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
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
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)
.(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
.
’
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
(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
’
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
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
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!
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
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!
=
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
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
.
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
`
’
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
+ 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
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
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
.
.
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
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
(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
(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
+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
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
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
·
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
2n
)
trong d
´o
−
1
2
k
(−1)
k
=(−1)
k
−
1
2
−
1
2
− 1
−
1
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
’
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)
.
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
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
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
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
+ 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
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)
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
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
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
+ 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
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
.
.
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
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
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
)
·
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
+ 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=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
)).
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
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 =
’
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
.
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.
+
(−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
(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)
−
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
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
.
.
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
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
.
’
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
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)
’
bo
.
’
imˆo
.
t trong c´ac k´yhiˆe
.
u
∂w
∂x
,
∂f(x, y)
∂x
,f
x
(x, y) ,w
x
.