xx
x
fx


;
0
lim
xx
x
gx


thì
0
lim
xx
x
f x g x



đưc gi là có
dạng vô định
.
Kh dạng vô đnh bằng phương pháp nhân lượng liên hp.
2. Bài tập minh họa
Trung Tâm Luyện Thi Đại Học Amsterdam Chương IV-i 2. Giới Hạn Hàm S
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Lớp Toán Thầy Diệp Tuân Tel: 0935.660.880
Bài toán 1. Giới hạn hữu hạn
lim lim .
xx
f x P x Q x
 
với
lim , lim
xx
P x L Q x
 

Phương pp:
Đặt mũ cao nhất ri da vào quy tc dấu để kết lun kết qu ca gii hn là

hoc
.
Bài tập 11. Tính giới các giới hạn sau:
a).
3
lim (2 3 )
x
xx

b).
2
lim 3 4
x
xx


c).
2
lim 2 1
x
xx


Lời giải
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Bài toán 2. Giới hn hữu hạn hữu tỉ
lim lim
xx
Px
fx
Qx
 
( bậc tử bé hơn hoc bằng bậc mẫu)
Phương pháp. ta đặt mũ cao nhất của tử và mẫu rồi rút gọn nhân tử chung.
Bài tập 12. Tìm các giới hạn sau:
a).
2
3
2 3 6
lim
4 3 5
x
xx
xx



. b).
4 6 7
5 9 3
2 3 5 3 6 2
lim
3 2 6 3 7 2
x
x x x
x x x

. c).
2
2
4 2 1 2
lim
9 3 2
x
x x x
x x x


d).
2
2
2 3 4 1
lim
4 1 2
x
x x x
xx

. e).
2
2
23
lim
4 1 2
x
x x x
xx


. f).
2
1
lim
1
x
xx
xx


g).
2
2
2 1 3
lim
5
x
xx
xx


. h).
3
3
2
5
lim
x
x x x
x x x



. k).
53
3
23
21
lim
21
x
xx
x x x



Lời giải
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Bài tập 13. Tìm giới hạn của các hàm số sau:
a).
2
2
3 2 1
lim
5 1 2
x
xx
x x x


b).
42
lim 1
21
x
x
x
xx


c).
2
1
lim
1
x
xx
xx


d).
2
23
lim
5
x
x
L
xx


e).
3
52
2
lim .
3
x
xx
x
xx


f).
42
21
lim
12
x
xx
x


.
Lời giải
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Bài tập 14. Tính các giới hạn sau:
a).
2
lim
10
x
x x x
x


b).
2
32
lim
31
x
x x x
x


c).
2
2
2 3 1
lim
4 1 1
x
x x x
xx

d).
2
2
2 3 1
lim
4 1 1
x
x x x
xx

e).
3
3 2 2 3 2 2
3
2
( 2 ) 2
lim
32
x
x x x x x x
xx

f).
2
3
23
lim
x
xx
x
xx






g).
42
4
4 1 2 4
lim
2
x
xx
x x x






Lời giải
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Bài toán 3. Giới hn vô cực
(bậc tử lớn hơn bậc mẫu:
)lim lim
0
xx
L
Px
fx
Qx
 



Phương pp.
Ta đặt mũ cao nhất của tử và mẫu rồi tách ra cùng bậc đưa v
lim
của tích
0
lim .
xx
x
f x g x



Bài tập 15. Tìm các giới hạn sau:
a).
2
52
lim
21
x
xx
x


. b).
42
21
lim
12
x
xx
x


.
Lời giải
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Bài toán 4. Giới hạn vô cực dạng vô định
0
1. Phương pháp.
Khi ta đặt mũ cao nhất ca t mu thì xut hin
;
0;
Kh dạng vô đnh bằng phương pháp nhân lượng liên hp rồi làm bình thường bài toán 1,2.
2. Bài tập minh họa
Bài tập 16. Tìm các giới hạn sau:
a).
2
lim ( )
x
x x x

. b).
2
lim( 3 2 )
x
x x x

. c).
2
lim ( )
x
x x x

d).
2
lim( 3 2 )
x
x x x

. e).
2
lim( 3 2 )
x
x x x

. f).
lim( 2 2)
x
xx

g).
22
lim 4 3 3 2
x
x x x x
. h).
22
4 3 3 2
x
lim x x x x

.
k).
4 2 2
lim 4 3 1 2
x
x x x

l).
3
3
lim 8 1 2 1
x
xx

m).
Lời giải
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Bài tập 17. Tìm các giới hạn sau
a).
22
lim 1 2
x
x x x

. b).
lim 3 3 3 3
x
x x x x




. c).
3
32
lim 6
x
x x x


d).
3
32
lim 3 1 2
x
xx

e).
lim 2 2 1
x
x x x

f).
Lời giải
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Bài toán 5. Giới hạn vô cực dạng vô định
0
1. Phương pháp.
Gi s cn tìm gii hn ca hàm s
.h x f x g x
khi
0
xx
hoc
x 
trong đó
0fx
gx 
. Ta thường biến đổi theo các hướng sau:
Nếu
0
xx
thì ta thường viết
.
1
fx
f x g x
gx
s đưa về dạng vô định
0
0
.
Nếu
x 
thì ta thường viết
.
1
gx
f x g x
fx
s đưa v v dng
.
Tuy nhiên nhiu bài toán gii hn loi này ta ch cn thc hin mt s biến đổi như đưa
tha s vào trong dấu căn thức, quy đồng mu s,... ta có th đưa về gii hn quen thuc.
2. Bài tập minh họa
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Bài tập 18. m các giới hạn sau:
a).
3
3
1 1 1
lim
3
3
x
x
x



. b).
3
1
lim 2
x
x
x
xx

.
Lời giải
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Dạng 4. m giới hạn của m số các hàm đặc biệt.
1. Phương pháp.
1 1 1 1
n
n
3 3 2 3
3 so hang
a b a b a ab b


1 2 3 2 2 1
so hang
n n n n n n n
n
a b a b a a b a b ab b


2. i tp minh họa
Bài tập 19. Dùng định nghĩa tìm các giới hạn sau:
a).
2
1
...
lim
1
n
x
x x x n
x
. b).
2
1
1
lim
1
n
x
x nx n
x
. c).
0
11
lim
n
x
ax
L
x

.
d).
0
11
lim
nm
x
ax bx
L
x
. e).
0
11
lim 0
11
n
m
x
ax
L ab
bx



f).
0
11
lim
11
nm
x
ax bx
L
x

Lời giải
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Bài tập 20. Dùng định nghĩa tìm các giới hạn sau:
a).
1
1
lim
1
m
n
x
x
L
x
. b).
23
23
1
lim
n
m
x
x x x x n
L
x x x x m
. c).
100
50
1
21
lim
21
x
xx
L
xx


.
d).
1
2
1
1
lim
1
n
x
x n x n
L
x
. e).
1
lim , , *
11
mn
x
mn
mn
xx





Lời giải
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3. u hỏi trc nghiệm
Mức độ 1. Nhn biết
Câu 38. Giá trị của giới hạn
3
lim 1
x
xx


là:
A.
1.
B.
.
C.
0.
D.
.
Lời giải.
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Câu 39. Giá trị của giới hạn
3
2
lim 2 3
x
x x x


là:
A.
0.
B.
.
C.
1.
D.

.
Lời giải.
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Câu 40. Giá trị của giới hạn
2
lim 1
x
xx


là:
A.
0.
B.
.
C.
2 1.
D.

.
Lời giải.
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Câu 41. Giá trị của giới hạn
3
32
lim 3 1 2
x
xx

là:
A.
3
3 1.
B.
.
C.
3
3 1.
D.

.
Lời giải.
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Câu 42. Giá trị của giới hạn
2
lim 4 7 2
x
x x x x


là:
A.
4.
B.
.
C.
6.
D.

.
Lời giải.
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Câu 43. Kết quả của giới hạn
2
2
2 5 3
lim
63
x
xx
xx



là:
A.
2.
B.
.
C.
3.
D.
2
.
Lời giải.
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Câu 44. Kết quả của giới hạn
32
2
2 5 3
lim
63
x
xx
xx



là:
A.
2.
B.
.
C.
.
D.
2
.
Lời giải.
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Câu 45. Kết quả của giới hạn
32
65
2 7 11
lim
3 2 5
x
xx
xx



là:
A.
2.
B.
.
C.
0.
D.
.
Lời giải.
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Câu 46. Kết quả của giới hạn
2
23
lim
1
x
x
xx


là:
A.
2.
B.
.
C.
3.
D.
1
.
Lời giải.
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Câu 47. (THPT Thái Tổ) Trong các mệnh đề sau mệnh đề nào sai?
A.
2
1
lim 1
2

x
x x x
. B.
2
1 2 1
lim
2 3 2





x
xx
x
.
C.
1
32
lim
1


x
x
x
. D.
32
lim 3
2


x
x
x
.
Lời giải
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Mức độ 2. Thông Hiu
Câu 48. Kết quả của giới hạn
2
41
lim
1
x
xx
x


là:
A.
2.
B.
1.
C.
2.
D.
.
Lời giải.
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Câu 49. Kết quả của giới hạn
2
2
4 2 1 2
lim
9 3 2
x
x x x
x x x


là:
A.
1
.
5
B.
.
C.
.
D.
1
5
.
Lời giải.
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Câu 50. Kết quả của giới hạn
3
32
2
21
lim
21
x
xx
x


là:
A.
2
.
2
B.
0.
C.
2
.
2
D.
1.
Lời giải.
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Câu 51. Giá trị của giới hạn
32
lim 2
x
xx

là:
A.
1.
B.
.
C.
1.
D.

.
Lời giải.
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Câu 52. Giá trị của giới hạn
2
lim 1 2
x
xx


là:
A.
0.
B.
.
C.
2 1.
D.

.
Lời giải.
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Câu 53. Giá trị của giới hạn
2
lim 1
x
xx


là:
A.
0.
B.
.
C.
1
.
2
D.

.
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Lời giải.
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Câu 54.(THPT Đn Tng) Tính
2
23
lim
1
x
x
xx

.
A.
0
. B.

. C.
1
. D.
1
.
Lời giải
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Câu 55.( THPT Thái Tổ) Giá tr
2
3 6 2
23
lim
x
x x x
x

bằng
A.
1
2
. B.
9
17
. C.
3
2
. D.
1
.
Lời giải
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Mức độ 3. Vn dng
Câu 56. Giá trị của giới hạn
22
lim 3 4
x
x x x x

là:
A.
7
.
2
B.
1
.
2
C.
.
D.
.
Lời giải.
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Câu 57. Giá trị của giới hạn
3
32
lim 3 1 2
x
xx

là:
A.
3
3 1.
B.
.
C.
3
3 1.
D.

.
Lời giải.
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Câu 58. Giá trị của giới hạn
33
lim 2 1 2 1
x
xx

là:
A.
0.
B.
.
C.
1.
D.

.
Lời giải.
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Câu 59. Kết quả của giới hạn
32
21
lim
32
x
x
x
xx


là:
A.
2
.
3
B.
6
.
3
C.
.
D.

.
Lời giải.
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Câu 60. Giá trị của giới hạn
3
2 3 2
lim
x
x x x x

là:
A.
5
.
6
B.
.
C.
1.
D.

.
Lời giải.
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Câu 61. (THPT Chuyên Vĩnh Phúc 2019)
3
2 3 2
lim 2 3
x
x x x x x

A.
1
2
. B.
0
. C.

. D.

.
Lời giải
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Câu 62. Tìm tất cả các giá trị của
a
để
2
lim 2 1
x
x ax


.
A.
2.a
B.
2.a
C.
2.a
D.
2.a
Lời giải.
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Câu 63. Biết rằng
2
2
4 2 1 2
lim 0
3
x
x x x
L
ax x bx



hữu hạn (với
,ab
tham số). Khẳng định
nào dưới đây đúng.
A.
0.a
B.
3
.L
ab

C.
3
.L
ba
D.
0.b
Lời giải.
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Câu 64. Biết rằng
2
23
1
ax
xx


giới hạn là

khi
x 
(với
a
là tham số). Tính giá trnhỏ
nhất của
2
2 4.P a a
A.
min
1.P
B.
min
3.P
C.
min
4.P
D.
min
5.P
Lời giải.
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Câu 65. Biết rằng
2
lim 5 2 5 5 .
x
x x x a b

Tính
5.S a b
A.
1.S
B.
1.S 
C.
5.S
D.
5.S 
Lời giải.
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Câu 66.(THPT Thái Tổ) Cho biết
2
1 4 5 2
lim
23
x
xx
ax

. Giá trị của
a
bằng
A. 3. B.
2
3
. C.
3
. D.
4
3
.
Lời giải
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Câu 67.(Sở GD và Đào Tạo Cần Thơ 2018) Cho biết
2
4 7 12 2
lim
17 3
x
xx
ax


. Giá trị của
a
bằng
A.
3
.
B.
3
. C.
6
. D.
6
.
Lời giải
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Câu 68.(THPT Lý Thái T) Cho
3
23
2
2 2 5 1
lim
1
x
x x x x a
xb





(
a
b
là phân số tối giản,
,ab
là
số nguyên). Tính tổng
22
L a b
.
A.
150
. B.
143
. C.
140
. D.
145
.
Lời giải
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Câu 69.(THPT Thái T Bắc Ninh) Có bao nhiêu giá trị
m
nguyên thuộc đoạn
20;20
để
2
lim 2 3
x
mx m x

A.21. B.22. C.20. D.41.
Lời giải
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Câu 70. (THPT ơng Thế Vinh Nội) Biết rằng
2
1
lim 5
2
x
x
ax b
x




. Tính tổng
ab
.
A.
6
. B.
7
. C.
8
. D.
5
.
Lời giải
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Câu 71. (THPT ơng Thế Vinh Hà Nội) Biết rằng
2
lim 2 3 1 2 2
x
a
x x x
b

, (
a
là số
nguyên,
b
là số nguyên dương,
a
b
tối giản). Tổng
ab
có giá trị là
A.
1
. B.
5
. C.
4
. D.
7
.
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Câu 72. (HSG Bc Ninh) Cho
2
lim 5 5
x
x ax x

. Khi đó giá trị
a
A.
10
. B.
6
. C.
6
. D.
10
.
Lời giải
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Câu 73.(THPT Nghèn Tĩnh 2018) Biết
2
lim 4 3 1 0
x
x x ax b

. Tính
4ab
ta được
A.
3
. B.
5
. C.
1
. D.
2
.
Lời giải
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Câu 74.(Sở GD và Đào Tạo Cần Thơ 2018) Cho
2
lim 5 5
x
x ax x

. Khi đó giá trị
a
A.
6
. B.
10
. C.
10
. D.
6
.
Lời giải
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Câu 75.(THPT Chuyên Thái Bình) Cho hai số thực
a
b
thoả mãn
2
4 3 1
lim 0
21
x
xx
ax b
x





. Khi đó
2ab
bằng:
A.
4
. B.
5
. C.
4
. D.
3
.
Lời giải
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Câu 76. (THPT Chuyên Vĩnh Phúc 2018) Cho
2
lim 5 5
x
x ax x

thì giá trị của
a
là một
nghiệm của phương trình nào trong các phương trình sau?
A.
2
11 10 0xx
. B.
2
5 6 0xx
. C.
2
8 15 0xx
. D.
2
9 10 0xx
.
Lời giải
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Câu 77.(Chuyên Lương Thế Vinh 2019) Giá tr ca s thc
m
sao cho
2
3
2 1 3
lim 6
47
x
x mx
xx



A.
3m 
. B.
3m
. C.
2m
. D.
2m 
.
Lời giải
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Câu 78.(THPT Quãng Xương 2019) Cho
2
1 2017 1
lim
2018 2
x
ax
x


;
2
lim 1 2
x
x bx x

. Tính
4P a b
.
A.
3P
. B.
1P 
. C.
2P
. D.
1P
.
Lời giải
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Câu 79.(THPT Xuân Trường 2018) Cho số thực
a
thỏa mãn
2
2 3 2017 1
lim
2 2018 2
x
ax
x


. Khi đó giá
trị của
a
A.
2
2
a
. B.
2
2
a
. C.
1
2
a
. D.
1
2
a 
.
Lời giải
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Câu 80.(Tạp chí Toán Học) Tìm giá trị dương của
k
để
2
3 1 1
lim 9 2
x
kx
f
x


với
2
ln 5f x x
:
A.
12k
. B.
2k
. C.
5k
. D.
9k
.
Lời giải
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Câu 81.(THPT Thái Tổ) Cho
,ab
các số dương. Biết
3
2 3 2
7
lim 9 27 5
27
x
x ax x bx

.
m giá trị lớn nhất của
.ab
A.
49
18
. B.
59
34
. C.
43
58
. D.
75
68
.
Lời giải
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Câu 82.(THPT Triệu Thị Trinh) Biết
2
3
1
2 7 1 2
lim
21
x
x x x a
c
b
x

với
a
,
b
,
c
a
b
phân số tối giản. Giá trị của
abc
bằng:
A.
5
. B.
37
. C.
13
. D.
51
.
Lời giải
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Câu 83.(THPT Yên Định 2018) Cho
fx
là một đa thức thỏa mãn
1
16
lim 24
1
x
fx
x
.
Tính
1
16
lim
1 2 4 6
x
fx
I
x f x
A. 24. B.
I 
. C.
2I
. D.
0I
.
Lời giải
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Câu 84.(THPT Chuyên Lương Thế Vinh 2018) Cho
1
10
lim 5
1
x
fx
x
.
Giới hạn
1
10
lim
1 4 9 3
x
fx
x f x
bằng
A.
1
. B.
2
. C.
10
. D.
5
3
.
Lời giải
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112
A. LÝ THUYT
I. CÁC ĐNH NGHĨA:
1. Gii hn bên phi:
Gi s
fx
hàm s xác định trên khong
0
;xb
, nếu vi mi dãy s
n
x
vi
0n
xx
lim
0n
xx
, ta đu :
lim
n
f x L
, thì lúc đó ta nói hàm số
fx
gii hn n phi s thc
L
khi dần đến
0
x
và được
hiu:
0
lim
xx
f x L
Ví d 1. m các giới hạn sau
a).
2
2
31
lim
2
x
xx
x


. b).
2
1
1 3 2
lim
1
x
xx
x
.
Li gii
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2. Gii hn bên trái:
Gi s
fx
hàm s xác định trên khong
0
;ax
, nếu vi mi dãy s
n
x
vi
0n
xx
lim
0n
xx
, ta đu có:
lim
n
f x L
, thì lúc đó ta nói hàm số
fx
gii hn bên trái s thc
L
khi dần đến
0
x
và được
hiu:
0
lim
xx
f x L
Ví d 2. m các giới hạn sau
a)
2
15
lim
2
x
x
x
. b).
2
3
1 3 2
lim
3
x
xx
x

.
c).
2
5
5
lim
25
x
x
x
. d).
2
5
5
lim
25
x
x
x
.
Li gii
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BI 3: GII HN MT BÊN
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3. Gii hn vô cc:
Gi s
fx
là hàm s xác định trên khong
0
;xb
, nếu vi mi dãy s
x
n
vi
0n
xx
0
lim
n
xx
, ta đều :
lim
n
xf 
, thì c đó ta nói hàm số
fx
gii hn bên phi
cc khi dần đến
0
x
và được
hiu:
0
lim
xx
fx

Các định nghĩa
0 0 0
lim , lim , lim ,
x x x x x x
f x f x f x
  
đưc phát biểu tương tự trên.
II. ĐỊNH LÝ V S TN TI CA GII HN.
Nếu
0
lim
xx
f x L
thì hàm s
f
có gii hn bên phi và gii hn bên trái tại điểm
0
x
.
00
lim lim
x x x x
f x f x L



Ngược li, nếu
00
lim lim
x x x x
f x f x L



thì hàm s
f
có gii hn tại điểm
0
x
0
lim
xx
f x L
.
Tương tự,
0
00
lim lim lim
xx
x x x x
f x f x f x


 
0
00
lim lim lim
xx
x x x x
f x f x f x


 
Ví d 3. Cho hàm số
3
3
2 2 1
3 1
x x x
fx
x x x


Tìm
11
lim ; lim .
xx
f x f x


Hàm số có giới hạn tại
1x
không? Vì sao?
Li gii
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B. PN DNG VÀ BÀI TP MINH HA.
Dạng 1. m giới hạn của m số bằng định nghĩa.
1. Phương pháp.
Khi
0
xx
thì
0
xx
suy ra
0
2
00
0xx
x x x
x x x x


Khi
0
xx
thì
0
xx
suy ra
0
2
00
0xx
x x x
x x x x

2. i tp minh họa.
Bài tập 1. Tìm các giới hạn sau:
Trung Tâm Luyện Thi Đại Học Amsterdam Cơng IV-Bài 3. Giới Hn Một n
114
a).
3
3
lim
5 15
x
x
x
b).
0
lim .
x
xx
xx
c).
0
2
lim
x
xx
xx
d).
2
32
1
43
lim
x
xx
xx


e).
3
2
1
44
lim
35
x
x
xx
f).
0
2
lim
x
xx
xx
Li gii
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Bài tập 2. Tìm các giới hạn sau:
a).
2
2
4
lim
2
x
x
x
b).
2
54
1
32
lim
x
xx
xx


c).
2
2
3
7 12
lim
9
x
xx
x

d).
22
2
11
lim
3 2 5 6
x
x x x x



e).
2
2
2
lim
2 5 2
x
x
xx

f).
2
2
32
lim
2
x
xx
x

g).
2
3
1 3 2
lim
3
x
xx
x

h).
2
2
3
2 5 3
lim
3
x
xx
x


k).
2
2
11
lim
24
x
xx




Li gii
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Bài tập 3. Tìm các giới hạn sau
( đưa ra đa thứckhỏi căn hoặc đưa vào căn)
a).
3
2
1
lim 1
1
x
x
x
x




b).
1
1
lim .
2 1 1
x
x
x
xx



c).
2
2
lim 2
4
x
x
x
x
d).
2
2
4
lim
2
x
x
x
e).
2
1
2
2
2 5 2
lim
1
4
x
xx
x

f).
2
1
5
lim 1
23
x
x
x
xx

Li gii
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3. u hi trc nghim.
Câu 1. Kết quả của giới hạn
2
15
lim
2
x
x
x
là:
A.
.
B.
.
C.
15
.
2
D.
1.
Lời giải.
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Câu 2.(THPT Lê Quý Đôn) Tính giới hạn
2
1
1
lim
1
x
x
x
.
A.
0
. B.

. C.

. D.
1
.
Lời giải
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Câu 3.(THPT Chuyên Biên Hòa 2018) Tính giới hạn
2
23
lim
2
x
x
x
.
A.

. B.
2
. C.

. D.
3
2
.
Lời giải
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Câu 4.(THPT Nguyn Trãi-Đà Nng-lần 1 năm 2017-2018) Tìm gii hn
1
43
lim
1
x
x
x
A.

. B.
2
. C.

. D.
2
.
Lời giải
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Câu 5.(THPT Chuyên Lương Thế Vinh) Kết quả của giới hạn
2
4
2
28
lim
2
x
xx
x

A.

. B.
0
. C.

. D.
1
.
Lời giải
Trung Tâm Luyện Thi Đại Học Amsterdam Chương IV-i 3. Giới Hạn Một Bên
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Lớp Toán Thầy Diệp Tuân Tel: 0935.660.880
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Câu 6.(THPT Chuyên Biên Hòa 2018) Tính giới hạn
2
23
lim
2
x
x
x
.
A.

. B.
2
. C.

. D.
3
2
.
Lời giải
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Câu 7. Kết quả của giới hạn
là:
A.
.
B.
.
C.
15
.
2
D. Không xác định.
Lời giải.
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Câu 8. Kết quả của giới hạn
2
36
lim
2
x
x
x

là:
A.
.
B.
3.
C.
.
D. Không xác định.
Lời giải.
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Câu 9. Kết quả của giới hạn
2
2
2
lim
2 5 2
x
x
xx

là:
A.
.
B.
.
C.
1
.
3
D.
1
.
3
Lời giải.
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Câu 10. Kết quả của giới hạn
2
2
3
13 30
lim
35
x
xx
xx



là:
A.
2.
B.
2.
C.
0.
D.
2
.
15
Lời giải.
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Câu 11. Giá trị của giới hạn
3
3
3
lim
27
x
x
x
là:
A.
1
.
3
B.
0.
C.
5
.
3
D.
3
.
5
Lời giải.
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Câu 12. Giá trị của giới hạn
2
2
0
lim
x
x x x
x

là:
A.
0.
B.
.
C.
1.
D.
.
Lời giải.
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Câu 13. Giá trị của giới hạn
2
2
11
lim
24
x
xx




là:
A.
.
B.
.
C.
0.
D.
1.
Lời giải.
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Câu 14. Kết quả của giới hạn
2
2
lim 2
4
x
x
x
x
là:
A.
1.
B.
.
C.
0.
D.

.
Lời giải.
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Câu 15. Kết quả của giới hạn
3
2
1
lim 1
1
x
x
x
x

là:
A.
3.
B.
.
C.
0.
D.

.
Lời giải.
Trung Tâm Luyện Thi Đại Học Amsterdam Chương IV-i 3. Giới Hạn Một Bên
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Lớp Toán Thầy Diệp Tuân Tel: 0935.660.880
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Câu 16.(Chuyên KHTN) Trong các mệnh đề sau, mệnh đề nào sai?
A.
2
3
lim 1 2
2
x
x x x

. B.
( 1)
32
lim
1
x
x
x


.
C.
2
lim 1 2
x
x x x


. D.
( 1)
32
lim
1
x
x
x


.
Lời giải
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Câu 17.(THPT Hoàng Hoa Thám 2018) Trong các mệnh đề sau, mệnh đề nào sai?
A.
0
1
lim
x
x

. B.
0
1
lim
x
x

. C.
5
0
1
lim
x
x

. D.
0
1
lim
x
x

.
Lời giải
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Câu 18. (THPT Trần Nhân ng 2018) Trong các mệnh đề sau mệnh đề nào sai
A.
2
3
lim 1 2
2
x
x x x

. B.
2
lim 1 2
x
x x x


.
C.
1
32
lim
1
x
x
x


. D.
1
32
lim
1
x
x
x


.
Lời giải
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120
Dạng 2. Chng minh s tn ti ca gii hn.
1. Phương pháp.
Hàm s có gii hn là
L
thì
0
00
lim lim lim
xx
x x x x
f x L f x f x L


Hàm s có gii hn là

thì
0
00
lim lim lim
xx
x x x x
f x f x f x


 
Hàm s có gii hn là

thì
0
00
lim lim lim
xx
x x x x
f x f x f x


 
2. Bài tp minh ha.
Bài tập 4. Cho hàm số
42
3
5 6 1
3 1
x x x x
fx
x x x

Tìm
11
lim ;lim .
xx
f x f x


Hàm số có giới hạn tại
1x
không? Vì sao?
Li gii
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Bài tập 5. Cho hàm số
2
23
1
1
1
1
8
x
x
x
y f x
x


a). Tìm
1
lim .
x
fx
So sánh
1
lim
x
fx
1f
b). Tìm
3
lim .
x
fx

So sánh
3
lim
x
fx

3.f
Li gii
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Bài tập 6. Cho hàm số
2
2 3 2
5 2
3 1 2
xx
f x x
xx



a). Tìm
22
lim ; lim .
xx
f x f x

b). Hàm số có giới hạn tại
2x
không? Tại sao?
Li gii
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Lớp Toán Thầy Diệp Tuân Tel: 0935.660.880
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Bài tập 7. Cho hàm số
3
2
2 1 8
0
1 2 0
4
2
2
xx
x
x
f x ax b x
x
x
x

Tìm a, b để hàm số cùng có giới hạn tại
2x 
0.x
Li gii
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Bài tập 8. Tìm các giới hạn sau :
a).
1
lim
x
fx
với
2
3, 1
13, 1
1 7 2, 1
x khi x
f x x khi x
x khi x

b).
2
lim
x
gx

với
32
, 2
1
10, 2
x
khi x
gx
x
x khi x

Li gii
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Bài tập 9. Tìm giới hạn của hàm số
31
, 0
1
10, 0
x
khi x
gx
x
x khi x

tại
0x
Li gii
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Bài tập 10. Tìm m để hàm số
3
22
1
, 1
1
, 1
x
khi x
hx
x
mx x m khi x

có giới hạn tại
1x 
.
Li gii
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Bài 11. Cho hàm số
2
2
32
khi 1
1
khi 1
2
xx
x
x
fx
x
x


. Tính các giới hạn sau
a).
1
lim
x
fx
. b).
1
lim
x
fx
. c).
1
lim
x
fx
, (nếu có).
Li gii
Trung Tâm Luyện Thi Đại Học Amsterdam Chương IV-i 3. Giới Hạn Một Bên
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Lớp Toán Thầy Diệp Tuân Tel: 0935.660.880
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Bài 12. Cho hàm số
3
2
khi 1
khi 3 12
x
fx
x
x
x


. Tìm
1
lim
x
fx

.
Li gii
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3. u hi trc nghim.
Câu 19. Cho hàm số
2
1
.
2
1
3 1 1
x
x
x
fx
xx

víi
víi
Khi đó
1
lim
x
fx
là:
A.
.
B.
2.
C.
4.
D.
.
Lời giải.
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Câu 20. Cho hàm số
2
.
1
1
1
2 2 1
x
x
fx
x
xx

víi
víi
Khi đó
1
lim
x
fx
là:
A.
.
B.
1.
C.
0.
D.
1.
Lời giải.
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Câu 21. Cho hàm số
2
3 2
1 2
.
xx
fx
xx


víi
víi
Khi đó
2
lim
x
fx
là:
A.
1.
B.
0.
C.
1.
D. Không tồn tại.
Lời giải.
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Câu 22. Cho hàm số
2 3 2
1
.
2
xx
fx
ax x
víi
víi
Tìm
a
để tồn tại
2
lim .
x
fx
A.
1.a
B.
2.a
C.
3.a
D.
4.a
Lời giải.
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Câu 23. Cho hàm số
2
2
2 3 3
1 .3
3 2 3
x x x
f x x
xx

víi
víi
víi
Khẳng định nào dưới đây sai?
A.
3
lim 6.
x
fx
B. Không tồn tại
3
lim .
x
fx
C.
3
lim 6.
x
fx
D.
3
lim 15.
x
fx

Lời giải.
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Câu 24.(THTT S 4 2018) Cho hàm số
42
khi 0
1
khi 0
4
x
x
x
fx
mx m x

,
m
là tham số . Tìm gtrị
của
m
để hàm số có giới hạn tại
0x
.
A.
1m
. B.
0m
. C.
1
2
m
. D.
1
2
m
.
Lời giải
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󰉪󰉗󰉭 󰉼󰉴-i 4. 󰉯󰉺
125
󰉵 󰉚 󰉪 Tel: 0935.660.880
A. 󰈸
I. Hàm s󰉯 liên t󰉺c t󰉗󰉨m:
󰉬Gi󰉘 s󰉿 hàm s󰉯
fx
󰉬nh trên kho󰉘ng
;ab
0
;.x a b
Hàm s󰉯
y f x
g󰉭i
là liên t󰉺c t󰉗󰉨m
0
x
n󰉦u:
0
0
.lim
xx
f x f x
Hàm s󰉯 không liên t󰉺c t󰉗󰉨m
0
x
g󰉭󰉗n t󰉗i
0
x
.
2. Nh󰉝n xét.
󰉨 xét tính liên t󰉺c t󰉗i m󰉳󰉨m
0
x
ta ti󰉦󰉼󰉵c sau.
󰉼󰉵c 1: Tính
0
fx
.
󰉼󰉵c 2: Tính
0
lim
xx
fx
.
N󰉦u
0
0
lim
xx
f x f x
thì hàm s󰉯
fx
liên t󰉺c t󰉗i
0
x
.
3. Ví d󰉺 minh h󰉭a.
󰉺. 󰉺󰉻󰉯󰉗󰉨
2x 
a).
2
4
2
x
fx
x
b).
2
4
2
2
4 2
x
x
gx
x
x

L󰉶i gi󰉘i
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󰉺. 󰉯
2
2
35
2
4
1
2
6
x
x
x
y f x
x



a). Tính
2
lim .
x
fx
b). 󰉺󰉻󰉯
fx
󰉗
2;x
L󰉶i gi󰉘i.
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4. HÀM S󰉁 LIÊN T󰉌C
󰉪󰉗󰉭 󰉼󰉴-Bài 4. 󰉯󰉺
126
II. Hàm s󰉯 liên t󰉺c trên m󰉳t kho󰉘ng, trên m󰉳󰉗n.
1. 󰉬
Gi󰉘 s󰉿 hàm s󰉯
f
󰉬nh trên kho󰉘ng
;ab
.Ta nói r󰉟ng hàm s󰉯
y f x
liên t󰉺c trên kho󰉘ng
;ab
n󰉦u nó liên t󰉺c t󰉗i m󰉭󰉨m c󰉻a kho󰉘
Hàm s󰉯
y f x
g󰉭i là liên t󰉺󰉗n
;ab
n󰉦u nó liên t󰉺c trên kho󰉘ng
;ab
lim
lim
xa
xb
f x f a
f x f b
2. Các tính ch󰉙t c󰉻a m s󰉯 liên t󰉺c.
󰉬nh 1.
Hàm s󰉯 󰉽c liên t󰉺c trên .
Hàm s󰉯 phân th󰉽c, các hàm s󰉯 󰉼󰉹ng giác liên t󰉺c trên t󰉾ng kho󰉘󰉬nh c󰉻a chúng.
󰉬nh 2. Gi󰉘 s󰉿
y f x
,
y g x
liên t󰉺c t󰉗󰉨m
0
x

Các hàm s󰉯
, , .y f x g x y f x g x y f x g x
liên t󰉺c t󰉗i
0
x
Hàm s󰉯
fx
y
gx
liên t󰉺c t󰉗i
0
x
n󰉦u
0
0gx
.
3. Ví d󰉺 minh h󰉭a.
󰉺. 󰉽󰉯󰉺
.
a).
42
2f x x x
b).
22
.sin 2cos 3f x x x x
c).
3
3
23
1
1
7
1
3
xx
x
x
fx
x



d).
2
43
1
1
5 1
xx
x
fx
x
xx

L󰉶i gi󰉘i.
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󰉪󰉗󰉭 󰉼󰉴-i 4. 󰉯󰉺
127
󰉵 󰉚 󰉪 Tel: 0935.660.880
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III. Ch󰉽󰉼󰉴󰉪m:
1. 󰉬nh lí 2
󰇛󰉬nh lí v󰉧 giá tr󰉬 trung gian c󰉻a hàm s󰉯 liên t󰉺c)
Gi󰉘 s󰉿m s󰉯
f
liên t󰉺󰉗n
;ab
.
N󰉦u
f a f b
thì v󰉵i m󰉲i s󰉯 th󰊁c
M
n󰉟m gi󰊀a
,f a f b
, t󰉰n t󰉗i ít nh󰉙t m󰉳󰉨m
;c a b
sao cho
.f c M
2. 󰉭c c󰉻󰉬nh
N󰉦u hàm s󰉯
f
liên t󰉺󰉗n
;ab
M
là m󰉳t s󰉯 th󰊁c n󰉟m gi󰊀a
,f a f b
󰉼󰉶ng
th󰉠ng
yM
c󰉞󰉰 th󰉬 c󰉻a hàm s󰉯
y f x
t󰉗i ít nh󰉙t m󰉳󰉨󰉳
;c a b
.
3. H󰉪 qu󰉘
N󰉦u hàm s󰉯
f
liên t󰉺󰉗n
;ab
.0f a f b
thì t󰉰n t󰉗i ít nh󰉙t m󰉳󰉨m
;c a b
sao
cho
0.fc
4. 󰉭c c󰉻a h󰉪 qu󰉘
N󰉦u hàm s󰉯
f
liên t󰉺󰉗n
;ab
.0f a f b
󰉰 th󰉬 c󰉻a hàm s󰉯
y f x
c󰉞t tr󰉺c
hoành ít nh󰉙t t󰉗i m󰉳󰉨󰉳
;c a b
.
5. Ví d󰉺 minh h󰉭a.
󰉺. 󰉽󰉟󰉼󰉴
32
4 8 1 0xx
󰉪󰉘
1;2
L󰉶i gi󰉘i.
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󰉺. 󰉽󰉼󰉴
42
4 2 3 0x x x
󰉙󰉪󰉳
1;1
.
L󰉶i gi󰉘i.
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󰉪󰉗󰉭 󰉼󰉴-Bài 4. 󰉯󰉺
128
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B. PN D󰈩󰉎󰉆󰉌 MINH H󰈿A.
󰈩󰉺󰉻󰉯󰉗󰉳󰉨
Ta xét d󰉗ng này g󰉰m hai bài toán t󰉱ng quát sau.
Bài toán 1. 󰉯
10
20
khi
khi
f x x x
fx
f x x x
.
󰉨󰉺󰉢󰉬󰉬󰉻󰉯󰉨󰉯󰉺󰉗󰉨
0
x
, ta 󰊁 󰉪
󰉼󰉵
1. 󰉼󰉴
B󰉼󰉵c 1. Tính gi󰉵i h󰉗n
00
1
lim lim
x x x x
f x f x L


.
B󰉼󰉵c 2. Tính
0 2 0
f x f x
.
B󰉼󰉵c 3. 󰉢c gi󰉘󰉼󰉴
20
L f x
, t󰉾 󰉼󰉦t lu󰉝n.
N󰉦u
20
L f x
thì k󰉦t lu󰉝n hàm s󰉯 liên t󰉺c t󰉗i
0
x
.
N󰉦u
20
L f x
thì k󰉦t lu󰉝n hàm s󰉯 không liên t󰉺c t󰉗i
0
x
hay b󰉬 󰉗n t󰉗i
0
x
.
2. 󰉝󰉭
󰉝 󰉺󰉻󰉯
2
2
khi 2
2
2 2 khi 2
x
x
fx
x
x
󰉗
2x
.
L󰉶i gi󰉘i
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󰉝. 󰉺󰉻󰉯
2
2
4
khi 2
2
2 khi 2
x
x
fx
xx
x
󰉗
2x
.
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-i 4. 󰉯󰉺
129
󰉵 󰉚 󰉪 Tel: 0935.660.880
󰉝. 󰉺󰉻󰉯
1 2 3
khi 2
2
1 khi 2
x
x
fx
x
x

󰉗
2x
.
L󰉶i gi󰉘i
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󰉝 󰉺󰉻󰉯
sin
khi 1
1
khi 1
x
x
fx
x
x

󰉗
1x
.
L󰉶i gi󰉘i
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󰉝 Tìm
m
󰉨󰉯
32
22
khi 1
1
3 khi 1
x x x
x
fx
x
x m x

󰉺󰉗
1x
.
L󰉶i gi󰉘i
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󰉝 Tìm
m
󰉨󰉯󰉺󰉗󰉨󰉫
2
1
sin khi 0
khi 0
xx
fx
x
mx
󰉗
0x
.
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 4. 󰉯󰉺
130
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󰉝 Tìm
m
󰉨󰉯󰉺󰉗󰉨󰉫
2
1 cos
khi
khi
x
x
x
fx
mx
󰉗
x
.
L󰉶i gi󰉘i
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󰉝 Tìm
m
󰉨󰉯
2
22
2 khi 1
1
11
khi 1
xx
mx x
x
fx
xx
x
x

󰉺󰉗
1x 
.
L󰉶i gi󰉘i
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Bài toán 2. Cho m s󰉯
10
20
khi
khi
f x x x
fx
f x x x
. 󰉨t tính liên t󰉺c ho󰉢󰉬nh giá tr󰉬 c󰉻a
tham s󰉯
m
󰉨 m s󰉯 liên t󰉺c t󰉗󰉨m
0
x
, ta làm 󰉼󰉵c
1. 󰉼󰉴
B󰉼󰉵c 1. Tính
0 2 0
f x f x
.
B󰉼󰉵c 2.
(Liên t󰉺c trái)
Tính gi󰉵i h󰉗n
00
11
lim lim
x x x x
f x f x L



.
󰉢c gi󰉘󰉼󰉴
1 2 0
L f x
, t󰉾 󰉼󰉦t lu󰉝n.
B󰉼󰉵c 3. (
Liên t󰉺c ph󰉘i)
Tính gi󰉵i h󰉗n
00
12
lim lim
x x x x
f x f x L



.
󰉢c gi󰉘󰉼󰉴
2 2 0
L f x
, t󰉾 󰉼󰉦t lu󰉝n.
B󰉼󰉵c 4. 󰉢c gi󰉘󰉼󰉴
12
LL
, t󰉾 󰉼󰉦t lu󰉝n.
N󰉦u
1 2 2 0
L L f x
thì k󰉦t lu󰉝n hàm s󰉯 liên t󰉺c t󰉗i
0
x
.
󰉪󰉗󰉭 󰉼󰉴-i 4. 󰉯󰉺
131
󰉵 󰉚 󰉪 Tel: 0935.660.880
N󰉦u
1 2 0
L f x
ho󰉢c
2 2 0
L f x
ho󰉢c
21
LL
k󰉦t lu󰉝n hàm s󰉯 không ln t󰉺c t󰉗i
0
x
hay b󰉬
󰉗n t󰉗i
0
x
.
2. Bài t󰉝p minh h󰉭a.
󰉝 󰉺󰉻󰉯
2
5
khi 5
2 1 3
5 3 khi 5
x
x
x
fx
xx

󰉗
5x
.
L󰉶i gi󰉘i
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󰉝 󰉺󰉻󰉯
1 cos khi 0
1 khi 0
xx
fx
xx


󰉗
0x
.
L󰉶i gi󰉘i
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󰉝 󰉺󰉻󰉯
3
3
khi 0
2
11
khi 0
11
xx
fx
x
x
x



󰉗
0x
.
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 4. 󰉯󰉺
132
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󰉝 Tìm
m
󰉨󰉯󰉺󰉗󰉨󰉫
2
khi 1
2 khi 1
1 khi 1
x x x
f x x
mx x



󰉗
1x
.
L󰉶i gi󰉘i
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󰉝 󰉯
2
32
khi 1
1
khi 1
xx
x
x
fx
ax

.
a). Tìm
a
󰉨󰉯󰉺󰉗󰉨
1x
.
b). Tìm
a
󰉨󰉯󰉺󰉘󰉗󰉨
1x
.
c). Tìm
a
󰉨󰉯󰉺󰉗󰉨
1x
.
L󰉶i gi󰉘i
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4. u h󰉮i tr󰉞c nghi󰉪m.
Câu 1. 󰉯
1
3
4
f x x
x
󰉺
A.
4;3 .
B.
4;3 .
C.
4;3 .
D.
; 4 3; . 
󰉶󰉘
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󰉪󰉗󰉭 󰉼󰉴-i 4. 󰉯󰉺
133
󰉵 󰉚 󰉪 Tel: 0935.660.880
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Câu 2. 󰉯
3
cos sin
2sin 3
x x x x
fx
x

󰉺
A.
1;1 .
B.
1;5 .
C.
3
;.
2




D.
.
󰉶󰉘
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Câu 3. 󰉯
fx
󰉬󰉺 󰉵
2
32
1
xx
fx
x

󰉵󰉭
1.x
Tính
1.f
A.
2.
B.
1.
C.
0.
D.
1.
󰉶󰉘
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Câu 4. 󰉯
fx
󰉬󰉺
m
󰉵
33xx
fx
x
󰉵
0x
. Tính
0f
.
A.
23
.
3
B.
3
.
3
C.
1.
D.
0.
󰉶󰉘
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Câu 5. 󰉯
fx
󰉬󰉺
4; 
󰉵
42
x
fx
x

󰉵
0x
.
Tính
0f
.
A.
0.
B.
2.
C.
4.
D.
1.
󰉶󰉘
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󰉪󰉗󰉭 󰉼󰉴-Bài 4. 󰉯󰉺
134
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Câu 6. 󰉬󰊁󰉻󰉯
m
󰉨󰉯
2
2
khi 2
2
khi 2
xx
x
fx
x
mx

󰉺󰉗
2.x
A.
0.m
B.
1.m
C.
2.m
D.
3.m
󰉶󰉘
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Câu 7. 󰉬󰊁󰉻󰉯
m
󰉨󰉯
32
22
khi 1
1
3 khi 1
x x x
x
fx
x
x m x

󰉺 󰉗
1.x
A.
0.m
B.
2.m
C.
4.m
D.
6.m
󰉶󰉘
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Câu 8. 󰉬󰊁󰉻󰉯
k
󰉨󰉯
1
khi 1
1
1 khi 1
x
x
y f x
x
kx


󰉺󰉗
1.x
A.
1
.
2
k
B.
2.k
C.
1
.
2
k 
D.
0.k
󰉶󰉘
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Câu 9. 󰉦󰉟󰉯
3
khi 3
12
khi 3
x
x
fx
x
mx

󰉺󰉗
3x
󰇛󰉵
m
󰉯󰇜󰉠
󰉬󰉼󰉵
A.
3;0 .m
B.
3.m 
C.
0;5 .m
D.
5; .m 
󰉶󰉘
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󰉪󰉗󰉭 󰉼󰉴-i 4. 󰉯󰉺
135
󰉵 󰉚 󰉪 Tel: 0935.660.880
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Câu 10. 󰉬󰊁󰉻󰉯
m
󰉨󰉯
2
1
sin khi 0
khi 0
xx
fx
x
mx
󰉺󰉗
0.x
A.
2; 1 .m 
B.
2.m 
C.
1;7 .m
D.
7; .m 
󰉶󰉘
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Câu 11. 󰉦󰉟
0
sin
lim 1.
x
x
x
󰉯
tan
khi 0
0 khi 0
x
x
fx
x
x
󰉺󰉘

?
A.
0; .
2



B.
0x
C.
;.
44




D.
;. 
󰉶󰉘
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Câu 12. 󰉦󰉟
0
sin
lim 1.
x
x
x
󰉬󰊁󰉻󰉯
m
󰉨󰉯
sin
khi 1
1
khi 1
x
x
fx
x
mx
󰉺󰉗
1.x
A.
.m

B.
.m
C.
1.m 
D.
1.m
󰉶󰉘
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󰉪󰉗󰉭 󰉼󰉴-Bài 4. 󰉯󰉺
136
Câu 13. 󰉦󰉟
0
sin
lim 1.
x
x
x
󰉬󰊁󰉻󰉯
m
󰉨󰉯
2
1 cos
khi
khi
x
x
x
fx
mx
󰉺󰉗
.x
A.
.
2
m
B.
.
2
m

C.
1
.
2
m
D.
1
.
2
m 
󰉶󰉘
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Câu 14. 󰉯
4
2
3 khi 1
khi 1, 0
1 khi 0
x
xx
f x x x
xx
x

󰉺󰉗
A. 󰉭󰉨󰉾
0, 1.xx
B. 󰉭󰉨
.x
C. 󰉭󰉨󰉾
1.x 
D. 󰉭󰉨󰉾
0.x
󰉶󰉘
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Câu 15. 󰉯󰉨󰉗󰉻󰉯
2
0,5 khi 1
1
khi 1, 1
1
1 khi 1
x
xx
f x x x
x
x

là:
A.
0.
B.
1.
C.
2.
D.
3.
󰉪󰉗󰉭 󰉼󰉴-i 4. 󰉯󰉺
137
󰉵 󰉚 󰉪 Tel: 0935.660.880
󰉶󰉘
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D󰈩NG 2: Xét tính liên t󰉺c c󰉻a m s󰉯 trên .
Ta xét d󰉗ng này g󰉰m bài toán t󰉱ng quát sau.
Bài toán 1. Cho m s󰉯
10
20
khi
khi
f x x x
fx
f x x x
. 󰉨t tính liên t󰉺c ho󰉢󰉬nh giá tr󰉬 c󰉻a
tham s󰉯 󰉨 hàm s󰉯 liên t󰉺c t󰉗󰉨m
0
x
, ta làm 󰉼󰉵c
1. 󰉼󰉴
󰉼󰉵c 1. Xét trên kho󰉘ng
0
xx
.
N󰉦󰉽󰉼󰉹ng giác, hàm phân th󰉽c h󰊀u t󰉫 thì k󰉦t lu󰉝n liên t󰉺c trên kho󰉘ng
0
;.x 
B󰉼󰉵c 2. Xét trên kho󰉘ng
0
xx
.
N󰉦u là 󰉽󰉼󰉹ng giác, hàm phân th󰉽c h󰊀u t󰉫 thì k󰉦t lu󰉝n liên t󰉺c trên kho󰉘ng
0
;.x 
B󰉼󰉵c 3. Xét liên t󰉺c t󰉗i
0
xx
.
󰉼󰉧 d󰉗ng toán 1: xét tinh liên t󰉺c t󰉗󰉨m
0
xx
g󰉰m hai bài toán 1 và 2.
B󰉼󰉵c 4. K󰉦t lu󰉝n.
2. Bài t󰉝p minh h󰉭a.
󰉝 󰉯
fx
󰉬󰉷
2
2 khi 3
3
khi 1 3
12
x x x
fx
x
x
x
.
󰉽󰉟󰉯󰉺󰉘
1;
.
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 4. 󰉯󰉺
138
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󰉝 󰉬
a
󰉨󰉯
2
1
khi 1
1
khi 1
x
x
fx
x
ax
󰉺󰉗
0;1
.
L󰉶i gi󰉘i
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󰉝 󰉯
1 3
3 5
7 5
x
f x ax b x
x
. 󰉬󰉨󰉯󰉺
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-i 4. 󰉯󰉺
139
󰉵 󰉚 󰉪 Tel: 0935.660.880
󰉝 󰉯󰉺󰉵
xR
󰉦󰉫󰉨
󰉗
a).
43
2 4 2 1f x x x x
b).
2
2
3 4 5
32
xx
fx
xx


c).
2
2 1 1
2 3 1 2
1
2
2
x
x
xx
fx
x



d).
32
2 6 3
3
3
19 3
x x x
x
fx
x
x


L󰉶i gi󰉘i
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󰉝 󰉯
3
2
8
2
4
3 2
3 5 3 2
x
khi x
x
f x khi x
x khi x

󰉘󰉿󰉘
󰉯󰇛󰇜󰉺
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 4. 󰉯󰉺
140
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󰉝 󰉯
2
32
2
2
2
xx
x
fx
x
ax

󰉵󰉬󰉻󰉯󰉺󰉗󰉨
2x
?
L󰉶i gi󰉘i
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󰉝 󰉯
2
2 7 6
khi x < 2
2
1
a + 2
2
xx
x
y f x
x
khi x
x


󰉬󰉨󰉯󰇛󰇜󰉺
󰉗
0
2x
.
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-i 4. 󰉯󰉺
141
󰉵 󰉚 󰉪 Tel: 0935.660.880
3. 󰉮󰉞󰉪.
Câu 16. 󰉬󰊁󰉻
m
󰉨󰉯
22
khi 2
1 khi 2
m x x
fx
m x x
󰉺 ?
A.
2
. B.
1
. C.
0
. D.
3
.
󰉶󰉘
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Câu 17. 󰉦 󰉟  󰉯
0;4
4;6
khi
1 khi
xx
fx
mx
󰉺 
0;6 .
󰉠󰉬   

A.
2.m
B.
2 3.m
C.
3 5.m
D.
5.m
󰉶󰉘
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Câu 18. 󰉬󰉯
a
󰉨󰉯
2
32
khi 1
1
khi 1
xx
x
x
fx
ax

󰉺
.
A.
1
. B.
2
. C.
0
. D.
3
.
󰉶󰉘
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󰉪󰉗󰉭 󰉼󰉴-Bài 4. 󰉯󰉺
142
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Câu 19. 󰉦󰉟
2
1
khi 1
1
khi 1
x
x
fx
x
ax
󰉺󰉗
0;1
󰇛󰉵
a
󰉯󰇜󰉠
󰉬󰉼󰉵󰉧󰉬
a

A.
a
󰉳󰉯 B.
a
󰉳󰉯󰉫
C.
5.a
D.
0.a
󰉶󰉘
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Câu 20.    󰉺 󰉻  󰉯
.
1
khi 1
21
2 khi 1
x
x
fx
x
xx

󰉠 󰉬  󰉼󰉵 

A.
fx
󰉺
.
B.
fx
󰉺
0;2 .
C.
fx
󰉗󰉗
1.x
D.
fx
󰉺
.
󰉶󰉘
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Câu 21. 󰉬󰉮󰉙󰉻
a
󰉨󰉯
2
2
56
khi 3
43
1 khi 3
xx
x
fx
xx
a x x


󰉺󰉗
3x
.
A.
2
3
. B.
2
.
3
C.
4
.
3
D.
4
.
3
󰉶󰉘
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󰉪󰉗󰉭 󰉼󰉴-i 4. 󰉯󰉺
143
󰉵 󰉚 󰉪 Tel: 0935.660.880
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Câu 22. 󰉬󰉵󰉙󰉻
a
󰉨󰉯
3
2
3 2 2
khi 2
2
1
khi 2
4
x
x
x
fx
a x x

󰉺󰉗
2.x
A.
max
3.a
B.
max
0.a
C.
max
1.a
D.
max
2.a
󰉶󰉘
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Câu 23. 󰉺󰉻󰉯
1 cos khi 0
1
.
khi 0
xx
x
fx
x
󰉠󰉬
A.
fx
󰉺󰉗
0.x
B.
fx
󰉺
;1 .
C.
fx
󰉺
.
D.
fx
󰉗󰉗
1.x
󰉶󰉘
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Câu 24. 󰉘󰉺󰉻󰉯
cos khi 1
2
1
.
khi 1
x
x
x
fx
x

󰉪󰉧 là sai?
A. 󰉯󰉺󰉗
1x 
.
B. 󰉯󰉺󰉘
.;, 1 1; 
C. 󰉯󰉺󰉗
1x
.
D. 󰉯󰉺󰉘
1,1
.
󰉶󰉘
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󰉪󰉗󰉭 󰉼󰉴-Bài 4. 󰉯󰉺
144
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Câu 25. 󰉯
fx
󰉰󰉬󰉼󰉺
󰉗󰉨󰉳
A.
0.x
B.
1.x
C.
2.x
D.
3.x
󰉶󰉘
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Câu 26. 󰉯
2
khi 1, 0
0 khi 0 .
khi 1
x
xx
x
f x x
xx


󰉯
fx
󰉺󰉗
A. 󰉭󰉨󰉳 . B. 󰉭󰉨󰉾
0x
.
C. 󰉭󰉨󰉾
1x
. D. 󰉭󰉨󰉾
0x
1x
.
󰉶󰉘
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Câu 27. 󰉯
2
1
khi 3, 1
1
4 khi 1
1 khi 3
x
xx
x
f x x
xx



󰉯
fx
󰉺󰉗
A. M󰉭󰉨󰉳 . B. M󰉭󰉨󰉾
1x
.
C. M󰉭󰉨󰉾
3x
. D. M󰉭󰉨󰉾
1x
3x
.
󰉶󰉘
󰉪󰉗󰉭 󰉼󰉴-i 4. 󰉯󰉺
145
󰉵 󰉚 󰉪 Tel: 0935.660.880
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Câu 28. 󰉯󰉨󰉗󰉻󰉯
2
2 khi 0
1 khi 0 2
3 1 khi 2
xx
h x x x
xx

là:
A. 1. B. 2. C. 3. D. 0.
󰉶󰉘
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Câu 29. 󰉱
S
󰉰󰉙󰉘󰉬
m
󰉨󰉯
2
2
khi 1
2 khi 1
1 khi 1
x x x
f x x
m x x



󰉺󰉗
1x
A.
1.S 
B.
0.S
C.
1.S
D.
2.S
󰉶󰉘
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Câu 30. 󰉯
2
3
cos khi 0
khi 0 1.
1
khi 1
x x x
x
f x x
x
xx

󰉯
fx
󰉺󰉗
A. 󰉭󰉨󰉳
.x
B. 󰉭󰉨󰉾
0.x
C. 󰉭󰉨󰉾
1.x
D. 󰉭󰉨󰉾
0; 1.xx
󰉶󰉘
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󰉪󰉗󰉭 󰉼󰉴-Bài 4. 󰉯󰉺
146
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D󰉗ng 3: Ch󰉽󰉼󰉴󰉪m.
Ta xét bài toán t󰉱ng quát sau.
Bài toán 1. 󰉼󰉴
0fx
. Ch󰉽󰉼󰉴trình có nghi󰉪m.
1. 󰉼󰉴
󰉼󰉵c 1: Bi󰉦󰉱󰉼󰉴󰉧 d󰉗ng
0fx
.
󰉼󰉵c 2: Hàm s󰉯 liên t󰉺󰉬nh trên t󰉝󰉬nh
hàm s󰉯 liên t󰉺c trên
;.ab
󰉼󰉵c 3: Tìm hai s󰉯
a
b
sao cho
.0f a f b
.
T󰉾 󰉼󰉴
0fx
có ít nh󰉙t m󰉳t nghi󰉪m thu󰉳c
;ab
.
Chú ý:
N󰉦u
.0f a f b
󰉼󰉴󰉙t m󰉳t nghi󰉪m thu󰉳c
;ab
N󰉦u hàm s󰉯
fx
liên t󰉺c trên
;a 
và có
. lim 0
x
f a f x

󰉼󰉴
0fx
ít nh󰉙t m󰉳t nghi󰉪m thu󰉳c
;a 
.
N󰉦u hàm s󰉯 f(x) liên t󰉺c trên
;a
và có
. lim 0
x
f a f x

󰉼󰉴
0fx
có ít nh󰉙t m󰉳t nghi󰉪m thu󰉳c
;a
.
󰉨 ch󰉽ng minh
0fx
có ít nh󰉙t
n
nghi󰉪m trên
;ab
󰉗n
;ab
thành
n
󰉗n
nh󰉮 r󰉶i nhau, r󰉰i ch󰉽ng minh trên m󰉲i kho󰉘󰉼󰉴󰉙t m󰉳t nghi󰉪m.
2. i t󰉝p minh h󰉭a.
󰉝 󰉽󰉟󰉼󰉴
a).
42
3 5 6 0x x x
󰉙󰉳󰉪󰉳󰉘
1;2
.
b).
3
10xx
󰉙󰉳󰉪󰉵󰉴
1
.
c).
2
cos sin 1 0x x x x
󰉙󰉳󰉪󰉳󰉘
0;
.
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-i 4. 󰉯󰉺
147
󰉵 󰉚 󰉪 Tel: 0935.660.880
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󰉝 󰉽󰉼󰉴
53
5 4 1 0x x x
󰉪
L󰉶i gi󰉘i
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󰉝 󰉽󰉟󰉼󰉴󰉪
a).
5
3 3 0xx
. b).
4 3 2
3 1 0x x x x
.
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 4. 󰉯󰉺
148
󰉝 󰉽󰉟󰉼󰉴
a).
42
4 2 3 0x x x
󰉙󰉪󰉪󰉳󰉘
1;1
.
b).
54
5 4 1 0x x x
󰉙󰉪󰉪󰉳󰉘
0;5
.
L󰉶i gi󰉘i
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󰉝 󰉽󰉟󰉼󰉴
a).
2
2 3 4 0xx
󰉪󰉪󰉳󰉘
3;1
.
b).
32
3 3 0xx
󰉪󰉪󰉳󰉘
1;3
.
c).
3
2 6 1 3xx
󰉪󰉪󰉳󰉘
7;9
.
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-i 4. 󰉯󰉺
149
󰉵 󰉚 󰉪 Tel: 0935.660.880
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Bài toán 2. Ch󰉽󰉼󰉴󰉽a tham s󰉯
m
luôn có nghi󰉪m v󰉵i m󰉭i
.m
1. 󰉼󰉴
󰉼󰉵c 1. Ch󰉭n hai s󰉯
a
b
th󰉮󰉼󰉶ng h󰉹p.
0fa
thì
0,f b m
󰇛󰉼󰉧 󰉼󰉴󰉦u khi bi󰉨u th󰉽c có ch󰉽a tham s󰉯
)m
. 0,f a f b m
󰉼󰉴󰉙t m󰉳t nghi󰉪m
.m
0fa
thì
0,f b m
󰇛󰉼󰉧 󰉼󰉴󰉦u khi bi󰉨u th󰉽c có ch󰉽a tham s󰉯
)m
. 0,f a f b m
󰉼󰉴󰉙t m󰉳t nghi󰉪m
.m
󰉼󰉵c 2. K󰉦t lu󰉝n
2. 󰉝󰉭
󰉝 󰉽󰉟󰉼󰉴󰉪
a).
3
22
1 1 3 0m x x x
. b).
2 7 5
5 1 0m m x x
.
c).
43
1 2 1 3 0m x x x x
. d).
cos cos2 0x m x
e).
2cos 2 2sin5 1m x x
. f).
11
cos sin
m
xx

L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 4. 󰉯󰉺
150
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󰉝 󰉽󰉟󰉦
3m 
󰉼󰉴
2 3 2
3 1 3 2 1 3 0m m x m x m x
󰉙󰉳󰉪󰉳󰉘
1;1
.
L󰉶i gi󰉘i
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Bài toán 2. Ch󰉽󰉼󰉴󰉽a tham s󰉯
m
luôn có nghi󰉪󰉼󰉴󰉢c nghi󰉪m âm
v󰉵i m󰉭i
.m
1. 󰉼󰉴
Ch󰉭n s󰉯
a
th󰉮󰉼󰉶ng h󰉹p.
󰉼󰉶ng h󰉹p 1. N󰉦u nghi󰉪󰉼󰉴󰉭n
0a
󰉪󰉗󰉭 󰉼󰉴-i 4. 󰉯󰉺
151
󰉵 󰉚 󰉪 Tel: 0935.660.880
hàm s󰉯
fx
liên t󰉺c trên
;a 
th󰉮a
00
lim 0
x
tính f
f
x

. lim 0
x
f a f x

t󰉼󰉴
0fx
có ít nh󰉙t m󰉳t nghi󰉪m 󰉼󰉴thu󰉳c
;a 
.
󰉼󰉶ng h󰉹p 2. N󰉦u nghi󰉪m âm ch󰉭n
0a
Thì hàm s󰉯
fx
liên t󰉺c trên
;a
th󰉮a
00
lim 0
x
tính f
f
x

. lim 0
x
f a f x

thì
󰉼󰉴
0fx
có ít nh󰉙t m󰉳t nghi󰉪m âm thu󰉳c
;a
.
2. i t󰉝p minh h󰉭a.
󰉝 󰉽󰉟󰉼󰉴
a).
32
10x mx
󰉳󰉪󰉼󰉴
b).
3
11x mx m
󰉳󰉪󰉵󰉴
1
.
c).
2 2 *
3 2 4 0,
n
m m x x n
luôn có ít nh󰉙t m󰉳t nghi󰉪m âm v󰉵i m󰉭i
.m
L󰉶i gi󰉘i
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󰉝 Tìm
m
󰉨󰉼󰉴
32
3 2 2 3 0x x m x m
󰉪󰉪
1 2 3
, , x x x
󰉮 mãn
1 2 3
1x x x
.
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 4. 󰉯󰉺
152
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󰉝 󰉽󰉼󰉴
4
30xx
󰉙󰉳󰉪
0
x
󰉮
󰉧󰉪
7
0
12x
.
L󰉶i gi󰉘i
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󰉝25. Cho
, , a b c
󰉯󰊁
0
󰉽󰉟󰉼󰉴
󰉪
a).
2
0ax bx c
󰉵
2 3 6 0.a b c
b).
2
0ax bx c
󰉵
0
21
a b c
m m m

0m
.
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-i 4. 󰉯󰉺
153
󰉵 󰉚 󰉪 Tel: 0935.660.880
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󰉝 󰉽󰉟󰉦
2 3 6 0a b c
󰉼󰉴
2
tan tan 0a x b x c
󰉙󰉳󰉪󰉘
;
4
kk




.
L󰉶i gi󰉘i
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󰉝 󰉽󰉟󰉭󰉼󰉴󰉝
32
0ax bx cx d
0a
luôn ít
󰉙󰉳󰉪
L󰉶i gi󰉘i
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󰉝 󰉽󰉟󰉭󰉼󰉴󰉝󰉯
4 3 2
0ax bx cx dx e
󰉵
.0ae
󰉙󰉳󰉪
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 4. 󰉯󰉺
154
󰉝 Cho
, , a b c
󰉯󰉼󰉴󰉪
󰉽󰉟󰉼󰉴
0a x b x c b x a x c c x a x b
󰉪󰉝󰉪
L󰉶i gi󰉘i
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4. 󰉮󰉞󰉪.
Câu 31. 󰉯
3
4 4 1.f x x x
󰉪󰉧
A. 󰉯󰉺
.
B. 󰉼󰉴
0fx
󰉪󰉘
;1 .
C. 󰉼󰉴
0fx
󰉪󰉘ng
2;0 .
D. 󰉼󰉴
0fx
󰉙󰉪󰉘
1
3; .
2



󰉶󰉘
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Câu 32. 󰉼󰉴
42
2 5 1 0.x x x
󰉪󰉧
A. 󰉼󰉴󰉪󰉘
1;1 .
B. 󰉼󰉴󰉪󰉘
2;0 .
C. 󰉼󰉴󰉫󰉳󰉪󰉘
2;1 .
D. 󰉼󰉴󰉙󰉪󰉘
0;2 .
󰉶󰉘
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󰉪󰉗󰉭 󰉼󰉴-i 4. 󰉯󰉺
155
󰉵 󰉚 󰉪 Tel: 0935.660.880
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Câu 33. 󰉯
3
31()f x x x 
󰉯󰉪󰉻󰉼󰉴
0fx
trên là:
A.
0.
B.
1.
C.
2.
D.
3.
󰉶󰉘
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Câu 34. 󰉯
fx
󰉺󰉗
1;4
sao cho
12f 
,
47f
󰉨󰉧
󰉯󰉪󰉻󰉼󰉴
5fx
󰉗
[ 1;4]
:
A. 󰉪 B. 󰉙󰉳󰉪
C. 󰉳󰉪 D. C󰉪
󰉶󰉘
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Câu 35. 󰉙󰉘󰉬󰉻 󰉯
m
󰉳󰉘
10;10
󰉨󰉼󰉴
trình
32
3 2 2 3 0x x m x m
󰉪󰉪
1 2 3
, , x x x
󰉮
1 2 3
1x x x
?
A.
19.
B.
18.
C.
4.
D.
3.
󰉶󰉘
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󰉪󰉗󰉭 󰉼󰉴-Bài 4. 󰉯󰉺
156
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