GIỚI HẠN HÀM SỐ LƯỢNG GIÁC
sinx
Dạng 4: lim
=1
x→ 0
x
Câu 1: Tìm các giới hạn sau:
tan2x
sin5x
1). lim
2). lim
x
→
0
x→ 0
3x
x
1− cosx
sin5x.sin 3x.sin x
4). lim
5). lim
2
x→ 0
x
→
0
x
45x3
sin7x − sin5x
lim
x→ 0

sinx
1− cos5x
1− cos2 2x
7). lim
8). lim
x→ 0 1− cos3x
x→ 0
x.sinx
x.sinax
L = lim
x→ 0 1− cosax
LỜI GIẢI
sin5x
1 sin5x 1
= lim ×
=
1). lim
x→ 0
x→ 0 5
x
5x
5
tan2x
2 tan2x 2
= lim ×
=
2). lim
x→ 0
x→ 0 3
3x

2x
3
x
2sin2
1− cosx
2 = limtan x = 0
= lim
3). lim
x→ 0
x
→
0
x
x x→ 0
sinx
2
2sin cos
2
2

3). lim
x→ 0

1− cosx
sinx

6).

9).


2

x

x
sin ÷
1− cosx
1
2
2÷ = 1
4). lim
= lim
= lim 
2
2
x→ 0
x
→
0
x
→
0
x
2 x ÷ 2
x

÷
 2 
sin 5x.sin3x.sin x
1 sin 5x sin 3x sinx 1

= lim ×
×
×
=
5). lim
3
x→ 0
x
→
0
3 5x
3x
x
3
45x
sin7x − sin5x
2cos6xsin x
= lim
= lim2cos6x = 2
6). lim
x→ 0
x→ 0
x→ 0
sinx
sinx
2sin2

2

5x


5x 
sin ÷
1− cos5x
25 
2
2 ÷
7). lim
= lim
= lim ×
x→ 0 1− cos3x
x→ 0
x→ 0 9
3x
5x

÷
2sin2
÷
2
 2 
5x
3x
sin
2 = 1, lim 2 = 1)
( Vì lim
x→ 0
x→ 0
5x
3x

sin
2
2
2
( 1− cos2x) ( 1+ cos2x)
1− cos 2x
8). lim
= lim
x→ 0
x
→
0
x.sinx
x.sinx
2sin2

2

 3x 

÷ 25
× 2 ÷ =
9
 sin 3x ÷
÷

2 


= lim


2sin2 x ( 1+ cos2x)
x.sin x

x→ 0

= lim2( 1+ cos2x)
x→ 0

sinx
=4
x

ax
ax
ax
ax
cos
cos
2
2 = lim x ×cos ax = lim 2 ×
2
x→ 0
x→ 0
ax
ax
ax
a
2
2sin2

sin
sin
2
2
2
2
ax
ax
cos
2 = 1 và lim
2 = 2 ). Vậy L = 2 .
(Vì lim
x→ 0
x→ 0
ax
a
a
a
sin
2
2
Câu 2: Tìm các giới hạn sau:
1− cosax
sin x.sin 2x....sin nx
1− cosax
1). lim
2). lim
3). lim
(a ≠ 0)
n

x→ 0 1− cosbx
x→ 0
x→ 0
n!x
x2
sinx − tan x
tanx − sin x
sin x − sina
4). lim
5). lim
6). lim
3
3
x→ 0
x→ 0
x→ a
x− a
x
sin x
cosx − cosb
1− 2x + 1
7). lim
8). lim
9).
x→ b
x→ 0
x− b
sin2x
cos(a + x) − cos(a − x)
lim

x→ 0
x
LỜI GIẢI
ax

ax bx 
2sin2
sin

1− cosax
a
2 = lim  ×
2× 2 ÷
=
÷
1). L = lim
x→ 0 1− cosbx
x
→
0
bx
 b ax sin bx ÷
2sin2
÷
2

2
2 
ax
bx

sin
a
2
= 1, lim 2 = 1. Vậy L =
Vì lim
x→ 0
x→ 0
ax
bx
b
sin
2
2
sin x.sin 2x....sin nx
2). L = lim
x→0
n!xn
sin x.sin 2x....sin nx
sinx sin 2x sin nx
= lim
= lim
×
×××
n
x→ 0
x
→
0
x
2x

nx
1.2.3....nx
sin x
sin 2x
sinnx
Vì lim
= 1, lim
= 1,×××, lim
=1
x→ 0
x→ 0
x→ 0
x
2x
nx
Vậy L = 1.
x.sinax
= lim
9). L = lim
x→ 0 1− cosax
x→ 0

x.2sin

2

3). L = lim
x→ 0

Vậy L =


1− cosax
= lim
x→ 0
x2

a2
.
4

ax
ax

ax 
sin
2  sin
÷
a
2 = 1).
2 = lim 
2 ÷ (vì lim
x→ 0
x→ 0 4
ax
ax
x2

÷
÷
2

 2 

2sin2


4).

sinx − tanx
L = lim
= lim
x→ 0
x→ 0
x3

sin x
cosx = sinx ( cosx − 1)
x3
x3 cosx

sinx −
2

x

x
−2sin2 sinx
 sin 2 ÷ sin x −1
2
= lim
= lim 

×
÷ ×
x→ 0
x→ 0
x 2cosx
x3 cosx
 x ÷
÷
 2 
x
sin
2 = 1, lim sinx = 1, lim −1 = − 1 .
Vì lim
x→ 0
x→ 0
x→ 0 2cosx
x
x
2
2
1
Vậy L = −
2
sin x
x
tanx − sin x
− sin x
2sin2
1
−

cosx
5). lim
2
x→ 0
= lim cosx 3
= lim
= lim
sin3 x
x→ 0
x→ 0 cosxsin2 x
x→ 0 cosxsin2 x
sin x
2


x
sin ÷
1
2÷
.
2 x ÷

÷
 2  =1
= lim
2
x→ 0
2
 sinx 
cosx.

÷
 x 
x+ a
x− a
x− a
sin
sin
x
+
a
2
2 = limsin
2 = sina
×
x→ a
x− a
x−a
2
2
x+ b
x− b
x− b
−2sin
sin
sin
cosx − cosb

x+ b
2
2

2 = − sin b
= lim
= lim  − sin
7). lim
÷.
x→ b
x→ b
x→ b
x− b
x− b
2  x− b

2
−1
1
1− 2x + 1 = lim 2x ×
=−
8). lim
x→ 0 sin 2x
x→ 0
2
1+ 2x + 1
sin2x
sin x − sina
= lim
6). lim
x→a
x→ a
x− a


2sin

cos(a + x) − cos(a − x)
−2sinasinx
sin x
= lim
= lim
.(−2sina)
x
→
0
x
→
0
x
x
x
sinx
(Vì lim
= 1 ). Vậy L = −2sina
x→ 0
x
Câu 2: Tìm các giới hạn sau:
tanx − tanc
1− cos3 x
sin2 x − sin2 a
1). lim
2). lim
3). lim
x→ c

x→ 0 xsin x
x→a
x− c
x2 − a2
cosαx − cosβx
sin5x − sin 3x
πx
4). lim
5). lim
6). lim ( 1− x) tan
2
x→ 0
x
→
0
x
→
1
sin x
x
2
9). L = lim
x→ 0


7). xlim
→−2
9). lim

x3 + 8

tan(x + 2)

8). lim
x→ 0

1− cosx.cos2x.cos3x
1− cosx

sin ( a + 2x) − 2sin ( a + x) + sina
2

x

x→ 0

10) lim

tan ( a + 2x) − 2tan ( a + x) + tana

x→ 0

x2

LỜI GIẢI
1). lim
x→ c

tanx − tanc
sin(x − c)
1

1
sin(x − c)
= lim
×
=
= 1 ).
(vì lim
2
x
→
c
x
→
c
x− c
x− c
cosxcosc cos c
x− c

(

)

1− cos3 x
( 1− cosx) 1+ cosx + cos2 x
= lim
x→ 0 xsin x
x→ 0
xsinx


2). lim

x
2
2 . 1+ cosx + cos x = 3 .
= lim
1+ cosx + cos x = lim
x→ 0
x→ 0
x
x
x
x
2
x.2sin cos
2cos
2
2
2
2
2sin2

3). lim
x→a

= lim
x→a

=


x
2

(

2

)

sin

( sin x − sina) ( sin x + sina)
sin2 x − sin2 a
= lim
2
2
x
→
a
x −a
( x − a) ( x + a)

2cos

x+ a
x− a
x−a
x+ a
sin
sin

cos
( sinx + sina)
2
2 ×sinx + sina = lim
2 ×
2
x→a
x− a
x−a
x+ a
x+ a
2.
2
2

2cosa.sina sin2a
=
.
2a
2a

x(α + β)
x(α − β)
−2sin
×sin
cosαx − cosβx
2
2
lim
= lim

x→ 0
x→ 0
x2
x2
x(α + β)
x(α − β)
sin
sin
2
2
2
2 .lim (α + β) (α − β) (−2) = β − α .
= lim
×lim
x→ 0
x(α + β) x→0 x(α − β) x→0 2
2
2
2
2
sin5x − sin 3x
2cos4xsinx
= lim
= lim(2cos4x) = 2
5). lim
x→ 0
x→ 0
x→ 0
sin x
sinx

πx
6). L = lim ( 1− x) tan
. Đặt t = x − 1, vì x → 1⇒ t → 0
x→1
2
π
π π 
π
L = lim(−t)tan ( t + 1) = lim(−t)tan  + t ÷ = limtcot t
t→ 0
t→ 0
t→ 0
2
2
2
2


4).

π
π
π
cos t
t cos t
2
2
2 =2
= limt.
= lim

×
t→ 0
t
→
0
π
π
π
π
sin t
sin t
2
2
2


7). xlim
→−2

(

)

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

→−
2
tan(x + 2) x→−2
tan(x + 2)
tan(x + 2)

( Vì xlim
→−2

(

x+ 2
= 1 ).
tan(x + 2)

1− cosx.cos2x.cos3x
1− cosx
1
−
cosx.
cos2x.cos3x
+ ( 1− cos2x) cos3x + ( 1− cos3x)
(
)

8). lim
x→ 0

lim


1− cosx
1
−
cosx.
cos2x.cos3x
(
)
( 1− cos2x) cos3x + lim 1− cos3x
= lim
+ lim
x→ 0
x→ 0
x→0 1− cosx
1− cosx
1− cosx
3x
2sin2
2sin2 xcos3x
2
= limcos2x.cos3x + lim
+ lim
x→ 0
x→0
x→ 0
x
x
2sin2
2sin2
2
2

x→ 0

2


3x

 sin 2
÷

÷
3x
x
x

÷
4sin2 cos2 cos3x
2  = 1+ 4 + 9 = 14
2
2
= 1+ lim
+ lim9. 
2
x→ 0
x→ 0
x

x

sin2

sin

÷
2
2

x ÷

÷

2 
sin ( a + 2x) − 2sin ( a + x) + sina
9). lim
x→ 0
x2
sin ( a + 2x) − sin ( a + x) + sina − sin ( a + x)
= lim
x→ 0
x2

3x 
x

x
x
2cos a + ÷sin − 2cos a + ÷sin
2
2
2
2





= lim
2
x→ 0
x
x

3x 

x 
x
x
2sin  cos a +
÷− cos a + 2 ÷
−4sin sin ( a + x) sin
2
2




2
2
= lim
= lim
x→ 0
x→0

x2
x2
2


x
 sin 2 ÷
= lim ( −1) 
÷ sin ( a + x) = − sina
x→ 0
 x ÷
÷
 2 
10). lim
x→ 0

tan ( a + 2x) − 2tan ( a + x) + tana
x2

)


= lim

(

)

tan ( a + 2x) − tan ( a + x) − tan ( a + x) − tana
2


x
sinx
sinx
−
cos(a + 2x)cos(a + x) cos(a + x)cosa
= lim
x→ 0
x2


sin x 
cosa − cos(a + 2x)
sinx 
2sinxsin(a + x)
= lim 2 
÷ = lim
÷
2 
x→ 0 x
x
→
0
x  cos(a + 2x)cos(a + x)cosa 
 cos(a + 2x)cos(a + x)cosa 
x→ 0

2

 2sina

 sinx  
2sin(a + x)
.
= lim 
÷=
÷
3
x→ 0
 x   cos(a + 2x)cos(a + x)cosa  cos a
Câu 3: Tìm các giới hạn sau:
sinax + tan bx
cos3x − cos5x.cos7x
(a + b ≠ 0)
1). lim
2). lim
x→ 0
x→ 0
(a + b)x
x2
cosax − cosbx.coscx
x→ 0
x2

4). lim
x→ 0

3). lim
5). lim
x→ 0


sin ( a + x) − sin ( a − x)

tan ( a + x) − tan ( a − x)

sin2 2x − sin x.sin 4x
x→ 0
x4

2x + 1 − 3 x2 + 1
sinx

6). lim

 1
1 
−
8). lim

÷
x→ 0 sin x
tanx 


1− cos5x.cos7x
x→ 0
sin2 11x
sinx − sin 2x
lim
x
→

0

x
9).
x 1− 2sin2 ÷
2

7). lim

10). lim
x→ 0

1+ x2 − cosx
x2

LỜI GIẢI
1). lim sinax + tan bx = lim
x→ 0
x→ 0
(a + b)x
= lim
x→ 0

sin bx
sin bx
cosbx = lim sinax + lim
x→ 0 (a + b)x
x→ 0 (a + b)x.cosbx
(a + b)x


sinax +

a sinax
b
sin bx
a
b
×
+ lim
×
=
+
=1
x
→
0
a + b ax
(a + b)cosbx bx
a+ b a+ b

2). lim
x→ 0

cos3x − 1+ ( 1− cos5x) cos7x + 1− cos7x
cos3x − cos5x.cos7x
= lim
2
x→ 0
x
x2


( 1− cos5x) cos7x + lim 1− cos7x
cos3x − 1
lim
2
x
→
0
x→ 0
x
x2
x2
3x
5x
7x
−2sin2
2sin2 cos7x
2sin2
2 + lim
2
2
= lim
+ lim
x→ 0
x→ 0
x→ 0
x2
x2
x2
= lim

x→ 0

2

2

2


3x 

5x 

7x 
sin ÷
sin ÷
sin ÷


−9 
25cos7x
49
2 ÷ + lim
2 ÷ + lim 
2 ÷ = − 9 + 25 + 49 = 65
= lim .

x→ 0 2
x→ 0
x

→
0
3x
5x
7x
2
2
2 2 2
2

÷

÷
÷
÷
÷

÷
 2 
 2 
 2 


cosax − 1− ( cosbx − 1) coscx + 1− coscx
cosax − cosbx.coscx
= lim
2
x→ 0
x→ 0
x

x2
ax
bx
cx
−2sin2
2sin2
coscx
2sin2
2
2
2
= lim
+ lim
+ lim
x→ 0
x→ 0
x→ 0
x2
x2
x2

3). lim

2

2

2



ax 

bx 

cx 
sin ÷
sin
sin ÷
2
2
2
÷
−a 
b2 coscx 
c2 
2
2
2 ÷ = −a + b + c
= lim
×

÷ + lim
÷ + lim ×
x→ 0 2
x→ 0
x→ 0 2
2
2
 ax ÷
 bx ÷

 cx ÷
÷
÷
÷
 2 
 2 
 2 
2

4).

lim
x→ 0

sin ( a + x) − sin ( a − x)

tan ( a + x) − tan ( a − x)

= lim
x→ 0

2cosasinx
sin 2x
cos(a + x)cos(a − x)
= lim
x→ 0

5). lim
x→ 0


cosacos(a + x)cos(a − x)
= cos3 a
cosx

2x + 1 − 3 x2 + 1
sinx

2x + 1 − 1+ 1− 3 x2 + 1
2x + 1 − 1
1− 3 x2 + 1
= lim
+ lim
x→ 0
x→ 0
x→0
sinx
sin x
sin x
2
2x
−x
= lim
+ lim
2
x→ 0
sin x 2x + 1 + 1 x→0 sinx 1+ 3 x2 + 1 + 3 x2 + 1 





x
2
x
−x
2
= lim
×
+ lim
×
=
+ 0= 1
2
x→ 0 sin x
x→ 0 sin x
1+ 1
2x + 1 + 1
1+ 3 x2 + 1 + 3 x2 + 1
lim

(

)

(

)

(

)


sin 2x − sin x.sin 4x
sin 2x − 2sin xsin2xcos2x
= lim
4
x
→
0
x
x4
sin 2x ( 2sinxcosx − 2sinxcos2x)
2

2

6). lim
x→ 0

= lim

x4

x→ 0

= lim
x→ 0

2sin 2x.sin x ( cosx − cos2x)
x4


= lim
x→ 0

4sin 2x.sinx.sin

3x
x
.sin
2
2

x4


3x  
x
sin ÷  sin ÷
 sin2x   sin x  
2
2 ÷= 6
= lim6×
÷×
÷×
÷×
x→ 0
 2x   x   3x ÷  x ÷

÷
÷
 2   2 

1− cos5x.cos7x
7). lim
x→ 0
sin2 11x
5x
7x
2sin2 cos7x
2sin2
1− cos5x) cos7x + 1− cos7x
(
2
2
= lim
= lim
+ lim
x→ 0
x→ 0
x→ 0 sin2 11x
sin2 11x
sin2 11x


2

2


5x 

7x 

 sin 2 ÷
 sin 2 ÷

÷ cos7x

÷
 5x ÷
 7x ÷
÷
÷
25
 2  × 49 = 25 + 49 = 37
= lim  2 
×
+
lim
2
x→ 0
484 x→0  sin11x  2 484 484 484 242
 sin11x 
 11x ÷
 11x ÷




x
2sin2
 1
1 

 1
cosx 
1− cosx
2
−
−
= lim
8). lim

÷ = lim 
÷ = lim
x→ 0 sinx
x→ 0
x
x
tanx  x→ 0  sin x sinx  x→0 sinx

2sin cos
2
2
x
= limtan = 0 .
x→ 0
2
3x
−x
x
3x
2cos sin
sin − cos

sinx − sin 2x
2
2 = lim
2×
2 = −1
= lim
9). lim
x→ 0
x→ 0
x→ 0
x
xcosx
cosx

2 x
x 1− 2sin ÷
2
2

10). lim
x→ 0

= lim
x→ 0

1+ x2 − cosx
1+ x2 − 1+ 1− cosx
1+ x2 − 1
1− cosx
= lim

= lim
+ lim
2
2
x→ 0
x→ 0
x→ 0
x
x
x2
x2
2

x2

(

x

)

1+ x2 + 1

+ lim
x→ 0

2sin2
x2

x

2 = lim
x→ 0

2


x
sin ÷
1
1 
2 ÷ = 1 + 1 = 1.
+ lim ×
x
→
0
2  x ÷ 2 2
1+ x2 + 1

÷
 2 

Câu 3: Tìm các giới hạn sau:
π

1+ tan x − 1+ sin x
1). limtan2x.tan
2). lim
 − x÷
π
4

x
→
0
x→


x3
4

cosx
π
x+
2
sin(x − 1)
lim 2
x→1 x − 4x + 3
2sinx − 1
7). limπ
2
x→ 4cos x − 3
4).

lim

π
x→−
2

6


5). lim
x→π

8). limπ
x→

4

1+ cosx

( x − π)

2

2sinx − 1
2cos2 x − 1

π

sin  − x ÷
6


lim
π 1− 2sin x
x→
6

LỜI GIẢI
π


2x.tan  − x ÷ . Đặt t = x − π , vì x → π ⇒ t → 0
1). L = limtan
π
4
x→


4
4
4

3). lim
x→1

x+ 3− 2
tan(x − 1)

6).

9).





π
L = lim  tan  2t + ÷(−1)tant = lim ( cot2t.tan t )
t→ 0
2



 t→ 0
cos2t sint
cos2t sin t
cos2t 1
= lim
= lim
= lim
=
t→ 0 sin2t cost
t→ 0 2sintcost cost
t→ 0 2cos2 t
2
sinx ( x − cosx)
tanx − sin x
= lim
1+ tan x − 1+ sin x = lim
x→ 0

 x→0 x3.A.cosx
2). lim
3
3
x→ 0
x  114+ 4
tan
sin
x
44x2+ 4 14+ 4

43x ÷

÷

A

2


x
x
sin ÷
 sin x  
1
1
2sin xsin2
2
= .
2 = lim

÷. x ÷ .
= lim 3
x→ 0
x
2A.cosx
4


÷
x→ 0 x .A.cosx


÷
 2 
x+ 3− 2
= lim
3). lim
x→1 tan(x − 1)
x→1

(

x + 3− 4

)

x + 3 + 2 tan(x − 1)

= lim
x→ 1

x−1
1
tan(x − 1) x + 3 + 2

x− 1
1
1
= 1, lim
= )
x→1

tan(x − 1)
x+ 3+ 2 4
1
Vậy L = .
4
cosx
L = lim
π
π
π
4).
π . Đặt t = x + , vì x → − ⇒ t → 0
x→−
2 x+
2
2
2
 π
cos t − ÷
 2  = lim sin t = 1 .
L = lim
t→ 0
t→ 0
t
t
1+ cosx
5). L = lim
2 . Đặt t = x − π , vì x → π ⇒ t → 0
x→π
( x − π)

(Vì lim
x→1

2

L = lim
t→ 0

6). L = lim
x→1
L = lim
t→ 0

1+ cos(t + π)
1− cost
= lim
= lim
2
t
→
0
t→ 0
t
t2

2sin2
t2

t


t
sin ÷

1
2 = lim 
2÷ = 1 .
t→ 0 2
2
 t ÷
÷
 2 

sin(x − 1)
sin(x − 1)
= lim
2
x
→
1
x
x − 4x + 3
( − 1) ( x − 3) . Đặt t = x − 1, vì x → 1⇒ t → 0

sin t
sint 1
1
= lim
×
=− .
t.(t − 2) t→0 t t − 2

2
2sin x − 1
2sin x − 1
2sinx − 1
= lim
= lim
2
2
π
π 1− 4sin2 x
x→ 4 1− sin x − 3
x→
x→ 4cos x − 3
6
6
6

7). L = limπ

(

)


= lim
π
x→
6

2sinx − 1

−1
1
= lim
=−
π
( 1− 2sinx) ( 1+ 2sin x) x→ 1+ 2sin x 2
6

8). L = limπ
x→

= lim
π
x→
4

( 1−

4

2sin x − 1
2sin x − 1
2sin x − 1
= lim
= lim
2cos2 x − 1 x→ π 2 1− sin2 x − 1 x→ π 1− 2sin2 x
4
4

(


2sin x − 1

)(

2sinx 1+ 2sinx

)

)

= lim
π
x→
4

−1

1
=− .
2
1+ 2sinx

π

π


π
sin  − x ÷

sin  − x ÷
sin  x − ÷
6
6
 = lim
6
 = lim

9). = limπ
π
π
1− 2sin x x→

1  x→ 
π
x→
6
6 −2 sinx −
6 2 sinx − sin

÷

2
6÷




x π 
x π 

x π 
2sin  − ÷cos − ÷
cos − ÷
 2 12 
 2 12  = lim 1
 2 12  = 3
= lim
π
π 2
3

x
π


x
π


x π
x→
x→
6 4cos
6
cos + ÷
 2 + 12 ÷sin  2 − 12 ÷





 2 12 
Câu 4: Tìm các giới hạn sau:
1− 2sinx
1). x→ π π
−x
6
6
lim

π

sin  − x ÷
4 
2). lim
π
x→ 1− 2sin x
4

4). lim

x2 + 1 − cos2x
x2

5). lim

7). limπ

sin x − 3cosx
sin3x


8). lim

x→ 0

x→

3

x→ 0

1+ 2x − cosx − x
x2

1− cosx cos2x
x→ 0
x2

3).

lim
π
x→
4

6). lim

3

x→ 0


2 − 2cosx

π
sin  x − ÷
4


2x + 1 − 1− x
sin2x

1− 3 cosx
x→ 0
tan2 x

9). lim

LỜI GIẢI

1

π
2 sin x − ÷
2 sin x − sin ÷
1− 2sinx
lim
2
6
2sin x − 1
 = lim 
1). x→ π π

= lim
= lim 
π
π
π
−x
6
π
π
π
x→
x→
x→
x−
x−
x−
6
6
6
6
6
6
6
x π 
x π 
x π 
4cos + ÷sin  − ÷
sin  − ÷
2
12

2
12



 = lim
 2 12  2cos x + π  = 3
= lim

÷
π
π
x π 
x π 
x→
x→
 2 12 
6
6
2 − ÷
− ÷

 2 12 
 2 12 
π


π
sin  − x ÷
sin  x − ÷

4
4

 = lim

= lim
2). limπ
π
x→ 1− 2sin x
x→
2sinx − 1 x→ π4
4
4


π

π
sin  x − ÷
sin  x − ÷
4
4

 = lim

π

2  x→ 4 2  sinx − sin π 
2  sin x −
÷



4÷


2 ÷




x π
x π
x π
2sin  − ÷cos − ÷
2cos − ÷
 2 8
 2 8  = lim
 2 8 = 2
= lim
π
π

x
π


x
π



x
π
x→
x→
4
4
2cos + ÷sin  − ÷
cos + ÷
 2 8
 2 8
 2 8

2

π
−2 cosx −
÷
−2 cosx − cos ÷

÷
2
4
 = lim 
3). lim 2 − 2cosx = lim 
π
π
π

π



π


π

x→
x→
x→
4 sin x −
4
4
sin  x − ÷
sin  x − ÷

4÷
4
4




x π
x π
x π
4sin  + ÷sin  − ÷
2sin  + ÷
2
8
2

8



 = lim
 2 8 = 2
= lim
π
π
x π
 x π  x→
x π
x→
4 2sin
4 cos
 2 − 8 ÷cos 2 − 8 ÷
 2 − 8÷






x2 + 1 − cos2x
. Đặt f ( x) =
x2

4). lim
x→ 0


x2 + 1 − cos2x
x2

x2 + 1 − 1+ 1− cos2x
x2 + 1 − 1
1− cos2x
=
lim
+ lim
2
2
x
→
0
x
→
0
x
x
x2

lim
x→ 0

• Tính lim
x→ 0

x2 + 1 − 1
x2 + 1− 1
1

1
= lim
= lim
=
2
x→0 2
x→0
2
2
2
x
x + 1+ 1
x
x + 1+ 1

)

(

2

2
• Tính lim 1− cos2x = lim 2sin x = 2lim  sin x ÷ = 2
x→ 0
x→ 0
x→0
x2
x2
 x 


1
5
+ 2=
2
2
1+ 2x − cosx − x
1+ 2x − cosx − x
5). L = lim
. Đặt f ( x) =
2
x→ 0
x
x2
Vậy limf ( x) =
x→ 0

1+ 2x − (1+ x) + 1− cosx
1+ 2x − (1+ x)
1− cosx
= lim
+ lim
2
2
x
→
0
x
→
0
x

x
x2

L = lim
x→ 0

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

• Tính lim
x→ 0
= lim
x→ 0

2

x

(

(

− x2


)

1+ 2x − (1+ x)

= lim
x→ 0

(

−1

)

1+ 2x + (1+ x)

=−
2

• Tính lim
x→ 0

1− cosx
= lim
x→ 0
x2

1
2


x

x
sin ÷
2 = lim 1 
2÷ = 1
x→ 0 2
x
2
x2

÷
÷
 2 

2sin2

1 1
Vậy limf ( x) = − + = 0 .
x→ 0
2 2

)


3

6). L = lim
x→ 0


L = lim

3

x→ 0

2x + 1 − 1− x
. Đặt f ( x) =
sin 2x

= lim
x→ 0

2x + 1 − 1− x
sin 2x

3
2x + 1 − 1+ 1− 1− x
2x + 1 − 1
1− 1− x
= lim
+ lim
x
→
0
x
→
0
sin2x
sin2x

sin2x
3

• Tính

3

lim
x→ 0

2x + 1 − 1
= lim
x→ 0
sin2x


2sinxcosx 


(


sin 2x 


(

2x + 1− 1

)


2

2x + 1 + 3 2x + 1 + 1

2x
x
= lim
×
2
 x→ 0 sin x

3
2x + 1 + 3 2x + 1 + 1
cosx 


3

)

1

(

3

)



2x + 1 + 3 2x + 1 + 1

2

1− 1− x
x
x
1
1
= lim
= lim
×
=
• Tính lim
x→ 0
x
→
0
x
→
0
sin2x
sin x 2cosx 1+ 1− x
4
2sinxcosx 1+ 1− x

(

Vậy limf ( x) =
x→ 0


)

(

)

1 1 7
+ =
3 4 12

sin2 x − 3cos2 x
sin x − 3cosx = lim sinx − 3cosx =
3
π
x→ 3sinx − 4sin x
sin3x
sinx 3− 4sin2 x sin x + 3cosx
x→
3
3

7). limπ
=

4sin x − 3

(

)(


sinx 3− 4sin x sinx + 3cosx
2

8). L = lim
x→ 0

)

=

−1

(

sin x sinx + 3cosx

1− cosx cos2x
1− cos2 xcos2x
=
lim
x→ 0
x→ 0 2
x2
x 1+ cosx cos2x

(

cos2 x + sin2 x − cos2 xcos2x


(

x→ 0

= lim
x→ 0

x2 1+ cosx cos2x
2sin xcos x + sin x
2

2

2

(

x 1+ cosx cos2x
2

)

=

)

−2
3

1− cosx cos2x

x2

L = lim

= lim

)(

(

2

)

)

cos x ( 1− cos2x) + sin2 x

= lim
x→ 0

= lim
x→ 0

)

2

(


x2 1+ cosx cos2x

)

sin x
2cos x + 1
3
×
=
2
x
1+ cosx cos2x 2
2

2

1− 3 cosx
x→ 0
tan2 x

9). L = lim

x
2sin2 cos2 x
2
L = lim
= lim
2
x→ 0
x→ 0

x
x



tan2 x 1+ 3 cosx + 3 cosx 
4sin2 cos2 1+ 3 cosx +
2
2


1− cosx

= lim

(

)

(

)

cos2 x


3
cosx 

Câu 5: Tìm các giới hạn sau:

x→ 0

x
2cos2 1+ 3 cosx +
2

2

=

1
6.

(

3

)

2

cosx 


=

1
3



tanx − 1
2sin2 x − 1
3

1). limπ
x→

4

2). lim
x→ 0

1− cosxcos2x
x2

3).

 2

lim 
− cotx ÷
x→ 0 sin 2x


4). lim

1− 2x + 1 + sinx

3x + 4 − 2 − x
1− sin2x − cos2x

L = lim
x→ 0 1+ sin2x − cos2x

7).

x2 + 1 − cosx

5). lim

x→ 0

x2

x→0

π

cos cosx ÷
cos3x + 2cos2x + 2
lim
2

8). lim
π
sin3x
x→
x→0
x
3
sin2

2

6).

9).

lim

x→0

1− cosx

( 1−

1− x

)

2

LỜI GIẢI
1). L = limπ
x→

4

tanx − 1
2sin2 x − 1
3


tanx − 1
2


sin2 x − cos2 x  3 tan x + 3 tan x + 1


sin x − cosx
= lim
2
π
 3

x→
tan
x
+ 3 tanx + 1
4 cosx ( sin x − cosx) ( sinx + cosx)


1
1
= lim
=
2
π
 3
 3
x→
3

tanx
+
tanx
+
1
4 cosx ( sin x + cosx)


1− cosxcos2x
2). lim
x→ 0
x2
cosx( cosx − cos2x)
sin2 x + cos2 x − cosxcos2x
sin2 x
= lim
= lim 2 + lim
2
x→ 0
x→ 0
x→ 0
x
x
x2
3x
x

3x  
x
2cosxsin sin

sin ÷  sin ÷

2
2 = 1+ lim 
2 ÷×
2 ÷×9 ×cosx = 1+ 9 = 17
= 1+ lim
2
x→ 0
x
→
0
3x
x
8
8 8
x

÷
÷
÷ 
÷
 2   2 
L = lim
π
x→
4

(


)(

)

(

(

)

)

 2

− cotx ÷
3). L = lim

x→ 0 sin2x



1
cosx 
1− cos2 x
sin2 x
L = lim 
−
=
lim
=

lim
= limtanx = 0
÷
x→ 0 sinxcosx
sinx  x→0 sin xcosx x→ 0 sin xcosx x→ 0

4). L = lim
x→ 0

1− 2x + 1 + sin x
3x + 4 − 2 − x


L = lim

1− 2x + 1 + sinx

x→ 0

= lim
x→ 0

= lim
x→ 0

−2x

( −x

2


(

2

(

3x + 4 − 2 − x

3x + 4 + 2 + x

( x + 1) ( 1+

5). lim

)

)

− x 1+ 2x + 1

)

3x + 4 − 2 − x

x→ 0

3x + 4 + 2 + x

)(


1− 2x + 1

= lim

)

2x + 1

+ lim

(

x→ 0

x→ 0

)

sinx
3x + 4 − 2 − x

3x + 4 + 2 + x sinx
−x − x
2

x→ 0

+ lim


+ lim

sin x 3x + 4 + 2 + x
×
= 4− 4 = 0
x
−x − 1

x2 + 1 − cosx
x2

x→0



2
 1+ x2 − 1 + (1− cosx)


÷
  1+ x − 1÷
1− cosx 



I = lim
= lim 
+

x→0

x→0 
x2
x2
x2 




x
x
2sin2 ÷
sin2 ÷


x2
1
2 ÷ = lim 
2 ÷ =1
=lim 
+
+
2
2 ÷
x→0  2
x→0 
2
2
x
x
÷

1+ x + 1 2
 x ( 1+ x + 1)
÷

÷


4 

1− sin2x − cos2x
1+ sin2x − cos2x
1− sin2x − cos2x
1− cos2x − sin 2x
2sin2 x − 2sin xcosx
L = lim
= lim
= lim
x→ 0 1+ sin2x − cos2x
x→ 0 1− cos2x + sin2x
x→ 0 2sin2 x + 2sin xcosx
2sinx ( sinx − cosx)
sinx − cosx
= lim
= lim
= −1
x→ 0 2sinx sin x + cosx
x→ 0 sinx + cosx
(
)


6). L = lim
x→ 0

7). limπ
x→

= lim
x→

π
3

3

cos3x + 2cos2x + 2
4cos3 x + 4cos2 x − 3cosx
= lim
π
sin3x
3sinx − 4sin3 x
x→
3

cosx ( 2cosx + 3) ( 2cosx − 1)
sinx ( 2cosx − 1) ( 2cosx + 1)

=

π


cos cosx ÷
2


8). lim
x→0
2x
sin
2

π π

π

x
2 x
sin  − cosx ÷
sin  2sin2 ÷
 sin  π sin ÷
2
2 
2 2
 = lim
2

= lim
= lim  π.
=π



x→0
x
→
0
x
→
0
x
x
x
2
2
2
sin
sin
π sin


2
2
2 



9).

lim

x→0


= lim
x→∞

(

1− cosx

( 1−

1− x

)

2

( 1− cosx) ( 1+
1− 1− x

)(
2

1− x

)

(

2

1+ 1− x


)

2

 x
2sin2  ÷ 1+ 1− x
 2
= lim
x→0

2

 x
4. ÷
 2

)

2

( 1+
= lim
x→0

1− x
2

)


2

=2