8.3. C´ac di
.
nh l´y co
.
ba
’
nvˆe
`
h`am kha
’
vi 95
21. lim
x→0+0

cotgx

tgx
.(DS. 1)
22. lim
x→0

5
2+
√
9+x

1/ sin x
.(DS. e
−1/30
)
23. lim

x→0

cos x

cotg
2
x
.(DS. e
−1/2
)
24. lim
x→0+0

ln2x

1/lnx
.(DS. 1)
25. lim
x→0

1 + sin
2
x

1/tg
2
x
.(DS. e)
26. lim
x→0+0


cotgx

1/lnx
.(DS. e
−1
)
27. lim
x→π/2

sin x

tgx
.(DS. 1)
28. lim
x→0
e
+x
− e
−x
− 2x
sin x −x
.(DS. −2)
29. lim
x→0
e
−x
− 1+x −
x
2

2
e
x
3
− 1
.(DS. −
1
6
)
30. lim
x→0
e
−x
− 1+x
4
sin 2x
.(DS. −
1
2
)
31. lim
x→0
2
x
− 1 − xln2
(1 − x)
m
− 1+mx
.(DS.
ln

2
2
m(m −1)
)
32. lim
x→0

2
π
arccosx

1/x
.(DS. e
−
2
π
)
33. lim
x→∞
lnx
x
α
, α>0. (DS. 0)
34. lim
x→∞
x
m
a
x
,0<a= 1. (DS. 0)

35. lim
x→0+0
ln sin x
ln(1 − cos x)
.(DS.
1
2
)
36. lim
x→0

1
x
2
− cotg
2
x

.(DS.
2
3
)
37. lim
x→
π
4

tgx

tg2x

.(DS. e
−1
)
38. lim
x→
π
2
−0

tgx

cotgx
.(DS. 1)
96 Chu
.
o
.
ng 8. Ph´ep t´ınh vi phˆan h`am mˆo
.
tbiˆe
´
n
8.3.3 Cˆong th´u
.
c Taylor
Gia
’
su
.
’

h`am f(x) x´ac di
.
nh trong lˆan cˆa
.
n n`ao d´ocu
’
adiˆe
’
m x
0
v`a n lˆa
`
n
kha
’
vi ta
.
idiˆe
’
m x
0
th`ı
f(x)=f(x
0
)+
f

(x
0
)

1!
(x − x
0
)+
f

(x
0
)
2!
(x − x
0
)
2
+ ···+
+
f
(n)
(x
0
)
n!
(x − x
0
)
n
+ o((x − x
0
)
n

)
khi x → x
0
hay:
f(x)=
n

k=0
f
(k)
(x
0
)
k!
(x −x
0
)
k
+ o((x − x
0
)
n
),x→ x
0
. (8.15)
Dath´u
.
c
P
n

(x)=
n

k=0
f
(k)
(x
0
)
k!
(x −x
0
)
k
(8.16)
d
u
.
o
.
.
cgo
.
il`adath´u
.
c Taylor cu
’
a h`am f(x)ta
.
idiˆe

’
m x
0
, c`on h`am:
R
n
(x)=f( x) − P
n
(x)
d
u
.
o
.
.
cgo
.
il`asˆo
´
ha
.
ng du
.
hay phˆa
`
ndu
.
th ´u
.
n cu

’
a cˆong th´u
.
c Taylor.
Cˆong th ´u
.
c (8.15) du
.
o
.
.
cgo
.
i l`a cˆong th´u
.
c Taylor cˆa
´
p n dˆo
´
iv´o
.
i h`am
f(x)ta
.
i lˆan cˆa
.
ncu
’
ad
iˆe

’
m x
0
v´o
.
i phˆa
`
ndu
.
da
.
ng Peano (n´o c˜ung c`on
d
u
.
o
.
.
cgo
.
i l`a cˆong th´u
.
c Taylor di
.
aphu
.
o
.
ng). Nˆe
´

u h`am f(x)c´oda
.
o h`am
dˆe
´
ncˆa
´
p n th`ı n´o c´o thˆe
’
biˆe
’
udiˆe
˜
n duy nhˆa
´
tdu
.
´o
.
ida
.
ng:
f(x)=
n

k=0
a
k
(x − x
0

)
k
+ o((x − x
0
)
n
),x→ x
0
v´o
.
ic´achˆe
.
sˆo
´
a
k
du
.
o
.
.
c t´ınh theo cˆong th´u
.
c:
a
k
=
f
(k)
(x

0
)
k!
,k=0, 1, ,n.
8.3. C´ac di
.
nh l´y co
.
ba
’
nvˆe
`
h`am kha
’
vi 97
Nˆe
´
u x
0
= 0 th`ı (8.15) c´o da
.
ng
f(x)=
n

k=0
f
(k)
(0)
k!

x
k
+ o(x
n
),x→ 0 (8.17)
v`a go
.
i l`a cˆong th´u
.
c Macloranh (Maclaurin).
Sau d
ˆay l`a cˆong th´u
.
c Taylor ta
.
i lˆan cˆa
.
nd
iˆe
’
m x
0
=0cu
’
amˆo
.
tsˆo
´
h`am so
.

cˆa
´
p
I. e
x
=
n

k=0
x
k
k!
+ o(x
n
)
II. sin x = x −
x
3
3!
+
x
5
5!
+ ···+
(−1)
n
x
2n+1
(2n + 1)!
+ o(x

2n+2
)
=
n

k=0
(−1)
k
x
2k+1
(2k + 1)!
+ o(x
2n+2
)
III. cos x =
n

k=0
(−1)
k
x
2k
(2k!)
+ o(x
2n+1
)
IV. (1 + x)
α
=1+
n


k=1
α(α −1) (α −k +1)
k!
x
k
+ o(x
n
)
=1+
n

k=1

α
k

x
k
+ o(x
n
)
α(α −1) (α − k +1)
k!
=












α
k


nˆe
´
u α ∈ R,
C
k
α
nˆe
´
u α ∈ N.
Tru
.
`o
.
ng ho
.
.
p riˆeng:
IV
1
.

1
1+x
=
n

k=0
(−1)
k
x
k
+ o(x
n
),
IV
2
.
1
1 − x
=
n

k=0
x
k
+ o(x
n
).
98 Chu
.
o

.
ng 8. Ph´ep t´ınh vi phˆan h`am mˆo
.
tbiˆe
´
n
V. ln(1 + x)=
n

k=1
(−1)
k−1
k
x
k
+ o(x
n
).
ln(1 − x)=−
n

k=1
x
k
k
+ o(x
n
).
Phu
.

o
.
ng ph´ap khai triˆe
’
n theo cˆong th´u
.
c Taylor
Nhu
.
vˆa
.
y, d
ˆe
’
khai triˆe
’
n h`am f(x) theo cˆong th´u
.
c Taylor ta pha
’
i´ap
du
.
ng cˆong th´u
.
c
f(x)=T
n
(x)+R
n+1

(x),
T
n
(x)=
n

k=0
a
k
(x − x
0
)
k
,
a
k
=
f
(k)
(x
0
)
k!
· (8.18)
1) Phu
.
o
.
ng ph´ap tru
.

.
ctiˆe
´
p: du
.
.
a v`ao cˆong th´u
.
c (8.18). Viˆe
.
csu
.
’
du
.
ng cˆong th´u
.
c (8.18) dˆa
˜
nd
ˆe
´
nnh˜u
.
ng t´ınh to´an rˆa
´
tcˆo
`
ng kˆe
`

nh m˘a
.
c
d`u n´o cho ta kha
’
n˘ang nguyˆen t˘a
´
cd
ˆe
’
khai triˆe
’
n.
2) Phu
.
o
.
ng ph´ap gi´an tiˆe
´
p: du
.
.
a v`ao c´ac khai triˆe
’
n c´o s˘a
˜
n I-V sau
khi d˜abiˆe
´
ndˆo

’
iso
.
bˆo
.
h`am d˜a cho v`a lu
.
u´ydˆe
´
n c´ac quy t˘a
´
c thu
.
.
chiˆe
.
n
c´ac ph´ep to´an trˆen c´ac khai triˆe
’
n Taylor.
Nˆe
´
u
f(x)=
n

k=0
a
k
(x − x

0
)
k
+ o((x − x
0
)
n
)
g(x)=
n

k=0
b
k
(x − x
0
)
k
+ o((x − x
0
)
n
)
th`ı
a) f(x)+g(x)=
n

k=0
(a
k

+ b
k
)(x − x
0
)
k
+ o((x − x
0
)
n
);
8.3. C´ac di
.
nh l´y co
.
ba
’
nvˆe
`
h`am kha
’
vi 99
b) f(x)g(x)=
n

k=0
c
k
(x − x
0

)
k
+ o((x − x
0
)
n
)
c
k
=
k

p=0
a
p
b
k−p
c) F (x)=f[g(x)] =
n

j=0
a
j

n

k=0
(b
k
(x −x

0
)
k
− x
0

j
+o


n

k=0
b
k
(x −x
0
)
k
− x
0

n

3) D
ˆe
’
khai triˆe
’
n c´ac phˆan th´u

.
ch˜u
.
uty
’
theo cˆong th´u
.
c Taylor thˆong
thu
.
`o
.
ng ta biˆe
’
udiˆe
˜
n phˆan th´u
.
cd
´odu
.
´o
.
ida
.
ng tˆo
’
ng cu
’
ad

ath´u
.
c v`a c´ac
phˆan th´u
.
cco
.
ba
’
n (tˆo
´
i gia
’
n !) rˆo
`
i´apdu
.
ng VI
1
,IV
2
.
4) Dˆe
’
khai triˆe
’
n t´ıch c´ac h`am lu
.
o
.

.
ng gi´ac thˆong thu
.
`o
.
ng biˆe
´
ndˆo
’
i
t´ıch th`anh tˆo
’
ng c´ac h`am.
5) Nˆe
´
u cho tru
.
´o
.
c khai triˆe
’
nd
a
.
o h`am f

(x) theo cˆong th´u
.
c Taylor
th`ı viˆe

.
c t`ım khai triˆe
’
n Taylor cu
’
a h`am f(x)d
u
.
o
.
.
c thu
.
.
chiˆe
.
nnhu
.
sau.
Gia
’
su
.
’
cho biˆe
´
t khai triˆe
’
n
f


(x)=
n

k=0
b
k
(x − x
0
)
k
+ o((x − x
0
)
n
),
b
k
=
f
(k+1)
(x
0
)
k!
·
Khi d´otˆo
`
nta
.

i f
(n+1)
(x
0
)v`adod´o h`am f(x) c´o thˆe
’
biˆe
’
udiˆe
˜
ndu
.
´o
.
i
da
.
ng
f(x)=
n+1

k=0
a
k
(x − x
0
)
k
+ o((x − x
0

)
n+1
)
= f(x
0
)+
n

k=0
a
k+1
(x − x
0
)
k+1
+ o((x − x
0
)
n+1
)
trong d
´o
a
k+1
=
f
(k+1)
(x
0
)

(k + 1)!
=
f
(k+1)
(x
0
)
k!
·
1
k +1
=
b
k
k +1
·
100 Chu
.
o
.
ng 8. Ph´ep t´ınh vi phˆan h`am mˆo
.
tbiˆe
´
n
Do d´o
f(x)=f(x
0
)+
n


k=0
b
k
k +1
(x − x
0
)
k+1
+ o((x − x
0
)
n+1
) (8.19)
trong d
´o b
k
l`a hˆe
.
sˆo
´
cu
’
adath´u
.
c Taylor dˆo
´
iv´o
.
i h`am f


(x).
C
´
AC V
´
IDU
.
V´ı du
.
1. Khai triˆe
’
n h`am f(x) theo cˆong th´u
.
c Maclaurin dˆe
´
nsˆo
´
ha
.
ng
o(x
n
), nˆe
´
u
1) f(x)=(x +5)e
2x
;2)f(x)=ln
3+x

2 − x
Gia
’
i 1) Ta c´o f(x)=xe
2x
+5e
2x
.
´
Ap du
.
ng I ta thu du
.
o
.
.
c
f(x)=x

n−1

k=0
2
k
x
k
k!
+ o(x
n−1
)


+5

n

k=0
2
k
x
k
k!
+ o(x
n
)

=
n−1

k=0
2
k
k!
x
k+1
+
n

k=0
5 · 2
k

k!
x
k
+ o(x
n
).
V`ı
n−1

k=0
2
k
x
k+1
k!
=
n

k=1
2
k−1
(k −1)!
x
k
nˆen ta c´o
f(x)=5+
n

k=1


2
k−1
(k −1)!
+
5 · 2
k
k!

x
k
+ o(x
n
)
=
n

k=0
2
k−1
k!
(k + 10)x
k
+ o(x
n
).
2) T`u
.
d˘a
’
ng th´u

.
c
f(x)=ln
3
2
+ln

1+
x
3

−ln

1 −
x
2

v`a V ta thu d
u
.
o
.
.
c
f(x)=ln
3
2
+
n


k=1
1
k

1
2
k
+
(−1)
k−1
3
k

x
k
+ o(x
n
). 
8.3. C´ac di
.
nh l´y co
.
ba
’
nvˆe
`
h`am kha
’
vi 101
V´ı du

.
2. Khai triˆe
’
n h`am f(x) theo cˆong th´u
.
c Taylor ta
.
i lˆan cˆa
.
nd
iˆe
’
m
x
0
= −1dˆe
´
nsˆo
´
ha
.
ng o((x +1)
2n
)nˆe
´
u
f(x)=
3x +3
√
3 − 2x − x

2
·
Gia
’
i. Ta c´o
f(x)=
3(x +1)

4 − (x +1)
2
=
3
2
(x +1)

1 −
(x +1)
2
4

−
1
2
.
´
Ap du
.
ng cˆong th´u
.
c IV ta thu du

.
o
.
.
c
f(x)=
3
2
(x +1)+
3
2
(x +1)
n−1

k=1


−
1
2
k


(−1)
k
(x +1)
2k
4
k
+ o((x +1)

2n
)
trong d
´o


−
1
2
k


(−1)
k
=(−1)
k

−
1
2

−
1
2
− 1



−
1

2
−(k − 1)

k!
=
(2k −1)!!
2
k
k!
·
Do d
´o
f(x)=
3
2
(x +1)+
n−1

k=1
3(2k −1)!!
2
3k+1
k!
(x +1)
2k+1
+ o((x +1)
2n
). 
V´ı du
.

3. Khai triˆe
’
n h`am f(x) theo cˆong th´u
.
c Taylor ta
.
i lˆan cˆa
.
nd
iˆe
’
m
x
0
=2dˆe
´
nsˆo
´
ha
.
ng o((x −2)
n
), nˆe
´
u
f(x)=ln(2x −x
2
+3).
Gia
’

i. Ta biˆe
’
udiˆe
˜
n
2x − x
2
+3=(3− x)(x + 1) = [1 − (x − 2)][3 + (x − 2)]
= 3[1 − (x − 2)]

1+
x − 2
3

.
102 Chu
.
o
.
ng 8. Ph´ep t´ınh vi phˆan h`am mˆo
.
tbiˆe
´
n
T`u
.
d
´o suy ra r˘a
`
ng

f(x) = ln3 + ln[1 −(x − 2)] + ln

1+
x − 2
3

v`a ´ap du
.
ng cˆong th´u
.
cVtathud
u
.
o
.
.
c
f(x) = ln3 −
n

k=1
1
k
(x − 2)
k
+
n

k=1
(−1)

k−1
(x − 2)
k
k3
k
+ o((x − 2)
n
)
= ln3 +
n

k=1

(−1)
k−1
3
k
− 1

(x − 2)
k
k
+ o((x − 2)
n
). 
V´ı du
.
4. Khai triˆe
’
n h`am f(x) = ln cos x theo cˆong th ´u

.
c Maclaurin
dˆe
´
nsˆo
´
ha
.
ng ch´u
.
a x
4
.
Gia
’
i.
´
Ap du
.
ng III ta thu du
.
o
.
.
c
ln(cos x)=ln

1 −
x
2

2
+
x
4
24
+ o(x
4
)

= ln(1 + t),
trong d´otad˘a
.
t
t = −
x
2
2
+
x
4
24
+ o(x
4
).
Tiˆe
´
p theo ta ´ap du
.
ng khai triˆe
’

nV
ln(cos x) = ln(1 + t)=t −
t
2
2
+ o(t
2
)
=

−
x
2
2
+
x
4
24
+ o(x
4
) −
1
2

−
x
2
2
+
x

4
4
+ o(x
4
)

2

+ o


−
x
2
2
+
x
4
24
+ o(x
4
)

2

= −
x
2
2
+

x
4
24
−
x
4
8
+ o(x
4
)=−
x
2
2
−
x
4
12
+ o(x
4
). 
V´ı du
.
5. Khai triˆe
’
n h`am f(x)=e
x cos x
theo cˆong th´u
.
c Maclaurin
d

ˆe
´
nsˆo
´
ha
.
ng ch´u
.
a x
3
.
Gia
’
i. Khai triˆe
’
ncˆa
`
n t`ım pha
’
ic´oda
.
ng
e
x cos x
=
3

k=0
a
k

x
k
+ o(x
3
).
8.3. C´ac di
.
nh l´y co
.
ba
’
nvˆe
`
h`am kha
’
vi 103
V`ı x cos x = x +0(x), (x cos x)
k
= x
k
+ o(x
k
), k =1, 2, nˆen
trong cˆong th´u
.
c
e
w
=
n


k=0
w
k
k!
+ o(w
n
),w= x cos x
ta cˆa
`
nlˆa
´
y n = 3. Ta c´o
w = x cos x = x −
x
3
2!
+ o(x
4
)
w
2
= x
2
+ o(x
3
),w
3
= x
3

+ o(x
3
)
v`a do d´o
e
x cos x
=
3

k=0
w
k
k!
+ o(w
3
)
=1+x −
x
3
2!
+ o(x
4
)+
1
2

x
2
+0(x
3

)

+
1
3!

x
3
+ o(x
3
)

+0(x
3
)
=1+x +
1
2
x
2
−
1
3
x
3
+ o(x
3
). 
V´ı du
.

6. Khai triˆe
’
n theo cˆong th´u
.
c Maclaurin d
ˆe
´
n o(x
2n+1
)dˆo
´
iv´o
.
i
c´ac h`am
1) arctgx, 2) arc sin x.
Gia
’
i. 1) V`ı
(arctgx)

=
1
1+x
2
=
n

k=0
(−1)

k
x
2k
+ o(x
2n+1
)
nˆen theo cˆong th´u
.
c (8.19) ta c´o
arctgx =
n

k=0
(−1)
k
x
2k+1
(2k +1)
+ o(x
2n+2
).
V´o
.
i n =2tathud
u
.
o
.
.
c

arctgx = x −
x
3
3
+
x
5
5
+ o(x
6
)
104 Chu
.
o
.
ng 8. Ph´ep t´ınh vi phˆan h`am mˆo
.
tbiˆe
´
n
2) Ta c´o
(arcsinx)

=
1
√
1 − x
2
=1+
n


k=1
(−1)
k


−
1
2
k


x
2k
+ o(x
2n+1
)
=1+
n

k=1
(−1)
k
(2k − 1)!!
2
k
k!
x
2k
+ o(x

2n+1
).
T`u
.
d´o ´ap du
.
ng cˆong th´u
.
c (8.19) ta c´o
arc sin x = x +
n

k=0
(2k −1)!!
2
k
k!(2k +1)
x
2k+1
+ o(x
2n+2
).
V´o
.
i n =2tathudu
.
o
.
.
c

arc sin x = x +
1
6
x
3
+
3
40
x
5
+ o(x
6
). 
V´ı du
.
7. Khai triˆe
’
n h`am f(x)=tgx theo cˆong th´u
.
c Maclaurin d
ˆe
´
n
o(x
5
).
Gia
’
i. Ta s˜e d`ung phu
.

o
.
ng ph´ap hˆe
.
sˆo
´
bˆa
´
tdi
.
nh m`a nˆo
.
i dung du
.
o
.
.
c
thˆe
’
hiˆe
.
n trong l`o
.
i gia
’
i sau dˆa y .
V`ıtgx l`a h`am le
’
v`a tgx = x + o(x)nˆen

tgx = x + a
3
x
3
+ a
s
x
5
+ o(x
6
).
Ta su
.
’
du
.
ng cˆong th´u
.
c sin x =tgx·cos x v`a c´ac khai triˆe
’
nIIv`aIII
ta c´o
x −
x
3
3
+
x
5
5

+ o(x
6
)=

x + a
3
x
3
+ a
5
x
5
+ o(x
6
)


1 −
x
2
2!
+
x
4
4!
+0(x
5
)

Cˆan b˘a

`
ng c´ac hˆe
.
sˆo
´
cu
’
a x
3
v`a x
5
o
.
’
hai vˆe
´
ta thu d
u
.
o
.
.
c





−
1

6
= −
1
2
+ a
3
1
5!
=
1
4!
−
a
3
2!
+ a
5
8.3. C´ac di
.
nh l´y co
.
ba
’
nvˆe
`
h`am kha
’
vi 105
T`u
.

d
´o suy ra a
3
=
1
3
, a
5
=
2
15
.Dod´o
tgx = x +
x
3
3
+
2x
5
15
+ o(x
6
). 
V´ı du
.
8.
´
Ap du
.
ng cˆong th´u

.
c Maclaurin dˆe
’
t´ınh c´ac gi´o
.
iha
.
n sau:
1) lim
x→0
sin x −x
x
3
, 2) lim
x→0
e
−
x
2
2
−cos x
x
3
sin x
·
Gia
’
i. 1)
´
Ap du

.
ng khai triˆe
’
ndˆo
´
iv´o
.
i h`am sin x v´o
.
i n = 2 ta c´o
lim
x→0
sin x −x
x
3
= lim
x→0
x −
x
3
3!
+ o(x
4
) − x
x
3
= −
1
3!
+ lim

x→0
o(x
4
)
x
3
= −
1
3!
+0=−
1
3!
·
2)
´
Ap du
.
ng c´ac khai triˆe
’
nba
’
ng dˆo
´
iv´o
.
i e
t
, cost, sin t cho tru
.
`o

.
ng
ho
.
.
p n`ay ta c´o
lim
x→0
e
−
x
2
2
− cos x
x
3
sin x
= lim
x→0
1 −
x
2
2
+
x
4
8
+ o(x
4
) −1+

x
2
2
−
x
4
24
+0(x
4
)
x
3
(x +0(x))
= lim
x→0
x
4
8
−
x
4
24
+0(x
4
)
x
4
+0(x
4
)

= lim
x→0
1
8
−
1
24
+
o(x
4
)
x
4
1+
0(x
4
)
x
4
=
1
8
−
1
24
+0
1+0
=
1
12

· 
B
`
AI T
ˆ
A
.
P
Khai triˆe
’
n c´ac h`am theo cˆong th´u
.
c Maclaurin d
ˆe
´
n o(x
n
) (1-8)
1. f(x)=
1
3x +4
.(DS.
n

k=0
(−1)
k
3
k
4

k+1
x
k
+ o(x
n
))
106 Chu
.
o
.
ng 8. Ph´ep t´ınh vi phˆan h`am mˆo
.
tbiˆe
´
n
2. f(x)=
1
√
1+4x
.(DS.
n

k=0
(−1)
k
2
k
(2k−1)!!
k!
x

k
+ o(x
n
))
3. f(x)=
1
(1 − x)
2
.(DS.
n

k=0
(k +1)x
k
+ o(x
n
))
4. f(x)=ln
2 − 3x
2+3x
.(DS. ln
2
3
+
n

k=1
(−4)
k
− 9

k
k6
k
x
k
+ o(x
n
))
5. f(x) = ln(x
2
+3x + 2).
(DS. ln2 +
n

k=1
(−1)
k−1
k
(1 + 2
−k
)x
k
+ o(x
n
))
6. f(x) = ln(2 + x −x
2
). (DS. ln2 +
n


k=1
(−1)
k−1
− 2
−k
k
x
k
+ o(x
n
))
7. f(x)=
1 − 2x
2
2+x − x
2
.
(DS.
1
2
+
n

k=1
(−1)
k+1
− 7 · 2
−(k+1)
3
x

k
+ o(x
n
))
8. f(x)=
3x
2
+5x −5
x
2
+ x − 2
.(D
S.
5
2
+
n

k=1

(−1)
k
2
−(k+1)
−1

x
k
+ o(x
n

)).
Khai trˆe
’
n h`am theo cˆong th´u
.
c Maclaurin d
ˆe
´
n0(x
2n+1
) (9-13)
9. f(x) = sin
2
x cos
2
x.(DS.
n

k=1
(−1)
k+1
2
4k−3
(2k)!
x
2k
+ o(x
2n+1
))
10. f( x) = cos

3
x.(DS.
n

k=0
3(−1)
k
4(2k)!
(3
2k−1
+1)x
2k
+ o(x
2n+1
))
Chı
’
dˆa
˜
n. cos
3
x =
1
4
cos 3x +
3
4
cos x.
11. f( x) = cos
4

x + sin
4
x.
(DS. 1 +
n

k=1
(−1)
k
4
2k
(2k)!
x
2k
+0(x
2k+1
))
Chı
’
dˆa
˜
n. Ch´u
.
ng minh r˘a
`
ng cos
4
x + sin
4
x =

3
4
+
1
4
cos 4x.
12. f( x) = cos
6
x + sin
6
x.
(DS. 1 +
n

k=1
3(−1)
k
4
2k−1
2(2k)!
x
2k
+ o(x
2n+1
))
8.3. C´ac di
.
nh l´y co
.
ba

’
nvˆe
`
h`am kha
’
vi 107
13. f(x) = sin x sin 3x.
(D
S.
n

k=0
(−1)
k
2
2k−1
(2k)!
(1 − 2
2k
)x
2k
+ o(x
2n+1
)).
Khai triˆe
’
n h`am theo cˆong th´u
.
c Taylor trong lˆan cˆa
.

ndiˆe
’
m x
0
dˆe
´
n
o((x − x
0
)
n
) (14-20)
14. f(x)=
√
x, x
0
= 1. (DS.
n

k=0


1
2
k


(x − 1)
k
+ o((x − 1)

n
))
15. f(x)=(x
2
− 1)e
2x
, x
0
= −1.
(D
S.
n

k=1
e
−2
2
k−2
(k −5)
(k −1)!
(x +1)
k
+ o((x +1)
n
))
16. f(x) = ln(x
2
−7x + 12), x
0
=1.

(D
S. ln6 −
n

k=1
2
−k
+3
−k
k
(x − 1)
k
+ o(x − 1)
n
))
17. f(x)=ln
(x − 1)
x−2
3 − x
, x
0
=2.
(D
S. (x − 2) +
n

k=2

1
k

+
(−1)
k
k −1

(x − 2)
k
+ o((x − 2)
n
))
18. f(x)=
(x − 2)
2
3 − x
, x
0
= 2. (DS.
n

k=2
(x − 2)
k
+ o((x − 2)
n
))
19. f(x)=
x
2
− 3x +3
x − 2

, x =3.
(D
S. 3 +
n

k=2
(−1)
k
(x − 3)
k
+ o((x − 3)
n
))
20. f(x)=
x
2
+4x +4
x
2
+10x +25
, x
0
=2.
(D
S.
n

k=2
(−1)
k

(k −1)
3
k
(x +2)
k
+ o((x +2)
n
)).
´
Ap du
.
ng cˆong th´u
.
c Maclaurin d
ˆe
’
t´ınh gi´o
.
iha
.
n
21. lim
x→0
e
x
−e
−x
− 2x
x −sin x
.(DS. 2)

22. lim
x→0
tgx + 2 sin x − 3x
x
4
.(DS. 0)
108 Chu
.
o
.
ng 8. Ph´ep t´ınh vi phˆan h`am mˆo
.
tbiˆe
´
n
23. lim
x→0
e
x
− e
−x
− 2
x
2
.(DS. 1)
24. lim
x→0

1
x

−
1
sin x

.(DS. 0)
25. lim
x→0
cos x − e
−
x
2
2
x
4
.(DS. −
1
12
)
26. lim
x→0
1 −
√
1+x
2
cos x
x
4
.(DS.
1
3

)
27. lim
x→0
ln

cos x +
x
2
2

x(sin x −x)
.(DS. −
1
4
)
28. lim
x→0
sin(sin x) −x
3
√
1 − x
2
x
5
.(DS.
19
90
)
Chu
.

o
.
ng 9
Ph´ep t´ınh vi phˆan h`am
nhiˆe
`
ubiˆe
´
n
9.1 D
-
a
.
oh`amriˆeng 110
9.1.1 D
-
a
.
o h`am riˆeng cˆa
´
p1 110
9.1.2 D
-
a
.
o h`am cu
’
a h`am ho
.
.

p 111
9.1.3 H`am kha
’
vi 111
9.1.4 D
-
a
.
o h`am theo hu
.
´o
.
ng 112
9.1.5 D
-
a
.
o h`am riˆeng cˆa
´
pcao 113
9.2 Vi phˆan cu
’
a h`am nhiˆe
`
ubiˆe
´
n 125
9.2.1 Vi phˆan cˆa
´
p1 126

9.2.2
´
Ap du
.
ng vi phˆan d
ˆe
’
t´ınh gˆa
`
nd´ung . . . . . 126
9.2.3 C´ac t´ınh chˆa
´
tcu
’
aviphˆan 127
9.2.4 Vi phˆan cˆa
´
pcao 127
9.2.5 Cˆong th´u
.
cTaylor 129
9.2.6 Vi phˆan cu
’
ah`amˆa
’
n 130
9.3 Cu
.
.
c tri

.
cu
’
a h`am nhiˆe
`
ubiˆe
´
n 145
110 Chu
.
o
.
ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe
`
ubiˆe
´
n
9.3.1 Cu
.
.
c tri
.
145
9.3.2 Cu
.
.
c tri
.
c´o diˆe
`

ukiˆe
.
n 146
9.3.3 Gi´a tri
.
l´o
.
n nhˆa
´
t v`a b´e nhˆa
´
tcu
’
a h`am . . . . 147
9.1 D
-
a
.
oh`am riˆeng
9.1.1 D
-
a
.
o h`am riˆeng cˆa
´
p1
Gia
’
su
.

’
w = f(M), M =(x, y) x´ac di
.
nh trong lˆan cˆa
.
n n`ao d´o c u
’
adiˆe
’
m
M(x, y). Ta
.
idiˆe
’
m M ta cho biˆe
´
n x sˆo
´
gia t`uy´y∆x trong khi vˆa
˜
ngi˜u
.
gi´a tri
.
cu
’
abiˆe
´
n y khˆong d
ˆo

’
i. Khi d´o h`am f(x, y) nhˆa
.
nsˆo
´
gia tu
.
o
.
ng
´u
.
ng l`a
∆
x
w = f(x +∆x, y) −f(x, y)
go
.
il`asˆo
´
gia riˆeng cu
’
a h`am f(x, y) theo biˆe
´
n x ta
.
idiˆe
’
m M(x, y).
Tu

.
o
.
ng tu
.
.
da
.
ilu
.
o
.
.
ng
∆
y
w = f(x, y +∆y) − f(x, y)
go
.
il`asˆo
´
gia riˆeng cu
’
a h`am f(x, y) theo biˆe
´
n y ta
.
id
iˆe
’

m M(x, y).
D
-
i
.
nh ngh˜ıa 9.1.1
1. Nˆe
´
utˆo
`
nta
.
i gi´o
.
iha
.
nh˜u
.
uha
.
n
lim
∆x→0
∆
x
w
∆x
= lim
∆x→0
f(x +∆x, y) − f(x, y)

∆x
th`ı gi´o
.
iha
.
nd´odu
.
o
.
.
cgo
.
il`ada
.
o h`am riˆeng cu
’
a h`am f(x, y) theo biˆe
´
n
x ta
.
id
iˆe
’
m(x, y)v`adu
.
o
.
.
cchı

’
bo
.
’
imˆo
.
t trong c´ac k´yhiˆe
.
u
∂w
∂x
,
∂f(x, y)
∂x
,f

x
(x, y) ,w

x
.