Bài tập toán cao cấp part 6 doc

8.2. Vi phˆan 79
C
´
AC V
´
IDU
.
V´ı du
.
1. T´ınh vi phˆan df nˆe
´
u
1) f(x) = ln(arctg(sin x)); 2) f(x)=x
√
64 −x
2
+64arcsin
x
8
.
Gia
’
i. 1)
´
Ap du
.
ng c´ac t´ınh chˆa
´
tcu
’
a vi phˆan ta c´o

df =
d[arctg(sin x)]
arctg(sin x)
=
d(sin x)
(1 + sin
2
x)arctg(sin x)
=
cos xdx
(1 + sin
2
x)arctg(sin x)
·
2)
df = d[x
√
64 − x
2
]+d

64arcsin
x
8

= xd
√
64 − x
2
+

√
64 −x
2
dx +64d

arcsin
x
8

= x
d(64 − x
2
)
2
√
64 − x
2
+
√
64 −x
2
dx +64·
d

x
8


1 −
x

2
64
=
−x
2
dx
√
64 −x
2
+
√
64 − x
2
dx +64
dx
√
64 − x
2
=2
√
64 − x
2
dx, |x| < 8. 
V´ı du
.
2. T´ınh vi phˆan cˆa
´
p2cu
’
a c´ac h`am

1) f(x)=xe
−x
,nˆe
´
u x l`a biˆe
´
ndˆo
.
clˆa
.
p;
2) f(x)=sinx
2
nˆe
´
u
a) x l`a biˆe
´
ndˆo
.
clˆa
.
p,
b) x l`a h`am cu
’
amˆo
.
tbiˆe
´
ndˆo

.
clˆa
.
pn`aod´o .
Gia
’
i. 1) Phu
.
o
.
ng ph´ap I. Theo d
i
.
nh ngh˜ıa vi phˆan cˆa
´
p 2 ta c´o
d
2
f = d[df ]=d[xde
−x
+ e
−x
dx]
= d(−xe
−x
dx + e
−x
dx)=−d(xe
−x
)dx + d(e

−x
)dx
= −(xde
−x
+ e
−x
dx)dx −e
−x
dx
2
= xe
−x
dx
2
− e
−x
dx
2
−e
−x
dx
2
=(x − 2)e
−x
dx
2
.
80 Chu
.
o

.
ng 8. Ph´ep t´ınh vi phˆan h`am mˆo
.
tbiˆe
´
n
Phu
.
o
.
ng ph´ap II.T´ınh d
a
.
o h`am cˆa
´
p hai f

(x) ta c´o
f

(x)=(xe
−x
)

=(e
−x
− xe
−x
)


= −e
−x
− e
−x
+ xe
−x
=(x − 2)e
−x
v`a theo cˆong th´u
.
c (8.6) ta c´o
d
2
f =(x − 2)e
−x
dx
2
.
2) a) Phu
.
o
.
ng ph´ap I. Theo di
.
nh ngh˜ıa vi phˆan cˆa
´
p hai ta c´o
d
2
f = d[d sin x

2
]=d[2x cos x
2
dx]=d[2x cos x
2
]dx
=

2 cos x
2
dx +2x(−sin x
2
)2xdx

dx
= (2 cos x
2
− 4x
2
sin x
2
)dx
2
.
Phu
.
o
.
ng ph´ap II.T´ınh da
.

o h`am cˆa
´
p hai f

xx
ta c´o
f

x
=2x cos x
2
,f

xx
= 2 cos x
2
− 4x
2
sin x
2
v`a theo (8.6) ta thu du
.
o
.
.
c
d
2
f = (2 cos x
2

− 4x
2
sin x
2
)dx
2
.
b) Nˆe
´
u x l`a biˆe
´
n trung gian th`ı n´oi chung d
2
x =0v`adod´o ta c´o
d
2
f = d(2x cos x
2
dx)=(2x cos x
2
)d
2
x +[d(2x cos x
2
)]dx
=2x cos x
2
d
2
x + (2 cos x

2
− 4x
2
sin x
2
)dx
2
. 
V´ı d u
.
3.
´
Ap du
.
ng vi phˆan d
ˆe
’
t´ınh gˆa
`
nd´ung c´ac gi´a tri
.
:
1)
5

2 −0, 15
2+0, 15
; 2) arcsin 0, 51; 3) sin 29
◦
.

Gia
’
i. Cˆong th´u
.
cco
.
ba
’
nd
ˆe
’
´u
.
ng du
.
ng vi phˆan d
ˆe
’
t´ınh gˆa
`
nd´ung l`a
∆f(x
0
) ≈ df (x
0
) ⇒ f(x
0
+∆x) − f(x
0
) ≈ f


(x
0
)∆x
⇒
f(x
0
+∆x) ≈ f(x
0
)+f

(x
0
)∆x
8.2. Vi phˆan 81
T`u
.
d
´o , d ˆe
’
t´ınh gˆa
`
nd´ung c´ac gi´a tri
.
ta cˆa
`
n thu
.
.
chiˆe

.
nnhu
.
sau:
1
+
Chı
’
ra biˆe
’
uth´u
.
c gia
’
it´ıchd
ˆo
´
iv´o
.
i h`am m`a gi´a tri
.
gˆa
`
nd
´ung cu
’
a
n´o cˆa
`
n pha

’
i t´ınh.
2
+
Cho
.
ndiˆe
’
m M
0
(x
0
) sao cho gi´a tri
.
cu
’
a h`am v`a cu
’
ada
.
o h`am cˆa
´
p
1cu
’
a n´o ta
.
idiˆe
’
mˆa

´
y c´o thˆe
’
t´ınh m`a khˆong d `ung ba
’
ng.
3
+
Tiˆe
´
pdˆe
´
n l`a ´ap du
.
ng cˆong th´u
.
cv`u
.
a nˆeu.
1) T´ınh gˆa
`
nd
´ung
5

2 −0, 15
2+0, 15
Sˆo
´
d

˜a cho l`a gi´a tri
.
cu
’
a h`am
y =
5

2 − x
2+x
ta
.
idiˆe
’
m x =0, 15. Ta d˘a
.
t x
0
=0;∆x =0, 15. Ta c´o
y

=
−4
5

2 −x
2+x
5(4 − x
2
)

= −
4y
5(4 −x
2
)
⇒ y

(x
0
)=y

(0) = −
1
5
·
Do d
´ov`ıy(0) = 1 nˆen
y(0, 15) ≈ y(0) + y

(0)∆x
=1−
1
5
· (0, 15) = 1 −0, 03 = 0, 97.
2) T´ınh gˆa
`
nd
´ung arcsin 0, 51.
X´et h`am y = arcsin x.Sˆo
´

cˆa
`
nt´ınh l`a gi´a tri
.
cu
’
a h`am ta
.
idiˆe
’
m
0, 51; t´u
.
cl`ay(0, 51).
D˘a
.
t x
0
=0, 5; ∆x =0, 01. Khi d´o ta c´o
arcsin(x
0
+∆x ≈ arcsinx
0
+ (arcsinx)

x=x
0
∆x
⇒ arcsin(0, 5+0, 01) ≈ arcsin0, 5 + (arcsinx)




x=0,5
· 0, 01
=
π
6
+
1

1 −(0, 5)
2
× (0, 01).
82 Chu
.
o
.
ng 8. Ph´ep t´ınh vi phˆan h`am mˆo
.
tbiˆe
´
n
C´o thˆe
’
t´ınh gˆa
`
nd´ung

1 −(0, 5)
2

=
√
0, 75 ≈ 0, 88 v`a do d´o
arcsin0, 51 ≈
π
6
+0, 011 ≈ 0, 513.
3) Sˆo
´
sin 29
◦
l`a gi´a tri
.
cu
’
a h`am y = sin x khi x =
π
180
×29. Ta d˘a
.
t
x
0
=
π
180
×30 =
π
6
; y


π
6

=
1
2
,y= cos x ⇒ y


π
6

= cos
π
6
=
√
3
2
·
D
˘a
.
t∆x = x − x
0
=
29π
180
−

π
6
= −
π
180
.Dod´o
sin 29
◦
≈ y

π
6

+ y


π
6

· ∆x =
1
2
+
√
3
2

−
π
180


≈ 0, 48. 
B
`
AI T
ˆ
A
.
P
T´ınh vi phˆan df nˆe
´
u:
1. f(x) = arctg
1
x
.(DS. df =
−dx
1+x
2
)
2. f(x)=2
tg
2
x
.(DS. 2
tg
2
x
ln2 ·2tgx ·
dx

cos
2
x
)
3. f(x) = arccos(2
x
). (DS. −
2
x
ln2dx
√
1 −e
2x
)
4. f(x)=x
3
lnx.(DS. x
2
(1 + 3lnx) dx)
5. f(x) = cos
2
(
√
x). (DS. −2 cos
√
x ·sin
√
x ·
dx
2

√
x
)
6. f(x)=(1+x
2
)arcotgx.(DS. (2xarccotgx − 1)dx)
7. f(x)=
arctgx
√
1+x
2
.(DS.
1 − xarctgx
(1 + x
2
)
3/2
dx)
8. f(x) = sin
3
2x.(DS. 3 sin 2x sin 4xdx)
9. f(x) = ln(sin
√
x). (DS.
cotg
√
x
2
√
x

dx)
8.2. Vi phˆan 83
10. f(x)=e
−
1
cos
x
.(DS.
−tgx ·e
−
1
cos x
cos x
dx)
11. f(x)=2
−x
2
.(DS. −2xe
−x
2
ln2dx)
12. f(x) = arctg
√
x
2
+ 1. (DS.
2xdx
2+x
2
)

13. f(x)=
√
xarctg
√
x.(DS.
1
2
√
x

arctg
√
x +
√
x
1+x

dx)
14. f(x)=
x
2
arcsinx
.(DS.
x

2arcsinx −
x
√
1 −x
2


(arcsinx)
2
dx).
T´ınh vi phˆan cˆa
´
ptu
.
o
.
ng ´u
.
ng cu
’
a c´ac h`am sau
15. f(x)=4
−x
2
; d
2
f ?(DS. 4
−x
2
2ln4(2x
2
ln4 −1)(dx)
2
)
16. f(x)=


ln
2
x − 4. d
2
f ?(DS.
4lnx −4 −ln
3
x
x
2

(lnx − 4)
3
(dx)
2
)
17. f(x) = sin
2
x. d
3
f ?(DS. −4 sin 2x(dx)
3
)
18. f(x)=
√
x −1, d
4
f ?(DS.
−15
16(x −1)

7/2
(dx)
4
)
19. f(x)=xlnx, d
5
f ?(DS. −
6
x
4
(dx)
5
, x>0)
20. f(x)=x sin x; d
10
f ?(DS. (10 cos x − x sin x)(dx)
10
)
Su
.
’
du
.
ng cˆong th´u
.
cgˆa
`
nd
´ung
∆f ≈ df

(khi f

(x) =0)dˆe
’
t´ınh gˆa
`
nd´ung c´ac gi´a tri
.
sau
21. y =
√
3, 98. (DS. 1,955)
22. y =
3
√
26, 19. (DS. 2,97)
23. y =

(2, 037)
2
− 3
(2, 037)
2
+5
.(DS. 0,35)
24. y = cos 31
◦
.(DS. 0,85)
84 Chu
.

o
.
ng 8. Ph´ep t´ınh vi phˆan h`am mˆo
.
tbiˆe
´
n
25. y = tg45
◦
10

.(DS. 0,99)
26. y = ln(10, 21). (DS. 1,009)
27. y = sin 31
◦
.(DS. 0,51)
28. y = arcsin0, 54. (D
S. 0,57)
29. y = arctg(1, 05). (DS. 0,81)
30. y =(1, 03)
5
.(DS. 1,15)
8.3 C´ac di
.
nh l´yco
.
ba
’
nvˆe
`

h`am kha
’
vi.
Quy t˘a
´
c l’Hospital. Cˆong th´u
.
cTay-
lor
8.3.1 C´ac di
.
nh l´yco
.
ba
’
nvˆe
`
h`am kha
’
vi
D
-
i
.
nh l´y Rˆon (Rolle). Gia
’
su
.
’
:

i) f(x) liˆen tu
.
ctrˆen doa
.
n [a, b].
ii) f(x) c´o da
.
o h`am h˜u
.
uha
.
n trong (a, b).
iii) f(a)=f(b).
Khi d
´otˆo
`
nta
.
idiˆe
’
m ξ : a<ξ<bsao cho f(ξ)=0.
D
-
i
.
nh l´y Lagr˘ang (Lagrange). Gia
’
su
.
’

:
i) f(x) liˆen tu
.
ctrˆen d
oa
.
n [a, b].
ii) f(x) c´o da
.
o h`am h˜u
.
uha
.
n trong (a, b).
Khi d´ot`ımdu
.
o
.
.
c ´ıt nhˆa
´
tmˆo
.
tdiˆe
’
m ξ ∈ (a, b) sao cho
f(b) −f(a)
b −a
= f


(ξ) (8.12)
hay l`a
f(b)=f(a)+f

(ξ)(b −a). (8.13)
Cˆong th´u
.
c (8.12) go
.
i l`a cˆong th´u
.
csˆo
´
gia h˜u
.
uha
.
n.
8.3. C´ac di
.
nh l´y co
.
ba
’
nvˆe
`
h`am kha
’
vi 85
D

-
i
.
nh l´y Cˆosi (Cauchy). Gia
’
su
.
’
:
i) f(x) v`a ϕ(x) liˆen tu
.
ctrˆen d
oa
.
n [a, b].
ii) f(x) v`a ϕ(x) c´o d
a
.
o h`am h˜u
.
uha
.
n trong (a, b).
iii) [f

(x)]
2
+[ϕ

(x)]

2
=0, ngh˜ıa l`a c´ac da
.
o h`am khˆong dˆo
`
ng th`o
.
i
b˘a
`
ng 0.
iv) ϕ(a) = ϕ(b).
Khi d
´ot`ımdu
.
o
.
.
cdiˆe
’
m ξ ∈ (a, b) sao cho:
f(b) − f(a)
ϕ(b) − ϕ(a)
=
f

(ξ)
ϕ

(ξ)

· (8.14)
D
i
.
nh l´y Lagrange l`a tru
.
`o
.
ng ho
.
.
p riˆeng cu
’
adi
.
nh l´y Cauchy v`ı khi
ϕ(x)=x th`ı t`u
.
(8.14) thu du
.
o
.
.
c (8.13). Di
.
nh l´y Rˆon c˜ung l`a tru
.
`o
.
ng

ho
.
.
p riˆeng cu
’
adi
.
nh l´y Lagrange v´o
.
idiˆe
`
ukiˆe
.
n f(a)=f(b).
C
´
AC V
´
IDU
.
V´ı du
.
1. Gia
’
su
.
’
P ( x)=(x + 3)(x + 2)(x − 1).
Ch´u
.

ng minh r˘a
`
ng trong khoa
’
ng (−3, 1) tˆo
`
nta
.
i nghiˆe
.
mcu
’
aphu
.
o
.
ng
tr`ınh P

(ξ)=0.
Gia
’
i. D
ath´u
.
c P(x) c´o nghiˆe
.
mta
.
ic´acdiˆe

’
m x
1
= −3, x
2
= −2,
x
3
= 1. Trong c´ac khoa
’
ng (−3, −2) v`a (−2, 1) h`am P (x) kha
’
vi v`a
tho
’
a m˜an c´ac diˆe
`
ukiˆe
.
ncu
’
adi
.
nh l´y Rˆon v`a:
P ( −3) = P(−2)=0,
P ( −2) = P(1) = 0.
Do d
´o theo di
.
nh l´y Rˆon, t`ım du

.
o
.
.
cd
iˆe
’
m ξ
1
∈ (−3, −2); ξ
2
∈ (−2, 1)
sao cho:
P

(ξ
1
)=P

(ξ
2
)=0.
Bˆay gi`o
.
la
.
i ´ap du
.
ng d
i

.
nh l´y Rˆon cho doa
.
n[ξ
1
,ξ
2
] v`a h`am P

(x), ta
la
.
it`ımdu
.
o
.
.
cdiˆe
’
m ξ ∈ (ξ
1
,ξ
2
) ⊂ (−3, 1) sao cho P

(ξ)=0.
86 Chu
.
o
.

ng 8. Ph´ep t´ınh vi phˆan h`am mˆo
.
tbiˆe
´
n
V´ı du
.
2. H˜ay x´et xem h`am f(x) = arcsinx trˆen doa
.
n[−1, +1] c´o
tho
’
a m˜an di
.
nh l´y Lagrange khˆong ? Nˆe
´
u tho
’
a m˜an th`ı h˜ay t`ım diˆe
’
m
ξ (xem (8.12)).
Gia
’
i. H`am f(x) x´ac d
i
.
nh v`a liˆen tu
.
ctrˆen[−1, +1]. Ta t`ım f


(x).
f

(x)=
1
√
1 −x
2
→ f

(x) < ∞,x∈ (−1, 1)
(Lu
.
u´yr˘a
`
ng khi x = ±1da
.
o h`am khˆong tˆo
`
nta
.
inhu
.
ng diˆe
’
ud´o
khˆong a
’
nh hu

.
o
.
’
ng dˆe
´
nsu
.
.
tho
’
am˜andiˆe
`
ukiˆe
.
ncu
’
adi
.
nh l´y Lagrange !).
Nhu
.
vˆa
.
y h`am f tho
’
am˜andi
.
nh l´y Lagrange.
Tat`ımd

iˆe
’
m ξ. Ta c´o:
arcsin1 − arcsin( −1)
1 − (−1)
=
1

1 −ξ
2
⇒
π
2
−

−
π
2

2
=
1

1 −ξ
2
⇒

1 −ξ
2
=

2
π
⇒ ξ
1,2
= ±

1 −
4
π
2
Nhu
.
vˆa
.
y trong tru
.
`o
.
ng ho
.
.
p n`ay cˆong th´u
.
c (8.12) tho
’
a m˜an dˆo
´
iv´o
.
i

hai d
iˆe
’
m.
V´ı du
.
3. H˜ay kha
’
o s´at xem c´ac h`am f(x)=x
2
− 2x +3v`a ϕ(x)=
x
3
− 7x
2
+20x − 5 c´o tho
’
a m˜an diˆe
`
ukiˆe
.
ndi
.
nh l´y Cauchy trˆen doa
.
n
[1, 4] khˆong ? Nˆe
´
uch´ung tho
’

a m˜an d
i
.
nh l´y Cauchy th`ı h˜ay t`ım diˆe
’
m
ξ.
Gia
’
i. i) Hiˆe
’
n nhiˆen ca
’
f(x)v`aϕ(x)liˆen tu
.
c khi x ∈ [1, 4].
ii) f(x)v`aϕ(x)c´od
a
.
o h`am h˜u
.
uha
.
n trong (1, 4).
iii) D
iˆe
`
ukiˆe
.
nth´u

.
iii) c˜ung tho
’
a m˜an v`ı:
g

(x)=3x
2
− 14x +20> 0,x∈ R.
iv) Hiˆe
’
n nhiˆen ϕ(1) = ϕ(4).
8.3. C´ac di
.
nh l´y co
.
ba
’
nvˆe
`
h`am kha
’
vi 87
Do d´o f(x)v`aϕ(x) tho
’
a m˜an di
.
nh l´y Cauchy v`a ta c´o
f(4) −f(1)
ϕ(4) −ϕ(1)

=
f

(ξ)
ϕ

(ξ)
hay
11 − 2
27 − 9
=
2ξ −2
3ξ
2
− 14ξ +20
,ξ∈ (1, 4).
T`u
.
d
´othudu
.
o
.
.
c ξ
1
=2,ξ
2
=4v`ao
.

’
d
ˆay chı
’
c´o ξ
1
=2l`adiˆe
’
m trong
cu
’
a(1, 4). Do d´o: ξ =2.
V´ı d u
.
4. Di
.
nh l´y Cauchy c´o ´ap du
.
ng du
.
o
.
.
c cho c´ac h`am f(x)=cosx,
ϕ(x)=x
3
trˆen doa
.
n[−π/2,π/2] hay khˆong ?
Gia

’
i. Hiˆe
’
n nhiˆen f(x)v`aϕ(x) tho
’
a m˜an c´ac diˆe
`
ukiˆe
.
n i), ii) v`a
iv) cu
’
adi
.
nh l´y Cauchy. Tiˆe
´
p theo ta c´o: f

(x)=−sin x; ϕ

(x)=3x
2
v`a ta
.
i x = 0 ta c´o: f

(0) = −sin 0 = 0; ϕ

(0) = 0 v`a nhu
.

vˆa
.
y
[ϕ

(0)]
2
+[f

(0)]
2
= 0. Do d´odiˆe
`
ukiˆe
.
n iii) khˆong du
.
o
.
.
c tho
’
a m˜an. Ta
x´et vˆe
´
tr´ai cu
’
a (8.14):
f(b) − f(a)
ϕ(b) −ϕ(a)

=
cos(π/2) −cos(−π/2)
(π/2)
3
−(−π/2)
3
=0.
Bˆay gi`o
.
ta x´et vˆe
´
pha
’
icu
’
a (8.14). Ta c´o:
f

(ξ)
ϕ

(ξ)
= −
sin ξ
3ξ
2
·
Nhu
.
ng d

ˆo
´
iv´o
.
ivˆe
´
pha
’
i n`ay ta c´o:
lim
ξ→0

−
sin ξ
3ξ
2

= lim
ξ→0
sin ξ
ξ
· lim
ξ→0

−
1
3ξ

= ∞.
Diˆe

`
ud´och´u
.
ng to
’
r˘a
`
ng c´ac h`am d˜a cho khˆong tho
’
a m˜an di
.
nh l´y
Cauchy.
B
`
AI T
ˆ
A
.
P
1. H`am y =1−
3
√
x
2
trˆen doa
.
n[−1, 1] c´o tho
’
am˜andiˆe

`
ukiˆe
.
ncu
’
adi
.
nh
l´y Rˆon khˆong ? Ta
.
i sao ? (Tra
’
l`o
.
i: Khˆong)
88 Chu
.
o
.
ng 8. Ph´ep t´ınh vi phˆan h`am mˆo
.
tbiˆe
´
n
2. H`am y =3x
2
− 5 c´o tho
’
am˜andi
.

nh l´y Lagrange trˆen doa
.
n[−2, 0]
khˆong ? Nˆe
´
u n´o tho
’
a m˜an, h˜ay t`ım gi´a tri
.
trung gian ξ. (Tra
’
l`o
.
i:
C´o)
3. Ch´u
.
ng minh r˘a
`
ng h`am f(x)=x +1/x tho
’
a m˜an d
i
.
nh l´y Lagrange
trˆen d
oa
.
n[1/2, 2]. T`ım ξ.(DS. ξ =1)
4. Ch´u

.
ng minh r˘a
`
ng c´ac h`am f(x) = cos x, ϕ(x) = sin x tho
’
a m˜an
d
i
.
nh l´y Cauchy trˆen doa
.
n[0,π/2]. T`ım ξ ?(DS. ξ = π/4)
5. Ch´u
.
ng minh r˘a
`
ng h`am f(x)=e
x
v`a ϕ(x)=x
2
/(1 + x
2
) khˆong
tho
’
a m˜an d
i
.
nh l´y Cauchy trˆen doa
.

n[−3, 3].
6. Trˆen du
.
`o
.
ng cong y = x
3
h˜ay t`ım diˆe
’
m m`a ta
.
id´otiˆe
´
p tuyˆe
´
nv´o
.
i
du
.
`o
.
ng cong song song v´o
.
i dˆay cung nˆo
´
idiˆe
’
m A(−1, −1) v´o
.

i B(2, 8).
(D
S. M(1, 1))
Chı
’
dˆa
˜
n. Du
.
.
a v`ao ´y ngh˜ıa h`ınh ho
.
ccu
’
a cˆong th´u
.
csˆo
´
gia h˜u
.
uha
.
n.
8.3.2 Khu
.
’
c´ac da
.
ng vˆo d
i

.
nh. Quy t˘a
´
c Lˆopitan
(L’Hospitale)
Trong chu
.
o
.
ng II ta d˜a d ˆe
`
cˆa
.
pdˆe
´
nviˆe
.
ckhu
.
’
c´ac da
.
ng vˆo di
.
nh. Bˆay gi`o
.
ta tr`ınh b`ay quy t˘a
´
c Lˆopitan - cˆong cu
.

co
.
ba
’
ndˆe
’
khu
.
’
c´ac da
.
ng vˆo
di
.
nh
Da
.
ng vˆo di
.
nh 0/0
Gia
’
su
.
’
hai h`am f(x)v`aϕ(x) tho
’
a m˜an c´ac diˆe
`
ukiˆe

.
n
i) lim
x→a
f(x) = 0; lim
x→a
ϕ(x)=0.
ii) f(x)v`aϕ(x) kha
’
vi trong lˆan cˆa
.
n n`ao d´o c u
’
adiˆe
’
m x = a v`a
ϕ

(x) = 0 trong lˆan cˆa
.
nd´o, c´o thˆe
’
tr `u
.
ra ch´ınh d
iˆe
’
m x = a.
iii) Tˆo
`

nta
.
i gi´o
.
iha
.
n(h˜u
.
uha
.
n ho˘a
.
cvˆoc`ung)
lim
x→a
f

(x)
ϕ

(x)
= k.
8.3. C´ac di
.
nh l´y co
.
ba
’
nvˆe
`

h`am kha
’
vi 89
Khi d´o
lim
x→a
f(x)
ϕ(x)
= lim
x→a
f

(x)
ϕ

(x)
·
Da
.
ng vˆo di
.
nh ∞/∞
Gia
’
su
.
’
f(x)v`aϕ(x) tho
’
a m˜an c´ac diˆe

`
ukiˆe
.
n ii) v`a iii) cu
’
adi
.
nh l´y
trˆen dˆay c`on diˆe
`
ukiˆe
.
ni)du
.
o
.
.
cthaybo
.
’
idiˆe
`
ukiˆe
.
n:
i)
∗
lim
x→a
f(x)=∞, lim

x→a
ϕ(x)=∞.
Khi d
´o:
lim
x→a
f(x)
ϕ(x)
= lim
x→a
f

(x)
ϕ

(x)
Ch´u´y. Nˆe
´
uthu
.
o
.
ng f

(x)/ϕ

(x)la
.
ic´oda
.

ng vˆo di
.
nh 0/0 (ho˘a
.
c
∞/∞)ta
.
id
iˆe
’
m x = a v`a f

, ϕ

tho
’
a m˜an c´ac diˆe
`
ukiˆe
.
n i), ii) v`a iii)
(tu
.
o
.
ng ´u
.
ng i)
∗
, ii) v`a iii)) th`ı ta c´o thˆe

’
chuyˆe
’
n sang da
.
o h`am cˆa
´
p hai,
C´ac da
.
ng vˆo d
i
.
nh kh´ac
a) Dˆe
’
khu
.
’
da
.
ng vˆo di
.
nh 0 ·∞

lim
x→a
f(x)=0, lim
x→a
ϕ(x)=∞


ta
biˆe
´
nd
ˆo
’
it´ıchf(x) · ϕ(x) th`anh:
i)
f(x)
1/ϕ(x)
(da
.
ng 0/0)
ii)
ϕ(x)
1/f(x)
(da
.
ng ∞/∞).
b) D
ˆe
’
khu
.
’
da
.
ng vˆo di
.

nh ∞−∞
Ta biˆe
´
nd
ˆo
’
i f(x) − ϕ(x) (trong d´o lim
x→a
f(x)=∞, lim
x→a
ϕ(x)=∞)
th`anh t´ıch
f(x) − ϕ(x)=f(x)ϕ(x)

1
ϕ(x)
−
1
f(x)

ho˘a
.
c th`anh t´ıch da
.
ng
f(x) − ϕ(x)=f(x)

1 −
ϕ(x)
f(x)


90 Chu
.
o
.
ng 8. Ph´ep t´ınh vi phˆan h`am mˆo
.
tbiˆe
´
n
ho˘a
.
c
f(x) −ϕ(x)=ϕ(x)

f(x)
ϕ(x)
− 1

.
c) Da
.
ng vˆo di
.
nh 0
0
, ∞
0
,1
∞

Khi t´ınh gi´o
.
iha
.
ncu
’
a h`am da
.
ng F (x)=[f(x)]
ϕ(x)
thˆong thu
.
`o
.
ng
ta g˘a
.
p c´ac da
.
ng vˆo di
.
nh 0
0
, ∞
0
ho˘a
.
c1
∞
. Trong nh˜u

.
ng tru
.
`o
.
ng ho
.
.
p
n`ay ta c´o thˆe
’
biˆe
´
nd
ˆo
’
i F (x)dˆe
’
du
.
avˆe
`
da
.
ng vˆo d
i
.
nh 0 ·∞ d˜a n´oi trong
1) nh`o
.

ph´ep biˆe
´
nd
ˆo
’
i
F (x)=[f(x)]
ϕ(x)
= e
ln[f(x)]
ϕ(x)
= e
ϕ(x)lnf(x)
v`a do t´ınh liˆen tu
.
ccu
’
ah`amm˜u ta s´e c´o:
lim
x→a
[f(x)]
ϕ(x)
= e
lim[ϕ(x)·lnf(x)]
Ch´u´y. Ta lu
.
u´yr˘a
`
ng m˘a
.

cd`u quy t˘a
´
c Lˆopitan l`a mˆo
.
t cˆong cu
.
ma
.
nh d
e
’
t´ınh gi´o
.
iha
.
nnhu
.
ng n´o khˆong thˆe
’
thay to`an bˆo
.
c´ac phu
.
o
.
ng
ph´ap t´ınh gi´o
.
iha
.

nd˜a x´et trong chu
.
o
.
ng II. Diˆe
`
ud´odu
.
o
.
.
cch´u
.
ng to
’
trong v´ıdu
.
7 sau d
ˆa y .
C
´
AC V
´
IDU
.
V´ı d u
.
1. T´ınh lim
x→1
x

2
− 1+lnx
e
x
−e
Gia
’
i. Ta c´o vˆo d
i
.
nh da
.
ng “0/0”.
´
Ap du
.
ng quy t˘a
´
c L’Hospital ta
thu d
u
.
o
.
.
c
lim
x→1
x
2

− 1+lnx
e
x
−e
= lim
x→1
(x
2
− 1+lnx)

(e
x
− e)

= lim
x→1
2x +
1
x
e
x
=
3
e
. 
V´ı d u
.
2. T´ınh lim
x→+∞
x

n
e
x
8.3. C´ac di
.
nh l´y co
.
ba
’
nvˆe
`
h`am kha
’
vi 91
Gia
’
i. Tac´ovˆodi
.
nh da
.
ng “∞/∞”.
´
Ap du
.
ng quy t˘a
´
c L’Hospital n
lˆa
`
n ta thu d

u
.
o
.
.
c
lim
x→∞
x
n
e
x
= lim
x→1
nx
n−1
e
x
= lim
x→1
n(n −1)x
n−2
e
x
= ···= lim
x→1
n(n −1) ···2 · 1
e
x
= lim

x→1
n!
e
x
=0. 
V´ı du
.
3. T´ınh lim
x→0+0
xlnx.
Gia
’
i. Ta c´o vˆo di
.
nh da
.
ng “0 ·∞”. Nhu
.
ng
xlnx =
lnx
1
x
v`a ta thu du
.
o
.
.
cvˆod
i

.
nh da
.
ng “∞/∞”. Do d´o
lim
x→0+0
xlnx = lim
x→0+0
(lnx)


1
x


= lim
x→0+0
1
x
−
1
x
2
= − lim
x→0+0
x =0. 
V´ı du
.
4. T´ınh lim
x→0+0

x
x
.
Gia
’
i. O
.
’
dˆay ta c´o vˆo di
.
nh da
.
ng “0
0
”. Nhu
.
ng
x
x
= e
xlnx
v`a ta thu du
.
o
.
.
cvˆodi
.
nh da
.

ng 0 ·∞o
.
’
sˆo
´
m˜u. Trong v´ıdu
.
3tad˜a t h u
du
.
o
.
.
c
lim
x→0+0
(xlnx)=0,
do d
´o
lim
x→0+0
x
x
= lim
x→0+0
e
xlnx
= e
lim
x→0+0

xlnx
= e
0
=1. 
V´ı du
.
5. T´ınh lim
x→0

1+x
2

1
e
x
−1−x
92 Chu
.
o
.
ng 8. Ph´ep t´ınh vi phˆan h`am mˆo
.
tbiˆe
´
n
Gia
’
i. O
.
’

dˆay ta c´o vˆo di
.
nh da
.
ng 1
∞
.Nhu
.
ng

1+x
2

1
e
x
−1−x
= e
ln
(1+x
2
)
e
x
−1−x
v`a o
.
’
sˆo
´

m˜ucu
’
al˜uy th`u
.
a ta thu du
.
o
.
.
cvˆodi
.
nh da
.
ng “0/0”.
´
Ap du
.
ng
quy t˘a
´
c L’Hospital ta thu d
u
.
o
.
.
c
lim
x→0
ln(1 + x

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

− 1)2x
=
2
1
=2. 
V´ı d u
.
6. T´ınh lim
x→
π
2

tgx

2 cosx
.
Gia
’
i. Tac´ovˆodi
.
nh da
.
ng “∞
0
”. Nhu
.
ng

tgx


2 cos x
= e
2 cosxln tgx
= e
2ln tgx
1/ cos x
v`a o
.
’
sˆo
´
m˜ucu
’
al˜uy th`u
.
a ta thu du
.
o
.
.
cvˆodi
.
nh da
.
ng “∞/∞”.
´
Ap du
.
ng
quy t˘a

´
c L’Hospital ta c´o
lim
x→
π
2
2ln tgx
1
cos x
= 2 lim
x→
π
2
1
cos
2
x ·tgx
+ sin x
cos
2
x
= 2 lim
x→
π
2
1
cos x
tg
2
x

= 2 lim
x→
π
2
−
sin x
cos
2
x
2tgx ·
1
cos
2
x
= lim
x→
π
2
cos x =0.
Do d
´o
lim
x→
π
2

tgx

2 cos x
= e

lim
x→
π
2
2 cos x·ln tgx
= e
0
=1. 
V´ı d u
.
7. Ch´u
.
ng minh r˘a
`
ng gi´o
.
iha
.
n
1) lim
x→0
x
2
sin(1/x)
sin x
=0
8.3. C´ac di
.
nh l´y co
.

ba
’
nvˆe
`
h`am kha
’
vi 93
2) lim
x→∞
x −sin x
x + sin x
=1
khˆong thˆe
’
t`ım d
u
.
o
.
.
c theo quy t˘a
´
c L’Hospital. H˜ay t´ınh c´ac gi´o
.
iha
.
n
d
´o .
Gia

’
i. 1) Quy t˘a
´
c L’Hospital khˆong ´ap du
.
ng d
u
.
o
.
.
cv`ıty
’
sˆo
´
c´ac da
.
o
h`am [2x sin(1/x) − cos(1/x)]/ cos x khˆong c´o gi´o
.
iha
.
n khi x → 0.
Ta t´ınh tru
.
.
ctiˆe
´
p gi´o
.

iha
.
n n`ay.
lim
x→0
x
2
sin(1/x)
sin x
= lim
x→0
x
sin x
· lim
x→0
x sin
1
x
=1· 0=0.
2) Quy t˘a
´
c L’Hospital khˆong ´ap du
.
ng d
u
.
o
.
.
cv`ıty

’
sˆo
´
c´ac da
.
o h`am
1 −cos x
1 + cos x
=tg
2
(x/2)
khˆong c´o gi´o
.
iha
.
n khi x →∞.
Ta t´ınh tru
.
.
ctiˆe
´
p gi´o
.
iha
.
n n`ay
lim
x→∞
x − sin x
x + sin x

= lim
x→∞
[1 −(sin x)/x]
[1 + (sin x)/x]
=1v`ı |sin x|  1.
Nhu
.
o
.
’
phˆa
`
ndˆa
`
ucu
’
atiˆe
´
t n`ay d˜a n´oi, quy t˘a
´
c L’Hospital l`a mˆo
.
t
cˆong cu
.
ma
.
nh d
ˆe
’

t`ım gi´o
.
iha
.
nnhu
.
ng d
iˆe
`
ud´o khˆong c´o ngh˜ıa l`a n´o c´o
thˆe
’
thay cho to`an bˆo
.
c´ac phu
.
o
.
ng ph´ap t`ım gi´o
.
iha
.
n. Cˆa
`
nlu
.
u´yr˘a
`
ng
quy t˘a

´
c L’Hospital chı
’
l`a d
iˆe
`
ukiˆe
.
ndu
’
dˆe
’
tˆo
`
nta
.
i gi´o
.
iha
.
n: lim
x→a
f(x)
g(x)
ch´u
.
khˆong pha
’
il`adiˆe
`

ukiˆe
.
ncˆa
`
n.
B
`
AI T
ˆ
A
.
P
´
Ap du
.
ng quy t˘a
´
c L’Hospital d
ˆe
’
t´ınh gi´o
.
iha
.
n:
1. lim
x→2
x
4
− 16

x
3
+5x
2
−6x −16
.(DS.
16
13
)
2. lim
x→a
x
m
− a
m
x
n
− a
n
.(DS.
m
n
a
m−n
)
94 Chu
.
o
.
ng 8. Ph´ep t´ınh vi phˆan h`am mˆo

.
tbiˆe
´
n
3. lim
x→0
e
2x
− 1
sin x
.(DS. 2)
4. lim
x→0
1 −cos ax
1 + cos bx
.(D
S.
a
2
b
2
)
5. lim
x→0
e
x
−e
−x
− 2x
x −sin x

.(DS. 2)
6. lim
x→0
ln(1 + x
2
)
cos 3x − e
−x
.(DS. 0)
7. lim
x→∞
e
1/x
2
− 1
2arctgx
2
− π
.(Ds. −
1
2
)
8. lim
x→∞
2x +1
3x
2
+ x − 1
.(DS. 0)
9. lim

x→∞
ln(1 + x
2
)
ln[(π/2) − arctgx]
.(DS. −2)
10. lim
x→∞
√
x
2
−1
x
.(DS. −1)
11. lim
x→∞
x
ln(1 + x)
.(D
S. +∞)
12. lim
x→+0
ln sin x
ln sin 5x
.(D
S. 1)
13. lim
x→a
arcsin
x −a

a
cotg(x −a). (DS. 1/a)
14. lim
x→∞
(π − 2arctgx)lnx.(DS. 0)
15. lim
x→∞
(a
1/x
− 1)x, a>0. (DS. lna)
16. lim
x→1
(2 −x)
tg
πx
2
.(DS. e
2/π
)
17. lim
x→1

1
lnx
−
x
lnx

.(DS. −1)
18. lim

x→∞
(x −x
2
ln(1 + 1/x)). (Ds. 1/2)
19. lim
x→0

1
x
2
− cotg
2
x

.(DS. 2/3)
20. lim
x→0
x
1/ln(e
x
−1)
.(DS. e)