GIỚI HẠN VÔ ĐỊNH DẠNG 0 chia 0

TÌM GIỚI HẠN VÔ ĐỊNH DẠNG
DẠNG 1: L = xlim
→x
0

0
0

P(x)
với P(x), Q(x) là các đa thức và P(x0) = Q(x0) = 0
Q(x)

Câu 1: Tìm các giới hạn sau:
x2 + x − 6
x→ 2
x2 − 4

a). lim

x3 − 3x + 2
x→1 x4 − 4x + 3

d). lim

x2 − 5x
x→ 5 x2 − 25

b). lim

x2 + 2x − 3
x→1 2x2 − x − 1

e). lim

x3 − 8
x→ 2 x2 − 3x + 2

c). lim

f). lim

x→−3

x+ 3
x + 2x − 3
2

LỜI GIẢI
a). lim
x→ 2

x2 + x − 6
(x + 3)(x − 2)
x+ 3 5
= lim
= lim
=
x→ 2 (x − 2)(x + 2)
x→ 2 x + 2
4
x2 − 4

b). lim
x→ 5

x2 − 5x
x(x − 5)
x
1
= lim
= lim
=
x2 − 25 x→ 5 (x − 5)(x + 5) x→ 5 x + 5 2

c). lim
x→ 2

x3 − 8
(x − 2)(x2 + 2x + 4)
x2 + 2x + 4
=
lim
=
lim
= 12
x→ 2
x−1
x2 − 3x + 2 x→ 2 (x − 1)(x − 2)

x3 − 3x + 2
x→1 x4 − 4x + 3

Phân tích x3 − 3x + 2 thành nhân tử bằng Hoocner:
1
0
-3
2
d). L = lim

1

1

1

-2

0

⇒ x3 − 3x + 2 = (x − 1)(x2 + x − 2)
Phân tích x4 − 4x + 3 thành nhân tử bằng Hoocner:

1

1

0

0

-4

3

1

1

1

-3

0

⇒ x4 − 4x + 3 = (x − 1)(x3 + x2 + x − 3)
Vậy L = lim
x→1

(x − 1)(x2 + x − 2)
x2 + x − 2
=
lim
(khi x → 1 thì ta thấy cả
(x − 1)(x3 + x2 + x − 3) x→1 x3 + x2 + x − 3

tử và mẫu đều dần về 0, có nghĩa vẫn còn vô định
tích thành nhân tử tiếp).
Phân tích x2 + x − 2 thành nhân tử bằng Hoocner:

0
, nên ta phải phân
0

1

1

1

-2

1

2

0

⇒ x2 + x − 2 = (x − 1)(x + 2)
Phân tích x3 + x2 + x − 3 thành nhân tử bằng Hoocner:

1

1

1

1

-3

1

2

3

0

⇒ x3 + x2 + x − 3 = (x − 1)(x2 + 2x + 3)
L = lim =
x→1

e). lim
x→1
f). xlim
→−3

(x − 1)(x + 2)
x+ 2
1
= lim 2
=
2
x
→
1
(x − 1)(x + 2x + 3)
x + 2x + 3 2

x2 + 2x − 3
(x − 1)(x + 3)

x+ 3 4
= lim
= lim
=
2
x
→
1
x
→
1
(x − 1)(2x + 1)
2x + 1 3
2x − x − 1

x+ 3
x+ 3
1
1
= lim
= lim
=−
4
x2 + 2x − 3 x→−3 (x − 1)(x + 3) x→−3 x − 1

Câu 2: Tìm các giới hạn sau :
(1+ x)3 − 1
(x + 3)3 − 27
a). lim
b). lim

x→ 0
x→ 0
x
x
x3 + 2 2
2
2 x −2

2x3 − 5x2 − 2x − 3
x→ 3 4x3 − 12x2 + 4x − 12
x4 − 27x
f). lim 2
x→ 3 2x − 3x − 9

c). lim

x2 + 4x − 5
x→1
x→−
x−1
LỜI GIẢI
3
3
2
(1+ x) − 1
x + 3x + 3x
a). lim
= lim
= lim(x2 + 3x + 3) = 3
x→ 0

x→ 0
x→ 0
x
x
3
3
2
(x + 3) − 27
x + 9x + 27x
b). lim
= lim
= lim(x2 + 9x + 27) = 27
x→ 0
x→ 0
x→ 0
x
x
3
2
2x − 5x − 2x − 3
c). L = lim 3
x→ 3 4x − 12x2 + 4x − 12
Phân tích 2x3 − 5x2 − 2x − 3 thành nhân tử bằng Hoocner:
2
-5
-2
-3
d). lim

3

2

e). lim

1

1

0

⇒ 2x3 − 5x2 − 2x − 3 = (x − 3)(2x2 + x + 1)
Phân tích 4x3 − 12x2 + 4x − 12 thành nhân tử bằng Hoocner:
4
-12
4
-12
3

4

0

4

0

⇒ 4x3 − 12x2 + 4x − 12 = (x − 3)(4x2 + 4)
(x − 3)(2x2 + x + 1)
2x2 + x + 1 2.9 + 3+ 1 11

= lim
=
=
.
2
x→ 3 (x − 3)(4x + 4)
x→ 3
20
4x2 + 4
4.32 + 4

L = lim


( )

x3 + 2
x3 + 2 2
d). lim
= lim
2
x→− 2 x − 2
x→− 2
x2 − 2
= lim

x2 − 2x + 2

=−

3

( x + 2) ( x − 2x + 2)
= lim
( x + 2) ( x − 2)
2

x→− 2

3 2
2

x− 2
x + 4x − 5
(x − 1)(x + 5)
e). lim
= lim
= lim(x + 5) = 6
x→1
x→ 1
x→1
x−1
x−1
4
3
x − 27x
x(x − 27)
x(x − 3)(x2 + 3x + 9)
= lim
= lim

f). lim
2
x→ 3 2x − 3x − 9
x→ 3 (x − 3)(2x + 3)
x→ 3
(x − 3)(2x + 3)
x→− 2

2

x(x2 + 3x + 9)
=9
x→ 3
2x + 3
Câu 3: Tìm các giới hạn sau:
x3 + x2 − 5x − 2
x4 − 16
a). lim
b).
lim
x→ 2
x→−2 x2 + 6x + 8
x2 − 3x + 2
= lim

x4 − x3 − x + 1
3
x→1 x − 5x2 + 7x − 3

e). lim

d). lim

x→1

x5 − 2x4 + x − 2
x→ 2
x2 − 4

c). lim

x+ 2 x − 3

3

f). lim

x− 5 x + 4
LỜI GIẢI

x→1

x2 − 23 x + 1
(x − 1)2

x3 + x2 − 5x − 2
x→ 2
x2 − 3x + 2
Phân tích x3 + x2 − 5x − 2 thành nhân tử bằng Hoocner:
1

1
-5
-2
a). lim

2

1

3

1

0

⇒ x3 + x2 − 5x − 2 = (x − 2)(x2 + 3x + 1)
Vậy lim
x→ 2

(x − 2)(x2 + 3x + 1)
x2 + 3x + 1 11
= lim
=
x→ 2
(x − 1)(x + 2)
x+ 2
4

(

)(

)

x2 − 4 x2 + 4
x4 − 16
(x − 2)(x + 2)(x2 + 4)
=
lim
= lim
2
x→−2 x + 6x + 8
x→−2
( x + 2) ( x + 4) x→−2 (x + 2)(x + 4)

b). lim

(x − 2)(x2 + 4)
= −16
x→−2
x+ 4
x5 − 2x4 + x − 2
c). lim
x→ 2
x2 − 4
Phân tích x5 − 2x4 + x − 2 thành nhân tử bằng Hoocner:
1
-2
0
0

1

-2

2

0

= lim

1

0

0

⇒ x5 − 2x4 + x − 2 = (x − 2)(x4 + 1)

0

1


Vậy lim
x→ 2

(x − 2)(x4 + 1)
x4 + 1 17
= lim
=

(x − 2)(x + 2) x→ 2 x + 2 4

x4 − x3 − x + 1
x→1 x3 − 5x2 + 7x − 3
Phân tích x4 − x3 − x + 1 thành nhân tử bằng Hoocner:
1
-1
0
-1
1
d). lim

1

1

0

0

-1

0

⇒ x4 − x3 − x + 1 = (x − 1)(x3 − 1)
Phân tích x3 − 5x2 + 7x − 3 thành nhân tử bằng Hoocner:
1
-5
7
-3

1

1

-4

3

0

⇒ x3 − 5x2 + 7x − 3 = (x − 1)(x2 − 4x + 3)
(x − 1)(x3 − 1)
x3 − 1
(x − 1)(x2 + x + 1)
x2 + x + 1 3
=
lim
=
lim
=
lim
= .
x→1 (x − 1)(x2 − 4x + 3)
x→1 x2 − 4x + 3
x→ 1
x →1
(x − 1)(x + 3)
x+ 3
4

lim

e). lim
x→1

x+ 2 x − 3
x− 5 x + 4

= lim
x→1

( x − 1)( x + 3)
( x − 1)( x − 4)

(

3

)

x −1

= lim
x→1

x+3
x−4

(

2

3

=

4
.
−3

)

x −1

2

1
1
x − 2 x + 1 = lim
= lim
= lim
= .
2
2
x
→
1
x
→
1

x
→
1
2
4
( x − 1)( x + 1)
( x + 1)
x→1


(x − 1)2


 x − 1÷


Câu 4: Tìm các giới hạn sau:
x3 − 5x2 + 3x + 9
2x4 + 8x3 + 7x2 − 4x − 4
a). lim
b).
lim
x→ 3
x→−2
x4 − 8x2 − 9
3x3 + 14x2 + 20x + 8
4x6 − 5x5 + x
x4 − 5x3 + 9x2 − 7x + 2
x5 + x4 + x3 + x2 + x − 5
lim

c). lim 4
d).
e).
lim
x→1
x→1 x − 3x3 + x2 + 3x − 2
x→1
(1− x)2
x2 − 1
3

2

3

f). lim

( )

LỜI GIẢI
x3 − 5x2 + 3x + 9
a). lim
x→ 3
x4 − 8x2 − 9
Phân tích x3 − 5x2 + 3x + 9 thành nhân tử bằng Hoocner:
1
-5
3
9
3

1

-2

-3

0

⇒ x3 − 5x2 + 3x + 9 = (x − 3)(x2 − 2x + 3)
(x − 3)(x2 − 2x + 3)
(x − 3)(x2 − 2x − 3)
x2 − 2x − 3
= lim 2
= lim 2
=0
2
2
x→ 3
x→ 3 (x + 1)(x − 3)(x + 3)
x→ 3 (x + 1)(x + 3)
(x + 1)(x − 9)

lim

2x4 + 8x3 + 7x2 − 4x − 4
x→−2 3x3 + 14x2 + 20x + 8

b). lim

Phân tích 2x4 + 8x3 + 7x2 − 4x − 4 thành nhân tử bằng Hoocner:
2
8
7
-4
-4
-2

2

4

-1

-2

0

⇒ 2x4 + 8x3 + 7x2 − 4x − 4 = (x + 2)(2x3 + 4x2 − x − 2)
Phân tích 3x3 + 14x2 + 20x + 8 thành nhân tử bằng Hoocner:

-2

3

14

20

8

3

8

4

0

⇒ 3x3 + 14x2 + 20x + 8 = (x + 2)(3x2 + 8x + 4)
(x + 2)(2x3 + 4x2 − x − 2)
2x3 + 4x2 − x − 2
=
lim
(Khi x → −2 ta thấy cả tử và
x→−2
x→−2
(x + 2)(3x2 + 8x + 4)
3x2 + 8x + 4
mẫu đều dần về 0, nên vẫn còn vô định. Do đó ta phân tích thành nhân
tử cả tử và mẫu tiếp để khử dạng vô định).
Phân tích 2x3 + 4x2 − x − 2 thành nhân tử bằng Hoocner:
2
4
-1
-2
lim

-2

2

0

-1

0

⇒ 2x3 + 4x2 − x − 2 = (x + 2)(2x2 − 1)
Phân tích 3x2 + 8x + 4 thành nhân tử bằng Hoocner:
3
8
4
-2

3

2

0

⇒ 3x2 + 8x + 4 = (x + 2)(3x + 2)
(x + 2)(2x2 − 1)
2x2 − 1
7
= lim
=−
x→−2 (x + 2)(3x + 2)
x→−2 3x + 2

4
lim

x4 − 5x3 + 9x2 − 7x + 2
x→1 x4 − 3x3 + x2 + 3x − 2
Phân tích x4 − 5x3 + 9x2 − 7x + 2 thành nhân tử bằng Hoocner:
1
-5
9
-7
2
c). L = lim

1

1

-4

5

-2

0

⇒ x4 − 5x3 + 9x2 − 7x + 2 = (x − 1)(x3 − 4x2 + 5x − 2)
Phân tích x4 − 3x3 + x2 + 3x − 2 thành nhân tử bằng Hoocner:

1

1

-3

1

3

-2

1

-2

-1

2

0


⇒ x4 − 3x3 + x2 + 3x − 2 = (x − 1)(x3 − 2x2 − x + 2)
(x − 1)(x3 − 4x2 + 5x − 2)
x3 − 4x2 + 5x − 2
=
lim
(Khi x → 1 ta thấy cả tử
x→1 (x − 1)(x3 − 2x2 − x + 2)
x→1 x3 − 2x2 − x + 2
và mẫu đều dần về 0, nên vẫn còn vô định. Do đó ta phân tích thành

nhân tử cả tử và mẫu tiếp để khử dạng vô định).
Phân tích x3 − 4x2 + 5x − 2 thành nhân tử bằng Hoocner:
1
-4
5
-2
L = lim

1

1

-3

2

0

⇒ x3 − 4x2 + 5x − 2 = (x − 1)(x2 − 3x + 2)
Phân tích x3 − 2x2 − x + 2 thành nhân tử bằng Hoocner:
1
-2
-1
2
1

1

-1

-2

0

⇒ x3 − 2x2 − x + 2 = (x − 1)(x2 − x − 2)
(x − 1)(x2 − 3x + 2)
x2 − 3x + 2
=
lim
=0
x→1 (x − 1)(x2 − x − 2)
x→1 x2 − x − 2

L = lim

x5 + x4 + x3 + x2 + x − 5
x→1
x2 − 1
Phân tích x5 + x4 + x3 + x2 + x − 5 thành nhân tử bằng Hoocner:
1
1
1
1
1
-5
d). lim

1

1

2

3

4

5

0

(x − 1)(x4 + 2x3 + 3x2 + 4x + 5)
x4 + 2x3 + 3x2 + 4x + 5 15
= lim
=
x→1
x→1
(x − 1)(x + 1)
x+ 1
2

lim

e). lim
x→1

4x6 − 5x5 + x
x(4x5 − 5x4 + 1)
=
lim

x→1
(1− x)2
(x − 1)2

Phân tích 4x5 − 5x4 + 1 thành nhân tử bằng Hoocner:
4
-5
0
0
0
1
1

4

-1

-1

-1

-1

0

⇒ 4x5 − 5x4 + 1 = (x − 1)(4x4 − x3 − x2 − x − 1)
x(x − 1)(4x4 − x3 − x2 − x − 1)
x(4x4 − x3 − x2 − x − 1)
=
lim

x→1
x →1
(x − 1)
(x − 1)2

lim

Phân tích 4x4 − x3 − x2 − x − 1 thành nhân tử bằng Hoocner:
4
-1
-1
-1
-1
1

4

3

2

1

0


⇒ 4x4 − x3 − x2 − x − 1 = (x − 1)(4x3 + 3x2 + 2x + 1)
x(x − 1)(4x3 + 3x2 + 2x + 1)
= limx(4x3 + 3x2 + 2x + 1) = 10 .
x→1

x→1
x− 1
Câu 5: Tìm các giới hạn sau:
 1
2 

1
1

−
+ 2
a). lim
b). lim

÷

÷
x→1 x − 1 x2 − 1
x→ 2 x2 − 5x + 6
x
−
3x
+
2




lim

 2x − 3 x − 26 
−
c). xlim

2 ÷
→−2
 x + 2 4− x 


1
1 
− 3
d). lim

÷
2
x→1 x + x − 2
x − 1

LỜI GIẢI

 1
2 
−
a). lim

÷
x→1 x − 1 x2 − 1



 1

2
x + 1− 2
x−1
1
1
= lim 
−
= lim
= lim
=
÷ = lim
x→1 x − 1 (x − 1)(x + 1)
x→1 (x − 1)(x + 1)
x→1 (x − 1)(x + 1)
x→ 1 x + 1
2



1
1

+ 2
b). lim

÷
x→ 2 x2 − 5x + 6
x − 3x + 2 






1
1
x − 1+ x − 3
= lim 
+
÷ = lim

÷
x→ 2 (x − 2)(x − 3)
x→ 2 (x − 2)(x − 3)(x − 1)
(x
−
1)(x
−
3)




= lim
x→ 2

2(x − 2)
2
= lim

= −2 .
x
→
2
(x − 2)(x − 3)(x − 1)
(x − 3)(x − 1)

 2x − 3 x − 26 
−
c). xlim

2 ÷
→−2
 x + 2 4− x 
 2x − 3
x − 26 
(2x − 3)(x − 2) + x − 26
= lim 
+
÷ = xlim
x→−2
→−2
x
+
2
(x
−
2)(x
+
2)

(x − 2)(x + 2)


2x2 − 6x − 20
2(x + 2)(x − 5)
2(x − 5) 7
= lim
= lim
= .
x→−2 (x − 2)(x + 2)
x→−2 (x − 2)(x + 2)
x→−2 x − 2
2

= lim


1
1 
− 3
d). lim

÷
x→1 x2 + x − 2
x − 1



1
1

x2 + x + 1− x − 2
= lim 
−
=
lim
÷
x→1 (x − 1)(x + 2)
(x − 1)(x2 + x + 1)  x→1 (x − 1)(x + 2)(x2 + x + 1)

x2 − 1
(x − 1)(x + 1)
x+ 1
1
= lim
= lim
= lim
=
x→1 (x − 1)(x + 2)(x2 + x + 1)
x→1 (x − 1)(x + 2)(x2 + x + 1)
x→1 (x + 2)(x2 + x + 1)
9
Câu 6: Tính các giới hạn sau:
x−3
x→ 9 9x − x2

a). lim

d). lim
x→1

2x − x2 − 1
x2 − x

x+ 3− 3
x− 6

x3 + 1 − 1
x→ 6
x→ 0
x2 + x
x2 − 9
x+ 5− 3
e). lim 2
f). lim
x→ 3
x→ 4 x − 3x − 4
x + 1− 2
LỜI GIẢI
b) lim

c). lim


a). lim
x→ 9

x−3
x− 9
1
5

= lim
lim
=−
2
x
→
9
x
→
9
4
9x − x
− x(x − 9)( x + 3)
− x( x + 3)

b).
x+ 3− 3
(x + 3 − 9)
x− 6
1
1
= lim
= lim
= lim
=
x→ 6
x→ 6
x→ 6
x− 6
6

(x − 6)( x + 3 + 3)
(x − 6)( x + 3 + 3)
x+ 3+ 3

lim
x→ 6

x3 + 1 − 1
x2 + x
x3 + 1− 1

c). lim
x→ 0

= lim
x→ 0

(x2 + x)

x→1

x→1

)

x3 + 1 + 1

2x − x2 − 1
x2 − x
2x − x2 − 1

d). lim
= lim

(

(x − x)
2

(

= lim
x→ 0

)

2x − x + 1
2

x(x + 1)

= lim
x→1

(

x3

)

x3 + 1 + 1

= lim

−(x − 1)2
x(x − 1)

(

x→ 0

)

2x − x + 1
2

(x + 1)

= lim
x →1

x

(

(

x2

)

=0

)

=0

x3 + 1 + 1

−(x − 1)

2x − x2 + 1

x+ 5− 3
x + 5− 9
x− 4
= lim
= lim
e). lim
x→ 4 x2 − 3x − 4
x→ 4
x→ 4
2
(x − 3x − 4) x + 5 + 3
(x + 1)(x − 4) x + 5 + 3

(

= lim
x→ 4

(x + 1)

f). lim
x→ 3

)

x + 1− 2

(

=

x+ 5+ 3

x2 − 9

= lim(x + 3)
x→ 3

(

1

= lim

)

(

1
30

(x2 − 9)

(

) = lim (x + 3(x − 3)(

x + 1+ 2

x + 1− 4

x→ 3

)

x + 1 + 2 = 24 .

x→1

3x + 1 − x + 3
x+ 8− 3
2(x − 1)

(x − 1)

(

(

c). lim

x→−1

7 − 2x + x − 2
x2 − 1
LỜI GIẢI

2
d). lim 4 + x + x − 2
x→−1
x+ 1

= lim

)

x + 1+ 2

x− 3

x→ 3

Câu 7: Tìm các giới hạn sau:
3x + 1 − x + 3
3− x
a). lim
b). lim

x→1
x→ 9
x+ 8− 3
x− 5− 2

a). lim
x→1

)

e). lim

x→−1

= lim
x→1

)

(3x + 1− x − 3)
(x + 8 − 9)

x+ 8+ 3

)

3x + 1 + x + 3

= lim
x→1

(

2

(

(

)

x+ 8+ 3

3x + 1 + x + 3

)

x+ 8+ 3

3x + 1 + x + 3

=3

)

3+ 2x − x + 2
3x + 3


3− x

b). lim
x→ 9

= lim

−

x− 5− 2

(

(

) = lim −(x − 9)( x − 5 + 2)
x)
(x − 9) ( 3+ x )

x− 5+ 2

(

x→ 9

(x − 5− 4) 3+

) = −2.
3

3+ x

3+ 2x − x + 2
3+ 2x − (x + 2)
= lim
x
→−
1
3x + 3
(3x + 3) 3+ 2x + x + 2

c). lim
x→−1

x→−1

x→ 9

(9 − x)

x− 5+ 2

x→ 9

= lim

= lim

(

3(x + 1)

(

x+ 1
3+ 2x + x + 2

= lim

)

x→−1

)

1

(

3+ 2x + x + 2

3

)

=

1
.
6

4 + x + x2 − 2
4 + x + x2 − 4
x(x + 1)
= lim
= lim
x
→−
1
x
→−
1
2
x+ 1
(x + 1) 4 + x + x + 2
(x + 1) 4 + x + x2 + 2

d). lim
x→−1

)

(

1
=− .
4
4+ x + x + 2

)

(

x

= lim

x→−1

2

(

)

2

7 − 2x − (x − 2)2
7 − 2x + x − 2
e). lim
=
lim
x→−1
x→−1
x2 − 1
(x2 − 1)  7 − 2x − (x − 2)


7 − 2x − (x2 − 4x + 4)
−x2 + 2x + 3
= lim

x→−1
(x2 − 1)  7 − 2x − (x − 2) x→−1 (x2 − 1)  7 − 2x − (x − 2)





= lim

= lim

x→−1

−(x + 1)(x − 3)
(x − 1)(x + 1)

(

= lim

)

x→−1

7 − 2x − x + 2

(x − 1)

(

−(x − 3)

=−

)

7 − 2x − x + 2

1
3

Câu 8: Tìm các giới hạn sau:
a). lim

3

5x − 3 + 2
x+ 1

3

x+ 7 − 2

x→−1

d). lim

3

a).

lim

3

e). lim

x −1

x→1

1− 3 1− x
x→ 0
x

b). lim

(

x2 − 1 − 2
x− 3
x2 − 1

x−2

f). lim
x→−1

2x + 3x2 + 1

)

2

5x − 3 − 23 5x − 3 + 4

5
5
= lim
= lim
=
2
x→−1
x
→−
1
 3
 12
3
(x + 1)  3 5x − 3 − 2.3 5x − 3 + 4 


 5x − 3 − 2 5x − 3 + 4
x→−1

(


(x + 1) 


5(x + 1)

3

x→ 3

2x + 9 − 5
LỜI GIẢI
5x − 3+ 8

x→ 8

5x − 3 + 2
= lim
x→−1
x+ 1

c). lim

3

)

(

)

1− 3 1− x
1− (1− x)
1

= lim
= lim
2
x→ 0
b). x→0
x
 3
 x→0 1+ 3 1− x +
3
x 1+ 1− x + 1− x 


lim

(

)

(

3

1− x

)

2

=

1
3


3

c).

lim

x2 − 1 − 2
= lim
x→ 3
x− 3

(

)

2

x2 − 1 + 23 x2 − 1 + 4

(x − 3)(x + 3)
x+ 3
1
= lim
= lim
=
2

2
x→ 3
x→ 3
2
3 2
3 2
3 2
3 2
x − 1 + 2 x − 1+ 4
x − 1 + 2 x − 1+ 4
x→ 3

x −1

x→1

x+ 7 − 2=

3

Ta có :
1

x −1

Vậy lim
x→1

(

=

(

x + 7− 8
3

)

x+ 7 + 2 x+ 7 + 4

x− 1

)

=

.

2

x−2

(

3

)

x−1

2

x + 7 + 23 x + 7 + 4

(

3

)

x +1
2

x + 7 + 23 x + 7 + 4

=

1
6


2x + 9 + 5 ÷
.
÷
2
3
x + 23 x + 4 2x + 9 − 25 ÷


( )

x− 8


2x + 9 + 5 ÷
.
÷ = lim
2
x→ 8

3
x + 23 x + 4 2(x − 8) ÷
2



( )

(x

3

x+1
x−1



= lim 
e). lim
x→ 8

x→ 8
2x + 9 − 5



f). lim
x→−1

2

x +1
= lim
x
− 1 x→1
3
x + 7 + 23 x + 7 + 4
3



= lim 
x→ 8



)

(

x+ 7 − 2

3

d). lim

x− 8

2x + 9 + 5

( x)
3

2


+ 23 x + 4


=

10
24

x2 − 1
2x + 3x2 + 1

2

lim

3

)

(

Và


(x − 3) 


x2 − 1− 8

)(

(

)

4x − 3x + 1
2

x→−1

) = lim ( x − 1) ( 2x −

− 1 2x − 3x2 + 1
2

2

) = lim 2x −
(

3x2 + 1

x −1
2

x→−1

x→−1

)

3x2 + 1 = −4

Câu 9: Tìm các giới hạn sau:
a). lim
x→ 0

3 3
2
x − a + x− a
x + 9 + x + 16 − 7
b). lim x + 7 − x + 3 c). lim
,(
x
→

a
x→1
x
x− 1
x2 − a2

a > 0 ) d). lim
x→ 2

3

8x + 11 − x + 7
x2 − 3x + 2

1+ 2x − 3 1+ 3x
x→ 0
x2
LỜI GIẢI

e). lim

x + 9 + x + 16 − 7
x + 9 − 3+ x + 16 − 7
= lim
x
→
0
x
x
x + 9− 9

x+ 9− 3
x + 16 − 4 = lim
+ lim
= lim
+ lim
x→ 0
x + 9 + 3 x x→0
x→ 0
x→ 0
x
x

a). lim
x→ 0

(

)

(

x + 16 − 16

)

x + 16 + 4 x


= lim
x→ 0

x

(

)

x+ 9+ 3 x

+ lim
x→ 0

(

x

)

x + 16 + 4 x

1

= lim

x+ 9+ 3

x→ 0

1

+ lim

x + 16 + 4

x→ 0

=

7
24

3 3
3 3
2
2
b). lim x + 7 − x + 3 = lim x + 7 − 2 + 2 − x + 3
x→1
x→1
x− 1
x−1
3

x3 + 7 − 2
2 − x2 + 3
+ lim
x→1
x→1
x− 1
x−1
3

x + 7− 8
2 − x2 − 3
= lim
2
x→1 
+ lim

3 3
3 3
x→1
(2 + x2 + 3)(x − 1)
 x + 7 + 2 x + 7 + 4 (x − 1)


(x − 1)(x2 + x + 1)
−(x − 1)(x + 1)
= lim
+ lim
2
x→1 
x→1

3 3
3 3
(2 + x2 + 3)(x − 1)
 x + 7 + 2 x + 7 + 4 (x − 1)


2
x + x+ 4

x+ 1
3
= lim
+ lim
=
2
x→1
x→1
2
3 3
2+ x + 3 4
x + 7 + 23 x3 + 7 + 4
= lim

(

)

(

)

)

(

x − a + x− a

c). lim
x→a

x2 − a2
x− a

= lim
x→a

x −a
2

2

(

= lim
x→a

x→a

(

x − a. x + a.
3

d). lim
x→ 2

x−a

+ lim

x− a

(a> 0 )
= lim
x→ a

x −a
2

)

2

x−a
(x − a)(x + a)

(

x+ a

)

+ lim
x→ a

x− a
(x − a)(x + a)

2

x+ a

)

+ lim
x→a

1
x+ a

= lim
x→ a

x−a

x+ a

(

x+ a

)

+ lim
x→a

1
x+ a

3

8x + 11 − x + 7
8x + 11 − 3 + 3− x + 7
=
lim
x→ 2
x2 − 3x + 2
x2 − 3x + 2

8x + 11 − 3
3− x + 7
+ lim 2
2
x
→
2
x − 3x + 2
x − 3x + 2
8x + 11− 27
9− x − 7
= lim
+ lim
2
x→ 2 
x
→
2
 2
3
3
3+ x + 7 (x2 − 3x + 2)

 8x + 11 + 3 8x + 11 + 9 (x − 3x + 2)
8(x − 2)
−(x − 2)
= lim
+ lim
2
x→ 2 
x→ 2

3
3
(3+ x + 7)(x − 2)(x − 1)
 8x + 11 + 3 8x + 11 + 9 (x − 2)(x − 1)
8
−1
7
= lim
+ lim
=
2
x→ 2 
x→ 2
54

3
3
3+ x + 7 (x − 1)
 8x + 11 + 3 8x + 11 + 9 (x − 1)
= lim

3

x→ 2

(

)

(

)

(

)

e). L = lim
x→ 0

1+ 2x − 3 1+ 3x
x2

(

(

)

)

=

1
2a


 

 1  1+ 2x − 1 3 1+ 3x − 1 
 1
÷
2
3
= lim  
−
−
÷ = lim  
÷
2
÷
x→ 0 x 
x→ 0 x
x
x
3
3
  1+ 2x + 1
 
 
1+ 3x + 1+ 3x + 1÷


 




2
 2 3 1+ 3x + 23 1+ 3x + 2 − 3 1+ 2x − 3
1

= lim 
2
x→ 0  x
 3
 
3
1+ 2x + 1  1+ 3x + 1+ 3x + 1÷ 

 1 4 4 4 4 4 4 4 2 4 4 4 4 4 4 43  

A

2
  3

3
1+ 3x − 1
 1
1+ 3x − 1
1+ 2x − 1÷

= lim   2
+2
−3
÷
x→ 0 A
x
x
x
 
÷

 
3
3
 1 3
1+ 3x − 1
1+ 3x − 1
1+ 2x − 1 
= lim   2 1+ 3x + 1
+2
−3
÷
÷
x→ 0 A 
x
x
x
 

 


6 3 1+ 3x + 1
 1
÷
6
6
= lim  
+
−
÷
2
2
x→ 0 A
1+ 2x + 1÷
  3 1+ 3x + 3 1+ 3x + 1 3 1+ 3x + 3 1+ 3x + 1

 
1
1
= ( 4 + 2 − 3) = .
6
2
CÁCH 2:
1+ 2x − (1+ x) + (1+ x) − 3 1+ 3x
1+ 2x − (1+ x)
(1+ x) − 3 1+ 3x
L = lim
=
lim
+

lim
x→ 0
x→ 0
x→ 0
x2
x2
x2
−x2
3x3 + 3x2
= lim
+ lim
2
x→ 0 2 
x
1+ 2x + (1+ x) x→0 x2 ( 1+ x) 2 + ( 1+ x) 3 1+ 3x + 3 1+ 3x 




−1
3(x + 1)
1
1
= lim
+ lim
= − + 1=
2
x→ 0
x→ 0
2

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

(

(
(

)

(

)

)

(

(

)(

)

)

(

)

)

(

)

(

(

) (

)

(

)

)

Câu 10: Tính các giới hạn sau:
xn − nx + n − 1
xn − 1
lim
c).
x→1

x→1 xm − 1
(x − 1)2
LỜI GIẢI
2
n
x + x + ... + x − n
(x − 1) + (x2 − 1) + ... + xn − 1
a). lim
= lim
x→1
x→1
x− 1
x−1
n−1
(x − 1) + (x − 1)(x + 1) + ... + (x − 1)(x + xn− x + ... + 1)
= lim
x→1
x−1
(x − 1) 1+ (x + 1) + ... + (xn−1 + xn− x + ... + 1)
n(n + 1)
= lim
= 1+ 2 + 3+ ... + n =
x→1
x−1
2
x + x2 + ... + xn − n
x→1
x− 1

a). lim

b). lim


b). lim
x→1

xn − 1
(x − 1)(xn−1 + xn − 2 + xn− 3 + ... + 1) 1+ 1+ 1+ ... + 1 n
=
lim
=
=
xm − 1 x→1 (x − 1)(xm−1 + xm− 2 + xm− 3 + ... + 1) 1+ 1+ 1+ ... + 1 m

c). lim
x→1

xn − nx + n − 1
(xn − 1) − n(x − 1)
=
lim
x→1
(x − 1)2
(x − 1)2

(x − 1)(xn−1 + xn − 2 + xn− 3 + ... + 1) − n(x − 1)
x→1
(x − 1)2

= lim

(x − 1)(xn−1 + xn − 2 + xn− 3 + ... + 1− n)
xn −1 + xn− 2 + xn− 3 + ... + 1− n
=
lim
x→1
x →1
x− 1
(x − 1)2

= lim

(xn −1 − 1) + (xn− 2 − 1) + (xn − 3 − 1) + ... + (x − 1) + (1− 1)
x→1
x− 1
(x − 1)(xn− 2 + xn− 3 + ... + 1) + (x − 1)(xn − 3 + xn− 4 + ... + 1) + ... + (x − 1)
= lim
x→1
x− 1
(x − 1) (xn − 2 + xn− 3 + ... + 1) + (xn− 3 + xn − 4 + ... + 1) + ... + 1
= lim
x→1
x−1
(n − 1).n
= (n − 1) + (n − 2) + ... + 1 =
.
2
Câu 10: Tìm các giới hạn sau:
= lim

2 1− x − 3 8 − x
x→ 0
x

1). lim

x→1

x→ 4

(

x→1

5− x − x + 7
x2 − 1
3

3). lim
lim

3

2). lim

2

2 − x2 − 12

)(

x2 + x − 19 − 1

4).
33 4x3 − 24 + x + 2 − 8 2x − 3
x→ 2
4 − x2

)

5). lim

x + 12 − 2

7). lim

x2 + 2x + 6 − 4x + 1
x3 − 2x + 1

6). lim

x2 + 3 + 2x2 + 4x + 19 − 3x2 + 46
x2 − 1

x→1

x→1

9). lim

x→ 2

3

3x − 2 − 3 4x2 − x − 2
x2 − 3x + 2

3x + 2 − 3x − 2
x− 2

8). lim
x→ 2

3

x + 6 − 4 7x + 2
x− 2

10). lim
x→1

6x + 3 + 2x2 − 5x

( x − 1)

LỜI GIẢI
2 1− x − 8 − x
2 1− x − 2 + 2 − 3 8 − x
2 1− x − 2
1). lim

= lim
= lim
x→ 0
x
→
0
x
→
0
x
x
x
2− 3 8− x
4(1− x) − 4
8 − (8 − x)
+ lim
= lim
+ lim
2
x→ 0
x→ 0
x
x 2 1− x + 2 x→0 x  4 + 23 8 − x + 3 8 − x 


3

(

)

(

)

2


= lim
x→ 0

−4
2 1− x + 2

2). lim
x→1

+ lim
x→ 0

1
4 + 23 8 − x +

(

3

8− x

)

2

=

−4 1
11
+
=−
4 12
12

3x − 2 − 3 4x2 − x − 2
x2 − 3x + 2

3x − 2 − 1+ 1− 3 4x2 − x − 2
3x − 2 − 1
1− 3 4x2 − x − 2
=
lim
+
lim
x→1 x2 − 3x + 2
x→ 1
x2 − 3x + 2
x2 − 3x + 2

lim
x→1

3x − 2 − 1
3x − 2 − 1
= lim
• Tính lim
x→1 x2 − 3x + 2
x→1
(x − 1)(x − 2) 3x − 2 + 1

(

= lim
x→1

3(x − 1)
(x − 1)(x − 2)

(

= lim

)

x→1

3x − 2 + 1

1− 3 4x2 − x − 2
= lim
x→1
• Tính x→1 x2 − 3x + 2

lim

(x − 2)

)

(

)

3x − 2 + 1

=

3
3
=−
−1.2
2

1− (4x2 − x − 2)

(x

2


− 3x + 2 1+ 3 4x2 − x − 2 +


)

−(x − 1)(4x + 3)

= lim

3

(

3

)

2

4x2 − x − 2 


)

(

2


(x − 1)(x − 2) 1+ 3 4x2 − x − 2 + 3 4x2 − x − 2 


−(4x + 3)

−7 7
= lim
=
=
2
x→1
 3 2
 −1.3 3
3
2
(x − 2) 1+ 4x − x − 2 + 4x − x − 2 


3 7 5
Vậy giới hạn cần tìm: − + =
2 3 6
CÁCH 2:
x→1

)

(

3x − 2 − 1 1− 3 4x2 − x − 2
+
3x − 2 − 1+ 1− 4x − x − 2
x−1
x−1
lim
=

lim
x→1
x →1
x2 − 3x + 2
x2 − 3x + 2
x−1
2
−4x − 3
+
2
3
2
3x − 2 + 1 1+ 4x − x − 2 + 3 4x2 − x − 2
5
= lim
=
x→1
x− 2
6
3

2

(

3) L = lim
x→1

L = lim
x→ 1

)

5− x3 − 3 x2 + 7
x2 − 1

5− x3 − 2 + 2 − 3 x2 + 7
5− x3 − 2
2 − 3 x2 + 7
=
lim
+
lim
x→1
x→1
x2 − 1
x2 − 1
x2 − 1

g Tính lim
x→1

(

)

( 1− x) 1+ x + x2
5− x3 − 2
5− x3 − 4
=

lim
=
lim
x→1
x→1
x2 − 1
x2 − 1 5 − x3 + 2
( x − 1) ( x + 1) 5− x3 + 2

(

)(

)

(

)


= lim
x→1

(

− 1+ x + x2

( x + 1) (

)

=

)

5− x + 2
3

−3
3
=−
2.4
8

2 − 3 x2 + 7
lim
g Tính x→1 x2 − 1 = lim
x→1
= lim

1− x2

(

(x

2


− 1  4 + 3 x2 + 7 +



)

)

(



− 1  4 + 3 x2 + 7 + 3 x2 + 7 ÷


3 1
11
=−
Kết luận L = − −
8 12
24
x→1

(x

2

4). lim
x→ 4

(

)

2 − x2 − 12

)(

x2 + x − 19 − 1

Ta có 2 − x2 − 12 =

= lim
x→ 4

= lim
x→ 4

2 + x − 12

x2 + x − 19 − 1

( 4− x) ( 4 + x) ×
2 + x2 − 12
− ( 4 + x)

2 + x2 − 12

×

= lim
x →1

3

)

2

x2 + 7 ÷

−1


3 2
 4+ x + 7 +


)

2

=

2

(

(

3

)


x2 + 7 ÷

2

=−

1
12

x + 12 − 2

4 − (x2 − 12)

1

Ta có

)

8 − x2 + 7

=

16 − x2
2 + x − 12
2

( 4− x) ( 4 + x)

=

2 + x2 − 12

x2 + x − 19 + 1
x2 + x − 19 + 1
=
=
2
x + x − 19 − 1
x2 + x − 20

x2 + x − 19 + 1
( x − 4) ( x + 5)

x2 + x − 19 + 1
1
×
( x − 4) ( x + 5) x + 12 − 2
x2 + x − 19 + 1
1
−8 2 1
2
×
=
× × =−
x+ 5
9

x + 12 − 2 4 9 2

33 4x3 − 24 + x + 2 − 8 2x − 3
33 4x3 − 24 + x + 2 − 8 2x − 3
.
Đặt
f
x
=
( )
x→ 2
4 − x2
4 − x2

5). lim

33 4x3 − 24 − 6 + x + 2 − 2 + 8 − 8 2x − 3
x→ 2
4 − x2

limf ( x) = lim
x→ 2

33 4x3 − 24 − 6
x+ 2− 2
8 − 8 2x − 3
+ lim
+ lim
2
2

x→ 2
x
→
2
x
→
2
4− x
4− x
4 − x2

= lim

3
3
3
3
• Tính : lim 3 4x − 24 − 6 = lim3× 4x − 24 − 2
2
x→ 2
x→ 2
4− x
4 − x2
4x3 − 24 − 8
= 3lim
4 x3 − 8
4x3 − 24 − 8
x→ 2
3
2  3

3
3

=
3lim
=
3lim
4− x
4x − 24 + 2 4x − 24 + 2
x→ 2
x→ 2


4 − x2 .A
4 − x2 .A
E5555555555555555555555
F

(

= 3lim
x→ 2

(

)

(

)

) = 3lim −4( x

4( x − 2) x2 + 2x + 4

( 2− x) ( 2 + x) .A

(

A

2

x→ 2

)

(

) = 3lim −4.12 = − 9

+ 2x + 4

( 2 + x) .A

x→ 2

4.8

2

(

)

)


• Tính lim
x→ 2
= lim
x→ 2

( 2+ x) (

• Tính lim
x→ 2
= 8lim
x→ 2

x+ 2− 2
x + 2− 4
x− 2
= lim
= lim
2
x
→
2
x

→
2
4− x
4 − x2
x+ 2+ 2
( 2 + x) ( 2− x) x + 2 + 2

)(

(

−1

)

x+ 2+ 2

=−

)

(

)

1
4.

(

)

8 1− 2x − 3
1− ( 2x − 3)
8 − 8 2x − 3
=
lim
= 8lim
2
2
x
→
2
x
→
2
4− x
4− x
4 − x2 1+ 2x − 3
2( 2 − x)

( 2− x) ( 2+ x) ( 1+

)

2x − 3

= 8lim
x→ 2

)(

(

2
= 8× = 2
4.2
( 2+ x) 1+ 2x − 3

(

2

)

)

9 1
11
Vậy giới hạn cần tìm : limf ( x) = − − + 2 = −
x→ 2
2 4
4
x2 + 2x + 6 − 4x + 1
x3 − 2x + 1

7). lim

x→1

x2 + 2x + 6 − ( 4x − 1)

lim

x3 − 2x + 1

x→1

= lim
x→1

(x

3

= lim
x→1

(x

x2 + 2x + 6 − ( 4x − 1)

3

)(

)(

− 2x + 1

x2 + 2x + 6 + ( 4x − 1)

− 2x + 1

−15x2 + 10x + 5

)

x2 + 2x + 6 + 4x − 1

2

= lim
x→1

( x − 1) ( x

2

)

−5( x − 1) ( 3x + 1)

)(

+ x−1

)

x2 + 2x + 6 + 4x − 1

Phân tích x3 − 2x + 1 = (x − 1)(x2 + x − 1) , bằng sơ đồ Hoocne sau:
1
= lim
x→1

(x

2

1

0

-2

1

1

1

-1

0

−5( 3x + 1)

)(

+ x−1

6). L = lim
x→1

L = lim
x→1

= lim
x→1

)

x + 2x + 6 + 4x − 1
2

=

−20
10
=−
1.6
3

x2 + 3 + 2x2 + 4x + 19 − 3x2 + 46
x2 − 1

x2 + 3 − 2 + 2x2 + 4x + 19 − 5+ 7 − 3x2 + 46
x2 − 1

x2 + 3 − 2
2x2 + 4x + 19 − 5
7 − 3x2 + 46
+
lim
+
lim
x→1
x →1
x2 − 1
x2 − 1
x2 − 1

• Tính lim
x→1

x2 + 3 − 2
x2 + 3− 4
=
lim
x→1
x2 − 1
x2 − 1 x2 + 3 + 2

)

)(

(

= lim
x→1

(x

2

x2 − 1

)(

−1

)

x + 3+ 2
2

= lim
x →1

1
x + 3+ 2
2

=

1
4

2x2 + 4x + 19 − 5
2x2 + 4x + 19 − 25
= lim
2
x→1
x −1
x2 − 1 2x2 + 4x + 19 + 5

• Tính lim
x→1
= lim
x→1

= lim
x→1

(x

2

)(

(

2x2 + 4x − 6

)(

−1

( x + 1) (

)

2x2 + 4x + 19 + 5
2( x + 3)

)

2x + 4x + 19 + 5
2

= lim
x →1

=

)

2( x − 1) ( x + 3)

( x − 1) ( x + 1) (

)

2x2 + 4x + 19 + 5

2.4 2
=

2.10 5

(

)

−3 x2 − 1
7 − 3x2 + 46
49 − (3x2 + 46)
= lim
= lim
• Tính lim
x→1
x→1
x→1
x2 − 1
x2 − 1 7 + 3x2 + 46
x2 − 1 7 + 3x2 + 46
= lim
x→1

( 7+

−3
3x2 + 46

Kết luận L =

)

=−

)

)(

(

(

)(

3
14

1 2 3
61
+ −
=
4 5 14 140

3
x + 6 − 4 7x + 2
x + 6 − 4 7x + 2
. Đặt f ( x) =
x→ 2
x− 2
x− 2
3
4

3
x + 6 − 2 + 2 − 7x + 2
x+ 6− 2
2 − 4 7x + 2
limf ( x) = lim
= lim
+ lim
x→ 2
x→ 2
x→ 2
x→ 2
x− 2
x− 2
x− 2
3
x+ 6− 2
x + 6− 8
lim
= lim
2
x→ 2
• Tính x→ 2 x − 2
( x − 2)  3 x + 6 + 2.3 x + 6 + 4


1
1
= lim
=
2

x→ 2 3
12
3
x + 6 + 2. x + 6 + 4

8). lim

3

(

)

(

• Tính

)

2 − 4 7x + 2
4 − 7x + 2
16 − (7x + 2)
= lim
= lim
x→ 2
x→ 2
x→ 2
4
x− 2
( x − 2) 2 + 7x + 2

( x − 2) 2 + 4 7x + 2 4 + 7x + 2

lim

(

= lim
x→ 2

( x − 2) ( 2 +

Vậy limf ( x) =
x→ 2

−7(x − 2)
4

)(

7x + 2 4 + 7x + 2

)

)

(

= lim
x→ 2

( 2+

)(

−7
4

)(

7x + 2 4 + 7x + 2

)

=−

7
32

1 7
13
−
=−
.
12 32
96

3
3x + 2 − 3x − 2
3x + 2 − 3x − 2
. Đặt f ( x) =

x→ 2
x− 2
x− 2
3
3
3x + 2 − 2 + 2 − 3x − 2
3x + 2 − 2
2 − 3x − 2
Có limf ( x) = lim
= lim
+ lim
x→ 2
x→ 2
x→ 2
x→ 2
x− 2
x− 2
x− 2

9). lim

3

)

)


3

lim

• Tính

3x + 2 − 2
= lim
x→ 2
x− 2

( x − 2) (

3x + 2 − 8

)

2

3x + 2 + 2.3 3x + 2 + 4


3(x − 2)
3
1
= lim
= lim
=
2
2
x→ 2
x

→
2
 3
 4
3
( x − 2)  3 3x + 2 + 2.3 3x + 2 + 4
 3x + 2 + 2. 3x + 2 + 4


x→ 2

(

• Tính lim
x→ 2

)

(

x→ 2

Vậy limf ( x) =

= lim

lim
x→ 2

(x − 2)(2 + 3x − 2)

= lim
x→ 2

−3
2 + 3x − 2

=−

3
4

3
6 − x − 3 x2 + 4 lim x + 2 − x + 20
1+ 4x − 3 1+ 6x
;
;
.
lim
4
x→7
x→ 0
x
x+ 9− 2
x2 − 4

6x + 3 + 2x2 − 5x

( x − 1)

2

6x + 3 − (x + 2)

( x − 1)

x→1

−3(x − 2)

1 3
1
− =− .
4 4
2

Tương tự: Tìm
10). lim
x→1

)

2 − 3x − 2
4 − (3x − 2)
= lim
x
→
2
x− 2
(x − 2)(2 + 3x − 2)

= lim

x→ 2

3

2

= lim

6x + 3 − (x + 2) + 2(x2 − 2x + 1)

( x − 1)

x→1

+ lim
x→1

2

2(x2 − 2x + 1)

( x − 1)

2

6x + 3− (x + 2)2
− x2 + 2x − 1
−(x − 1)2

+ 2 = lim
+ 2 = lim
+ 2 = −1+ 2 = 1 .
2
2
x→1
x→1
x→1 (x − 1)2
(x − 1)
(x − 1)

= lim

Câu 10: Tìm các giới hạn sau:
2x4 − 5x3 + 3x2 + x − 1
1). lim
x→1
3x4 − 8x3 + 6x2 − 1
3). lim

n

x→ 0

n

lim

a+ x − a
x

n

2). lim

( 1+ x) ( 1+ 2x) ( 1+ 3x) − 1
x

x→ 0

4).

( 1+ 2x) ( 1+ 3x) ( 1+ 4x) − 1
x
x + 2 − 2x

x→ 0

5). lim

x − 1 − 3− x

x→ 2

x − 1 + x4 − 3x3 + x2 + 3

7). lim

2x − 2

x→ 2

9). lim
x→ 0

5

1+ 5x − 1
x

6). lim
x→1
8). lim
x→1

x−1
x + 3 + x3 − 3x
2

2x − 1 + x2 − 3x + 1
3

x − 2 + x2 − x + 1

20). lim
x→1

4

4x − 3 − 1
x− 1

LỜI GIẢI
2x4 − 5x3 + 3x2 + x − 1
1). L = lim
x→1
3x4 − 8x3 + 6x2 − 1
Phân tích 2x4 − 5x3 + 3x2 + x − 1 = (x − 1)(2x3 − 3x2 + 1) , bằng sơ đồ Hoocne sau:


1

2

-5

3

1

-1

2

-3

0

1

0

Phân tích 3x4 − 8x3 + 6x2 − 1 = (x − 1)(3x3 − 5x2 + x + 1) , bằng sơ đồ Hoocne sau:
1

3

-8

6

0

-1

3

-5

1

1

0

(x − 1)(2x3 − 3x2 + 1)
2x3 − 3x2 + 1
= lim 3
(thay x = 0 vào tử và mẫu
3
2

x→1 (x − 1)(3x − 5x + x + 1)
x→1 3x − 5x2 + x + 1

L = lim

vẫn còn dạng vô định

0
, nên tiếp tục phân tích đa thức thành nhân tử,
0

cả tử và mẫu).
Phân tích 2x3 − 3x2 + 1 = (x − 1)(2x2 − x − 1) , bằng sơ đồ Hoocne sau:
1

2

-3

0

1

2

-1

-1

0

Phân tích 3x3 − 5x2 + x + 1 = (x − 1)(3x2 − 2x − 1) , bằng sơ đồ Hoocne sau:
3

-5

1

1

1 3

-2

-1

0

(x − 1)(2x2 − x − 1)
2x2 − x − 1
=
lim
(thay x = 0 vào tử và mẫu vẫn còn
x→1 (x − 1)(3x2 − 2x − 1)
x→1 3x2 − 2x − 1

L = lim

dạng vô định

0
, nên tiếp tục phân tích đa thức thành nhân tử, cả tử và
0

mẫu).
L = lim
x→1

(x − 1)(2x + 1)
2x + 1 3
= lim
=
x
→
1
(x − 1)(3x + 1)
3x + 1 4

2). L = lim

( 1+ x) ( 1+ 2x) ( 1+ 3x) − 1 = lim x(1+ 2x)(1+ 3x) + lim 2x(1+ 3x) + lim 3x

x→ 0
x
x
= lim(1+ 2x)(1+ 3x) + lim2(1+ 3x) + 3 = 1+ 2 + 3 = 6
x→ 0

x→ 0

n

x→ 0

= lim

x→ 0 n

x

x→ 0

x→ 0

Tương tự: Tìm
3). L = lim

x→ 0

( 1+ x) ( 1+ 2x) ( 1+ 3x) ( 1+ 4x) − 1
L = lim
x

x→ 0

a + x − n a = lim
x→ 0
x
x

x

(

n

(a + x)n−1 + n (a + x)n− 2 .n a + ×××+ n an −1

1
n−1

(a + x)

n− 2 n

+ (a + x)
n

n

n− 1

. a + ×××+ a

=

1
n an−1
n

)

x


4). L = lim

n

( 1+ 2x) ( 1+ 3x) ( 1+ 4x) − 1 .
x

x→ 0

Đặt t =

n
n

Và lim t

( 1+ 2x) ( 1+ 3x) ( 1+ 4x) . Ta có x → 0 ⇒ t → 1
( 1+ 2x) ( 1+ 3x) ( 1+ 4x) − 1 = 9
−1
= lim

x
t−1
tn − 1
9

L
=
lim
=
lim
=
Vậy
n−1
n− 2
x→ 0 x
x→ 0
x t + t + ×××+ t + 1 n
x→ 0

x

x→ 0

(

5). L = lim
x→ 2

= lim
x→ 2

x + 2 − 2x

(

2(x − 2)

x + 2 + 2x

x−1

x + 2 + 2x

x→ 2

x − 1 + 3− x

(

x + 2 − 2x

= lim

x − 1 − 3− x

−(x − 2)

6). lim
x→1

)

)

) = lim − (

x→ 2

×

x − 1 + 3− x
x − 1− (3 − x)

x − 1 + 3− x

(

x + 2 + 2x

2

)

) =−1
4

x−1

= lim
x→1

x2 + 3 + x3 − 3x
x2 + 3 − 2 + x3 − 3x + 2
x− 1
1
x− 1

= lim
= lim
=2
x→1
x→1
2
3
x+ 1
2
x + 3 − 2 x − 3x + 2
+
x
+
x
−
2
+
x2 + 3 + 2
x− 1
x−1

7). L = lim

x − 1 + x4 − 3x3 + x2 + 3
2x − 2

x→ 2

= lim

x − 1 − 1+ x − 3x3 + x2 + 4
4

2x − 2

x→ 2

x − 1− 1

Tính M: lim
x→ 2
Tính N: lim
x→ 2

2x − 2

( x − 1) ( x

x − 1− 1

x4 − 3x3 + x2 + 4
2x − 2

x→ 2

2x + 2
= lim
2x
− 4 x→ 2 2
x − 1+ 1

x→ 2

)(

− x2 − x − 2

Vậy L = 1+ 0 = 1

×

(

2x + 2

)

x − 1+ 1

x→−1

3

3

2

x→ 2

2

3
3
x + x2 + x + 1 lim 2x − 1 − x
,
x→1
x+1
x −1

2x − 1 + x2 − 3x + 1
3

) = lim ( x − x − x − 2) (

2x + 2

2x − 4

Tương tự: Tìm lim

x→1

= lim

2x − 2

x→ 2

8). L = lim

2x − 2

x→ 2

+ lim

=M +N

=1

x4 − 3x3 + x2 + 4
3

= lim

x − 1− 1

= lim

x − 2 + x2 − x + 1

= lim
x→1

2x − 1 − 1+ x2 − 3x + 2
3

x − 2 + 1+ x2 − x

) =0

2x + 2


2x − 1 − 1 x2 − 3x + 2
+
= lim
x− 1
x−1
= lim
x→1
3
2
x→1
x− 2+1 x − x
+
x−1
x−1
9). L = lim

5

x→ 0

2

(

2x − 1 + 1
1

3

)

+ x− 2

2

x− 2 − 3 x− 2+ 1

= 0 . Vậy L = 0

+x

1+ 5x − 1
x

t5 − 1
. Ta có khi x → 0 thì t → 1
5
5(t − 1)
5(t − 1)
5
= lim
= lim 4 3 2
=1
Vậy L = lim
t→1 t5 − 1
t→1 (t − 1)(t4 + t3 + t2 + t + 1)
t→1 (t + t + t + t + 1)

Đặt t = 5 1+ 5x ⇒ t5 = 1+ 5x ⇒ x =

10). lim

4

x→1

= lim
x→1

(

4x − 3 − 1
4x − 3 − 1 = lim
= lim
x→1
(x − 1) 4 4x − 3 + 1 x→1 (x − 1)
x−1

(

4

4

)(

)

4x − 3 + 1

4x − 3 + 1

=

)

3). L = lim

2− x − 1
x−1

3

1+ x − 1− 2x
x + x2

4). L = lim

1+ 4x − 3 1+ 6x
x2

6). lim

x→ 0

5). L = lim
x→ 0

7). lim

3

(x

7

x→1

2

2). lim

)

2

+ 2004

4

2x − 1 + x − 2
x−1

7

)

4x − 3 + 1

x→1

5

x 2x − 1 + 3 3x − 2 − 2
x2 − 1

x→1

8). lim

2

4x + 4 + 9 − 6x − 5
x2

x→ 0

x +x

1− 2x − 2004

x

x→ 0

4

1+ x2 − 4 1− 2x

x→0

)(

4x − 3 + 1

4
=1
4

Câu 10: Tìm các giới hạn sau:
1). L = lim

(

4(x − 1)
4

LỜI GIẢI
2− x − 1
. Đặt t = 7 2 − x ⇒ t7 = 2 − x ⇒ x = 2 − t7
x−1
Ta có x → 1⇒ t → 1
t−1
t−1
−1
1
= lim
= lim

=−
Vậy L = lim
2
3
7
t→1 1− t7
t →1
t →1 1+ t + t2 + t3 + ×××+ t7
8
( 1− t) 1+ t + t + t + ×××+ t
1). L = lim

7

x→1

(x

2

2). lim
x→ 0

)

+ 2004

(

7

1− 2x − 2004

x

x→ 0

x

= limx.7 1− 2x + lim
x→ 0

(

x . 1− 2x + 2004
2 7

= lim

x→ 0

)

7

) = lim x . 1− 2x + lim 2004(

1− 2x − 1

(

2004

2 7

x→ 0

7

) = lim 2004(

1− 2x − 1
x

x

x→0

x→ 0

7

)

1− 2x − 1
x

7

)

1− 2x − 1
x


Đặt t = 7 1− 2x ⇒ t7 = 1− 2x ⇒ x =
Ta có khi x → 0 thì t → 1
Vậy lim

(

2004

7

= lim
t→1

) = lim 2.2004( t − 1) = lim

4008( t − 1)

1− 2x − 1
x

x→ 0

1− t7
2

1− t

7

t→1

t →1

(1− t)(1+ t + t2 + ×××+ t7 )

−4008
−4008
=
= −501
2
7
8
1+ t + t + ×××+ t

(x

2

Tương tự: lim

)

+ 2001 9 1− 5x − 2001
x

x→ 0

3

3). L = lim
x→ 0

L = lim
x→ 0

3

1+ x − 1− 2x
x + x2
2

4

3
1+ x2 − 1+ 1− 4 1− 2x
1+ x2 − 1
1− 4 1− 2x
=
lim
+ lim
2
2
x→ 0
x→ 0
x+ x

x+ x
x + x2

3
2
• Tính lim 1+ x − 1
x→ 0
x + x2
x2
= lim
= lim
2
x→ 0
 3
 x→0
3
2
2
x( 1+ x)  1+ x + 1+ x + 1
( 1+ x) 




)

(

(

x
3

1+ x

2

)

2


+ 1+ x + 1

3

=0

2

4
• Tính lim 1− 1− 2x
x→ 0
x + x2

1− 1− 2x

= lim

( x + x ) ( 1+

= lim

( 1+ x) ( 1+

x→ 0

x→ 0

2

4

1− 2x
2

4

)

= lim
x→ 0

(

2x

)(

x ( 1+ x) 1+ 1− 2x 1+ 1− 2x

)(

1− 2x 1+ 1− 2x

)

=

4

)

1
2

1 1
=
2 2
4
2x − 1 + 5 x − 2
4). L = lim
x→1
x− 1
4
4
2x − 1 − 1+ 1+ 5 x − 2
2x − 1 − 1
1+ 5 x − 2
L = lim

= lim
+ lim
=M +N
x→1
x→1
x→1
x− 1
x−1
x−1
Tính M:
4
2x − 1 − 1
2x − 1 − 1
2(x − 1)
lim
= lim
= lim
x→1
x
→
1
x
→
1
4
4
x− 1
( x − 1) 2x − 1 + 1
( x − 1) 2x − 1 + 1 2x − 1 + 1
Vậy L = 0 +

(

= lim
x→1

(

2

4

)(

2x − 1 + 1

)

)

2x − 1 + 1

=

1
2

(

)(

)


1+ 5 x − 2
x→1
x− 1

Tính N: lim

Đặt t = 5 x − 2 ⇒ t5 = x − 2 ⇒ x = t5 + 2
Ta có x → 1⇒ t → −1
1+ t
t+1
1
1
N = lim 5
= lim
= lim 4 3 2
=
t→−1 t + 1
t→−1 (t + 1)(t4 − t3 + t2 − t + 1)
t→−1 t − t + t − t + 1
5
Vậy L =

1 1 7
+ =
2 5 10

Tương tự tính: lim
x→ 0

5). L = lim
x→ 0

6
x + 1 + 3 x − 1 lim x − 6x + 5
, x→1
2
( x − 1)
x

1+ 4x − 3 1+ 6x
1+ 4x − (1+ 2x) + (1+ 2x) − 3 1+ 6x
=
lim
x→ 0
x2
x2

1+ 4x − (1+ 2x)
(1+ 2x) − 3 1+ 6x
+
lim
x→ 0
x2
x2

= lim

x→ 0

• Tính lim
x→ 0

(

)

1+ 4x − 1+ 4x + 4x2
1+ 4x − (1+ 2x)
−4
=
lim
= lim
= −2
x→0
x→0
2
x2
1+ 4x + 1+ 2x
x
1+ 4x + 1+ 2x

(

)

3
• Tính lim (1+ 2x) − 1+ 6x

x→ 0
x2

( 1+ 2x) − ( 1+ 6x)
3

= lim
x→ 0

= lim
x→ 0

= lim
x→ 0


x2 (1+ 2x)2 + (1+ 2x) 3 1+ 6x +

8x3 + 12x2

x2 (1+ 2x)2 + (1+ 2x) 3 1+ 6x +

8x + 12
(1+ 2x)2 + (1+ 2x) 3 1+ 6x +

(

3

(

3

(

3

)

2

1+ 6x 


)

2

1+ 6x 


1+ 6x

)

2

=4

Vậy L = −2 + 4 = 2

x 2x − 1 + 3 3x − 2 − 2
x 2x − 1 − 1+ 3 3x − 2 − 1
6). Ta có lim
=
lim
x→1
x→1
x2 − 1
x2 − 1
 x 2x − 1 − 1 3 3x − 2 − 1
= lim 
+

2
x→1 
x2 − 1 
 x −1




3
2
2x
−
x
−
1
3x
−

3

= lim 
+
2

x→1 
2


x − 1 x 2x − 1 + 1
x2 − 1  3 ( 3x − 2) + 3 3x − 2 + 1 


 


(

)(

) (

)







2
2x + x + 1
3

= lim 
+

x→1 
2


3
x
+
1
x
2x
−
1
+
1
)
(
( x + 1)  3 ( 3x − 2) + 3x − 2 + 1 

 


(

=

)

4 3 3
+ = .
4 6 2

7). lim

3

1+ x2 − 4 1− 2x
x2 + x

x→0

3

1+ x2 − 1 1− 4 1− 2x
+
1+ x − 1+ 1− 1− 2x
x
x
= lim
=
lim
2
2
x→0

x
→
0
x +x
x +x
x
3

lim

x→0

=

2

4

x

+ lim

2

 3 1+ x2  + 3 1+ x2 + 1

÷


lim ( x + 1)

x→ 0 x

1− 1− 2x

( 1+

4

1− 2x

)

x→ 0

0 + lim

x→0

=
8). lim

( 1+

2

4

)(

1− 2x 1+ 1− 2x

1
4x + 4 + 9 − 6x − 5

) = 1.
2

x2
4x + 4 − ( x + 2) + ( x − 3) + 9 − 6x

x→ 0

= lim

x→0

= lim

x→0

4x + 4 − ( x + 2)
x2

x2

+ lim

( x − 3) +

x→0

9 − 6x

x2

 4x + 4 − ( x + 2)   4x + 4 + ( x + 2) 
( x − 3) + 9 − 6x  ( x − 3) − 9 − 6x 






= lim
+ lim 
2
2
x→0
x
→
0
x
4x + 4 + ( x + 2) 
x ( x − 3) − 9 − 6x 




4x + 4 − (x + 2)2

(x + 3)2 − (9 − 6x)
+ lim
x→0 2 
x
4x + 4 + ( x + 2)  x→0 x2 ( x − 3) − 9 − 6x 





= lim

= lim

− x2

+ lim

x2

x2  4x + 4 + ( x + 2)  x→0 x2 ( x − 3) − 9 − 6x 




−1
1
1 1
5
= lim

+ lim
=− − =−
x→0 4x + 4 + x + 2
( ) x→0 ( x − 3) − 9 − 6x 4 6 12 .
x→0