1. NGUYˆE ˜
N THUY’ THANH
B`AI TˆA.
P
TO´AN CAO CˆA ´
P
Tˆa.
p 2
Ph´ep t´ınh vi phˆan c´ac h`am
NH`A XUˆA ´
T BA ’
N DA.
C QUˆO ´
C GIA H`A NˆO.
I HO.
I
5. 4 Chu.o.ng 7. Gi´o.i ha.
n v`a liˆen tu.c cu’ a h`am sˆo´
7.1 Gi´o.i ha.
n cu’a d˜ay sˆo´
H`am sˆo´ x´ac di.
nh trˆen tˆa.
.p N du.o.
p ho.
.c go.
i l`a d˜ay sˆo´ vˆo ha.n. D˜ay sˆo´
thu.`o.ng du.o.
.c viˆe´t du .
´o.i da.
ng:
a1, a2, . . . ,an, . . . (7.1)
ho˘a.
c {an}, trong d´o an = f(n), n 2 N du.o.
.c go.
i l`a sˆo´ ha.ng tˆo’ng qu´at
cu’ a d˜ay, n l`a sˆo´ hiˆe.
u cu’a sˆo´ ha.ng trong d˜ay.
Ta cˆa `n lu .
u ´y c´ac kh´ai niˆe.
m sau dˆay:
i) D˜ay (7.1) du.o.
.c go.
n nˆe´u 9M 2 R+ : 8 n 2 N ) |an| 6
i l`a bi. ch˘a.
n nˆe´u: 8M 2 R+ : 9 n 2 N ) |an|>M.
M; v`a go. i l`a khˆong bi. ch˘a.
ii) Sˆo´ a du.o.
.c go.
i l`a gi´o.i ha.
n cu’ a d˜ay (7.1) nˆe´u:
8" > 0, 9N(") : 8 n > N ) |an − a| < ". (7.2)
iii) Sˆo´ a khˆong pha’ i l`a gi´o.i ha.
n cu’ a d˜ay (7.1) nˆe´u:
9" > 0, 8N : 9 n > N ) |an − a| > ". (7.3)
iv) D˜ay c´o gi´o.i ha.
n du.o.
.c go.
i l`a d˜ay hˆo.
, trong tru.`o.ng ho.
i tu.
.p ngu.o.
.c
la. i d˜ay (7.1) go. i l`a d˜ay phˆan k`y.
v) D˜ay (7.1) go. i l`a d˜ay vˆo c`ung b´e nˆe´u lim
n!1
an = 0 v`a go.
i l`a d˜ay
vˆo c`ung l´o.n nˆe´u 8A > 0, 9N sao cho 8n > N ) |an| > A v`a viˆe´t
lim an = 1.
n cˆa`n dˆe’ d˜ay hˆo.
vi) Diˆe`u kiˆe.
i tu.
l`a d˜ay d´o pha’i bi. ch˘a.
n.
Ch´u ´y: i) Hˆe.th´u.c (7.2) tu.o.ng du.o.ng v´o.i:
−" < an −a < " , a − " < an < a+ ". (7.4)
6. 7.1. Gi´o.i ha.
n cu’ a d˜ay sˆo´ 5
Hˆe.
th´u.c (7.4) ch´u.ng to’ r˘a`ng mo.i sˆo´ ha.ng v´o.i chı’ sˆo´n > N cu’ a d˜ay
hˆo.
i tu.
d`ˆeu n`a˘m trong khoa’ng (a − ", a + "), khoa’ng na`y go. i la` "-laˆn
n cu’a diˆe’m a.
Nhu. vˆa.
cˆa.
y, nˆe´u d˜ay (7.1) hˆo.
i tu.
dˆe´n sˆo´ a th`ı mo.i sˆo´ ha.ng cu’ a n´o tr`u.
t sˆo´ h˜u.u ha.
ra mˆo.
n sˆo´ ha.ng dˆe`u n˘a`m trong "-lˆan cˆa.
n bˆa´t k`y b´e bao
nhiˆeu t`uy ´y cu’a diˆe’m a.
ii) Ta lu.u ´y r˘a`ng d˜ay sˆo´ vˆo c`ung l´o.n khˆong hˆo.
i tu.
v`a k´y hiˆe.
u
lim an = 1 (−1) chı’ c´o ngh˜ıa l`a d˜ay an l`a vˆo c`ung l´o.n v`a k´y hiˆe.
u d´o
ho`an to`an khˆong c´o ngh˜ıa l`a d˜ay c´o gi´o.i ha.
n.
7.1.1 C´ac b`ai to´an liˆen quan t´o.i di.
nh ngh˜ıa gi´o.i
ha.n
Dˆe’ ch´u.ng minh liman = a b˘a`ng c´ach su. ’ du.ng di.
nh ngh˜ıa, ta cˆa `n tiˆe´n
h`anh theo c´ac bu.´o.c sau dˆay:
p biˆe’
u th´u.c |an − a|
i) Lˆa.
.i) sao cho |an − a| 6 bn 8 n v`a
ii) Cho.n d˜ay bn (nˆe´u diˆe`u d´o c´o l o.
v´o.i " du’ b´e bˆa´t k`y bˆa´t phu .
o.ng tr`ınh dˆo´i v´o .
i n:
bn < " (7.5)
t c´ach dˆe˜ d`ang. Gia’ su. ’ (7.5) c´o nghiˆe.m l`a n > f("),
c´o thˆe’ gia’i mˆo.
f(") > 0. Khi d´o ta c´o thˆe’ lˆa´y n l`a [f(")], trong d´o [f(")] l`a phˆa ` n
nguyˆen cu’ a f(").
C´AC V´I DU.
V´ı du.
1. Gia’ su’. an = n(−1)n. Ch´u.ng minh r˘a`ng:
i) D˜ay an khˆong bi. ch˘a.
n.
ii) D˜ay an khˆong pha’i l`a vˆo c`ung l´o.n.
Gia’ i. i) Ta ch´u.ng minh r˘a`ng an tho’a m˜an di.
nh ngh˜ıa d˜ay khˆong
bi. ch˘a.
n. Thˆa.
y, 8M >0 sˆo´ ha.ng v´o.i sˆo´ hiˆe.
t vˆa.
u n = 2([M] + 1) b˘a`ng
n v`a l´o.n ho .
n M. Diˆe`u d´o c´o ngh˜ıa l`a d˜ay an khˆong bi. ch˘a.
n.
7. 6 Chu.o.ng 7. Gi´o.i ha.
n v`a liˆen tu.c cu’ a h`am sˆo´
ii) Ta ch´u.ng minh r˘a`ng an khˆong pha’ i l`a vˆo c`ung l´o.n. Thˆa.
t vˆa.
y,
ta x´et khoa’ng (−2, 2). Hiˆe’n nhiˆen mo.i sˆo´ ha.ng cu’ a d˜ay v´o.i sˆo´ hiˆe.
u le’
dˆe`u thuˆo.
c khoa’ng (−2, 2) v`ı khi n le’ th`ı ta c´o:
n(−1)n
= n−1 = 1/n 2 (−2, 2).
Nhu. vˆa.
y trong kho’ng (−2, 2) c´o vˆo sˆo´ sˆo´ ha.ng cu’ a d˜ay. T`u. d´o,
nh ngh˜ıa suy ra an khˆong pha’ i l`a vˆo c`ung l´o.n. N
theo di.
V´ı du.
nh ngh˜ıa gi´o.i ha.
2. D`ung di.
n d˜ay sˆo´ dˆe’ ch´u.ng minh r˘a`ng:
1) lim
n!1
(−1)n−1
n
= 0. 2) lim
n!1
n
n + 1
= 1.
Gia’ i. Dˆe’ ch´u.ng minh d˜ay an c´o gi´o.i ha.
n l`a a, ta cˆa `n ch´u.ng minh
r˘a`ng dˆo´i v´o .
i mˆo˜
i sˆo´ " > 0 cho tru.´o.c c´o thˆe’
t`ım du.o.
.c sˆo´ N (N phu.
thuˆo.
c ") sao cho khi n >N th`ı suy ra |an − a| < ". Thˆong thu.`o.ng ta
c´o thˆe’ chı’ ra cˆong th´u.c tu .
`o.ng minh biˆe’u diˆe˜n N qua ".
1) Ta c´o:
|an − 0| =
13. =
1
n
·
Gia’ su’. " la` sˆo´ du.o.ng cho tru.o´.c tu`y y´. Khi do´:
1
n
< " , n >
1
"
·
. nhiˆen n`ao d´o tho’a m˜an diˆe`u kiˆe.
V`ı thˆe´ ta c´o thˆe’ lˆa´y N l`a sˆo´ tu.
n:
N >
1
"
)
1
N
< ".
(Ch˘a’
ng ha.n, ta c´o thˆe’ lˆa´y N = [1/"], trong d´o [1/"] l`a phˆa ` n nguyˆen
cu’a 1/").
Khi d´o 8 n > N th`ı:
|an − 0| =
1
n 6 1
N
< ".
14. 7.1. Gi´o.i ha.
n cu’ a d˜ay sˆo´ 7
Diˆe`u d´o c´o ngh˜ıa l`a lim
n!1
(−1)n
n
= 0.
. nhiˆen N(") sao cho 8n >
2) Ta lˆa´y sˆo´ " > 0 bˆa´t k`y v`a t`ım sˆo´ tu.
N(") th`ı:
26. =
1
n + 1 6 1
N + 1
< " ) lim
n!1
n
n + 1
= 1. N
V´ı du.
3. Ch´u.ng minh r˘a`ng c´ac d˜ay sau dˆay phˆan k`y:
1) an = n, n 2 N (7.6)
2) an = (−1)n, n2 N (7.7)
3) an = (−1)n +
1
n
· (7.8)
Gia’ i. 1) Gia’ su. ’ d˜ay (7.6) hˆo.
i tu.
v`a c´o gi´o.i ha.n l`a a. Ta lˆa´y " = 1.
nh ngh˜ıa gi´o.i ha.
Khi d´o theo di.
n tˆo`n ta.
i sˆo´ hiˆe.u N sao cho 8n > N th`ı
ta c´o |an−a| < 1 ngh˜ıa l`a |n−a| < 1 8n >N. T`u. d´o −1 < n−a < 1
8n > N , a − 1 < n < a+ 1 8n > N.
Nhu.ng bˆa´t d˘a’
ng th´u.c n < a+ 1, 8n > N l`a vˆo l´y v`ı tˆa.
.p c´ac
p ho.
. nhiˆen khˆong bi. ch˘a.
sˆo´ tu.
n.
2) C´ach 1. Gia’ su. ’ d˜ay an hˆo.
i tu.
v`a c´o gi´o.i ha.
n l`a a. Ta lˆa´y lˆan
cˆa.
n
a −
1
2
, a +
1
2
cu’a diˆe’m a. Ta viˆe´t d˜ay d˜a cho du.´o.i da.
ng:
{an} = −1, 1,−1, 1, . . . . (7.9)
27. 8 Chu.o.ng 7. Gi´o.i ha.
n v`a liˆen tu.c cu’ a h`am sˆo´
V`ı dˆo.
d`ai cu’ a khoa’ng
a −
1
2
, a +
1
2
l`a b˘a`ng 1 nˆen hai diˆe’m −1
v`a +1 khˆong thˆe’ dˆo`ng th`o.i thuˆo.
c lˆan cˆa.
n
a −
1
2
, a +
1
2
cu’a diˆe’m a,
a.
v`ı khoa’ng
cach ´giu.˜a −1 v`a +1 ba`˘ng 2. Di`ˆeu do´ co´ ngh˜ıa la` o’. ngoa`i
lan ˆcˆn
a −
1
2
, a +
1
2
c´o vˆo sˆo´ sˆo´ ha.ng cu’a d˜ay v`a v`ı thˆe´ (xem ch´u
´y o. ’ trˆen) sˆo´ a khˆong thˆe’ l`a gi´o.i ha.
n cu’ a d˜ay.
Ca´ch 2. Gia’ su’. an ! a. Khi do´ 8 0 (lˆa´y =
1
2
) ta c´o
|an − a|
1
2
8 n N.
V`ı an = ±1 nˆen
|1 − a|
1
2
, | − 1 − a|
1
2
)2 = |(1 − a) + (1 + a)| 6 |1 − a| + |a+ 1| 6 1
2
+
1
2
= 1
)2 1, vˆo l´y.
3) Lu.u ´y r˘a`ng v´o.i n = 2m ) a2m = 1+
1
2m
. Sˆo´ ha.ng kˆe` v´o.i n´o
c´o sˆo´ hiˆe.
u le’ 2m+1 (hay 2m − 1) v`a
a2m+1 = −1 +
1
2m + 1
0 (hay a2m−1 = −1 +
1
2m − 1 6 0).
T`u. d´o suy r˘a`ng
|an − an−1| 1.
Nˆe´u sˆo´ a n`ao d´o l`a gi´o.i ha.
n cu’ a d˜ay (an) th`ı b˘a´t dˆa ` u t`u. sˆo´ hiˆe.
u n`ao
ng th´u.c |an − a|
d´o (an) tho’a m˜an bˆa´t d˘a’
1
2
. Khi d´o
|an − an+1| 6 |an − a| + |an+1 − a|
1
2
+
1
2
= 1.
Nhu.ng hiˆe.
u gi˜u.a hai sˆo´ ha.ng kˆe ` nhau bˆa´t k`y cu’a d˜ay d˜a cho luˆon luˆon
l´o.n ho .
n 1. Diˆe`u mˆau thuˆa˜n n`ay ch´u.ng to’ r˘a`ng khˆong mˆo.
t sˆo´ thu.
.c
n`ao c´o thˆe’ l`a gi´o.i ha.
n cu’ a d˜ay d˜a cho. N
28. 7.1. Gi´o.i ha.
n cu’ a d˜ay sˆo´ 9
B`AI TˆA.
P
H˜ay su. ’ du.ng di.
nh ngh˜ıa gi´o.i ha.
n dˆe’ ch´u.ng minh r˘a`ng
1. lim
n!1
an = 1 nˆe´u an =
2n − 1
2n + 2
2. lim
n!1
an =
3
5
nˆe´u an =
3n2 + 1
5n2 − 1
B˘a´t dˆa ` u t`u. sˆo´ hiˆe.
u N n`ao th`ı:
|an − 3/5| 0, 01 (DS. N = 5)
3. lim
n!1
an = 1 nˆe´u an =
3n + 1
3n .
4. lim
n!1
cos n
n
= 0.
5. lim
n!1
2n + 5 · 6n
3n + 6n = 5.
6. lim
n!1
3 p
n2 sin n2
n + 1
= 0.
7. Ch´u.ng minh r˘a`ng sˆo´ a = 0 khˆong pha’ i l`a gi´o.i ha.
n cu’ a d˜ay an =
n2 − 2
2n2 − 9
.
8. Ch´u.ng minh r˘a`ng
lim
n!1
n2 + 2n + 1 + sinn
n2 + n + 1
= 1.
9. Ch´u.ng minh r˘a`ng d˜ay: an = (−1)n + 1/n phˆan k`y.
10. Ch´u.ng minh r˘a`ng d˜ay; an = sin n0 phˆan k`y.
11. T`ım gi´o.i ha.
n cu’ a da˜y: 0, 2; 0, 22; 0, 222; . . . , 0, |22{.z. . 2}
n
, . . .
Chı’ dˆa˜n. Biˆe’u diˆe˜n an du.´o.i da.
ng
an = 0, 22 . . .2 =
2
10
+
2
10
2
+ · · · +
2
10n (DS. lim an = 2/9)
29. 10 Chu.o.ng 7. Gi´o.i ha.
n v`a liˆen tu.c cu’ a h`am sˆo´
12. T`ım gi´o.i ha.
n cu’ a d˜ay sˆo´:
0, 2; 0, 23; 0, 233; 0, 2333; . . . , 0, 2 3|3{.z. . }3
n
, . . .
Chı’ dˆa˜n. Biˆe’u diˆe˜n an du.´o.i da.
ng
an =
2
10
+
3
102 +
3
103 + · · · +
3
10n
(DS. 7/30)
13. Ch´u.ng minh r˘a`ng nˆe´u d˜ay an hˆo.
i tu.
dˆe´n a, c`on d˜ay bn dˆa `n dˆe´n
1 th`ı d˜ay an/bn dˆa `n dˆe´n 0.
14. Ch´u.ng minh r˘a`ng
i) lim
n!1
n
2n = 0.
ii) lim
n!1
n
an = 0 (a 1).
Chı’ dˆa˜n. i) Su. ’ du.ng hˆe.
th´u.c:
2n = (1 + 1)n = 1+n +
n(n − 1)
2
+ · · · + 1 n +
n(n − 1)
2
n2
2
·
v`a u.´o.c lu .
.ng |an − 0|.
o.
ii) Tu.o.ng tu.
. nhu. i). Su’. du.ng hˆe.
th´u.c:
an = [1+(a − 1)]n
n(n − 1)
2
(a − 1).
15. Ch´u.ng minh r˘a`ng
lim an = 2 nˆe´u an = 1+
1
2
+ · · · +
1
2n
Chı’ dˆa˜n. ´Ap du.ng cˆong th´u.c t´ınh tˆo’ng cˆa´p sˆo´ nhˆan dˆe’ t´ınh an rˆo`i
u.´o.c lu .
.ng |an − 2|.
o.
16. Biˆe´t r˘a`ng d˜ay an c´o gi´o.i ha.
n, c`on d˜ay bn khˆong c´o gi´o.i ha.
n. C´o
thˆe’ n´oi g`ı vˆe` gi´o.i ha.n cu’ a d˜ay:
i) {an + bn}.
ii) {anbn}.
(DS. i) lim{an + bn} khˆong tˆo`n ta.
i. H˜ay ch´u.ng minh.
30. 7.1. Gi´o.i ha.
n cu’ a d˜ay sˆo´ 11
ii) C´o thˆe’ g˘a.
p ca’ hai tru.`o.ng ho.
.p c´o gi´o.i ha.
n v`a khˆong c´o gi´o.i ha.
n,
u.
v´ı d:
an =
n − 1
n
, bn = (−1)n; an =
1
n
, bn = (−1)n.
7.1.2 Ch´u.ng minh su.
. hˆo.
i tu.
cu’a d˜ay sˆo´ du.
.a trˆen
c´ac di.
nh l´y vˆe` gi´o.i ha.
n
Dˆe’ t´ınh gi´o.i ha.
n cu’ a d˜ay sˆo´, ngu.`o.i ta thu.`o.ng su. ’ du.ng c´ac di.
nh l´y v`a
kh´ai niˆe.
m sau dˆay:
Gia’ su’. liman = a, limbn = b.
i) lim(an ± bn) = liman ± lim bn = a ± b.
ii) lim anbn = lim an · lim bn = a · b.
iii) Nˆe´u b6= 0 th`ı b˘a´t dˆa ` u t`u. mˆo.
t sˆo´ hiˆe.
u n`ao d´o d˜ay an/bn x´ac
nh (ngh˜ıa l`a 9N : 8 n N ) bn6= 0) v`a:
di.
lim
an
bn
=
lim an
lim bn
=
a
b
·
iv) Nˆe´u liman = a, limbn = a v`a b˘a´t dˆa ` u t`u. mˆo.
t sˆo´ hiˆe.
u n`ao d´o
an 6 zn 6 bn th`ı lim zn = a (Nguyˆen l´y bi. ch˘a.
n hai phi´a).
v) T´ıch cu’ a d˜ay vˆo c`ung b´e v´o.i d˜ay bi. ch˘a.
n l`a d˜ay vˆo c`ung b´e.
vi) Nˆe´u (an) l`a d˜ay vˆo c`ung l´o.n v`a an6= 0 th`ı d˜ay
1
an
l`a d˜ay vˆo
c`ung b´e; ngu.o.
.c la.
i, nˆe´u n l`a d˜ay vˆo c`ung b´e v`a n6= 0 th`ı d˜ay
1
n
l`a vˆo c`ung l´o.n.
n x´et. Dˆe’ ´ap du. ng d´ung d˘a´n c´ac di.
Nhˆa.
nh l´y trˆen ta cˆa `n lu .
u ´y mˆo.
t
sˆo´ nhˆa.
n x´et sau dˆay:
nh l´y (iii) vˆe` gi´o.i ha.
i) Di.
n cu’a thu .
o.ng s˜e khˆong ´ap du.ng du.o.
.c nˆe´u
tu. ’ sˆo´ v`a mˆa˜u sˆo´ khˆong c´o gi´o.i ha.
n h˜u.u ha.
n ho˘a.
c mˆa˜
u sˆo´ c´o gi´o.i ha.
n
b˘a`ng 0. Trong nh˜u.ng tru.`o.ng ho.
.p d´o nˆen biˆe´n dˆo’i so .
bˆo.
d˜ay thu.o.ng,
ch˘a’
ng ha.n b˘a`ng c´ach chia ho˘a.
c nhˆan tu. ’ sˆo´ v`a mˆa˜u sˆo´ v´o.i c`ung mˆo.
t
biˆe’u th´u.c.
31. 12 Chu.o.ng 7. Gi´o.i ha.
n v`a liˆen tu.c cu’ a h`am sˆo´
ii) Dˆo´i v´o .
nh l´y (i) v`a (ii) c˜ung cˆa ` n pha’ i thˆa.
i di.
n tro.ng khi ´ap du. ng.
Trong tru.`o.ng ho.
.p n`ay ta cˆa ` n pha’i biˆe´n dˆo’i c´ac biˆe’u th´u.c an ± bn v`a
an · bn tru.´o.c khi t´ınh gi´o.i ha.
n (xem v´ı du.
1, iii).
iii) Nˆe´u an = a const 8 n th`ı lim
n!1
an = a.
C´AC V´I DU.
V´ı du.
1. T`ım lim an nˆe´u:
.
o 1) an = (1 + 7n+2)/(3 − 7n)
2) an = (2 + 4 +6+· · · + 2n)/[1 + 3 + 5+· · · + (2n + 1)]
3) an = n3/(12 + 22 + · · · + n2)
Gia’ i. Dˆe’ gia’ i cac ´b`ai toan ´n`ay ta d`ung ly ´thuyˆe´t caˆ´p sˆo´
1) Nhan ˆtu. ’ soˆ´ v`a maˆ˜u sˆo´ phan ˆthu.´c v´i 7−n ta co:
´an =
1 + 7n+2
3 − 7n =
7−n + 72
3 · 7−n − 1
Do d´o
lim an = lim
7−n + 72
3 · 7−n − 1
= −49 v`ı lim 7−n = 0, n!1.
2) Tu. ’ sˆo´ v`a mˆa˜u sˆo´ dˆe`u l`a cˆa´p sˆo´ cˆo.
ng nˆen ta c´o:
2 + 4 + 6+· · · + 2n =
2 + 2n
2
· n;
1+3 + 5+· · · + (2n + 1) =
1 + (2n + 2)
2
(n + 1).
Do d´o
an =
n
n + 1
) lim an = 1.
3) Nhu. ta biˆe´t:
12 + 22 + · · · + n2 =
n(n + 1)(2n + 1)
6
32. 7.1. Gi´o.i ha.
n cu’ a d˜ay sˆo´ 13
v`a do d´o:
lim an = lim
6n3
n(n + 1)(2n + 1)
= lim
6
(1 + 1/n)(2 + 1/n)
= 3. N
V´ı du.
2. T`ım gi´o.i ha.
n
lim
1 +
1
2
+
1
4
+ · · · +
1
2n
1 +
1
3
+
1
9
+ · · · +
1
3n
Gia’ i. Tu’. soˆ´ va` maˆ˜u sˆo´ d`ˆeu la` caˆ´p sˆo´ nhaˆn nˆen
1 +
1
2
+ · · · +
1
2n =
2(2n − 1)
2n ,
1 +
1
3
+ · · · +
1
3n =
3(3n − 1)
2 · 3n
v`a do d´o:
lim an = lim
2(2n − 1)
2n
·
2 · 3n
3(3n − 1)
= 2 lim
2n − 1
2n
·
2
3
lim
3n
3n − 1
= 2 lim[1 − (1/2)n] ·
2
3
lim
1
1 − (1/3)n = 2· 1 ·
2
3
·1 =
4
3
· N
V´ı du.
3.
1) an =
p
n2 + n − n
2) an = 3 p
n + 2 − 3 p
n
3) an = 3 p
n2 − n3 + n
Gia’ i.
1) Ta biˆe´n dˆo’i an b˘a`ng c´ach nhˆan v`a chia cho da.
i lu .
.ng liˆen ho.
o.
.p
an =
(
p
n2 + n − n)(
p
n2 + n + n)
p
n2 + n + n
=
n
p
n2 + n + n
=
1 p
1 + 1/n + 1
Do d´o
lim an =
1
lim
n!1
(
p
1 + 1/n + 1)
=
1
2
·
33. 14 Chu.o.ng 7. Gi´o.i ha.
n v`a liˆen tu.c cu’ a h`am sˆo´
2) Biˆe´n dˆo’i an tu.o.ng tu.
. nhu. 1) ta c´o:
an =
3 p
3
n + 2
−
3 p
3
n
3 p
2
n + 2
+ 3 p
n + 2 · 3 p
n +
3 p
2
n
an =
2
3 p
2
n + 2
+ 3 p
n + 2 · 3 p
n +
3 p
2
n
Biˆe’u th´u.c mˆa˜
u sˆo´ b˘a`ng:
n2/3 3 p
1 + 2/n
2 + 3 p
1 + 2/n + 1
!1
khi n!1v`a do d´o lim an = 0.
3) Ta c´o thˆe’ viˆe´t n = 3 p
n3 v`a ´ap du.ng cˆong th´u.c:
a3 + b3 = (a + b)(a2 − ab + b2)
suy ra
an =
3 p
3 p
n2 − n3 + n
n2 − n3
2
− n 3 p
n2 − n3 + n2
3 p
n2 − n3
2
− n 3 p
n2 − n3 + n2
=
n2
3 p
n2 − n3
2
− n 3 p
n2 − n3 + n2
=
1
[1/n − 1]2/3 − [1/n − 1]1/3 + 1
suy ra lim an =
1
3
· N
V´ı du.
4. T`ım gi´o.i ha.
n cu’ a c´ac d˜ay sau
an =
n
p
n2 + n
, bn =
n
p
n2 + 1
,
cn =
1
p
n + 1
+
1
p
n2 + 2
+ · · · +
1
p
n2 + n
·
Gia’ i. Dˆa `u tiˆen ta ch´u.ng minh lim an = 1. Thˆa.
t vˆa.
y:
lim an = lim
n
p
n
1 + 1/n
= lim
1 p
1 + 1/n
= 1.
34. 7.1. Gi´o.i ha.
n cu’ a d˜ay sˆo´ 15
Tu.o.ng tu.
. lim bn = 1.
Dˆe’ t`ım gi´o.i ha.
n cu’ a cn ta s˜e ´ap du.ng Nguyˆen l´y bi. ch˘a.
n hai ph´ıa.
Mˆo.
t m˘a.
t ta c´o:
cn
1
p
n2 + 1
+
1
p
n2 + 1
+ · · · +
1
p
n2 + 1
=
n
p
n2 + 1
= bn
nhu.ng m˘a.
t kh´ac:
cn
1
p
n2 + n
+
1
p
n2 + n
+ · · · +
1
p
n2 + n
= an.
Nhu. vˆa.
y an cn bn v`a lim
n!1
an = lim
n!1
bn = 1. T`u. d´o suy ra
lim
n!1
cn = 1. N
V´ı du.
5. Ch´u.ng minh r˘a`ng d˜ay (qn) l`a: 1) d˜ay vˆo c`ung l´o.n nˆe´u
|q| 1; 2) d˜ay vˆo c`ung b´e khi |q| 1.
Gia’ i. 1) Gia’ su’. |q| 1. Ta lˆa´y sˆo´ A 0 baˆ´t ky`. Tu`. da˘’ng thu´.c
|q|n Ata thu du.o.
.c n log|q|A. Nˆe´u ta lˆa´y N = [log|q|A] th`ı 8n N
ta c´o |q|n A. Do d´o d˜ay (qn) l`a d˜ay vˆo c`ung l´o.n.
2) Gia’ su’. |q| 1, q6= 0. Khi do´ qn =
h1
q
ni−1
. V`ı
40. 1 nˆen
d˜ay
1
q
n
l`a d˜ay vˆo c`ung l´o.n v`a do d´o d˜ay
h1
q
n
i−1
l`a vˆo c`ung
b´e, t´u.c l`a d˜ay (qn) l`a d˜ay vˆo c`ung b´e khi |q| 1.
3) Nˆe´u q = 0 th`ı qn = 0, |q|n 8 n v`a do d´o (qn) l`a vˆo c`ung b´e.
N
B`AI TˆA.P
T`ım gi´o.i ha.
n lim
n!1
an nˆe´u
1. an =
n2 − n
n −
p
n
. (DS. 1)
2. an = n2(n −
p
n2 + 1). (DS. −1)
41. 16 Chu.o.ng 7. Gi´o.i ha.
n v`a liˆen tu.c cu’ a h`am sˆo´
3. an =
1 + 2 + 3+· · · + n
p
9n4 + 1
. (DS. 1/6)
4. an =
p
n cos n
n + 1
. (DS. 0)
5. an =
5n
n + 1
+
sin n
n
. (DS. 5)
6. an =
n3
n2 + 1
−
3n2
3n + 1
. (DS. 1/3)
7. an =
n
n + 11
−
cos n
10n
. (DS. 1)
8. an =
n3 + 1
n2 − 1
(DS. 1)
9. an =
cos n3
n
−
3n
6n + 1
. (DS. −
1
2
)
10. an =
(−1)n
5
p
n + 1
. (DS. 0)
11. an =
p
n2 + 1+
p
n
3 p
n3 + n −
p
n
. (DS. +1)
12. an = 3 p
1 − n3 + n. (DS. 0)
13. an =
p
n2 + 4n
3 p
n3 − 3n2
. (DS. 1)
14. an =
(n + 3)!
2(n + 1)! − (n + 2)!
. (DS. −1)
15. an =
2 + 4+· · · + 2n
n + 2
− 2. (DS. −1)
16. an = n − 3 p
n3 − n2. (DS.
1
3
)
17. an =
1 − 2 + 3 − 4 + 5 − ·· ·−2n
p
n2 + 1+
p
4n2 + 1
. (DS. −
1
3
)
18. an =
1
1 · 2
+
1
2 · 3
+ · · · +
1
n(n + 1)
.
Chı’ dˆa˜n. ´Ap du. ng
1
n(n + 1)
=
1
n
−
1
n + 1
(DS. 1)
42. 7.1. Gi´o.i ha.
n cu’ a d˜ay sˆo´ 17
19. an = 1−
1
3
+
1
9
−
1
27
+ · · · +
(−1)n−1
3n−1 . (DS.
3
4
)
20. an =
2n+1 + 3n+1
2n + 3n . (DS. 3)
21. an =
n + (−1)n
n − (−1)n. (DS. 1)
22. an =
1
p
n
1
p
1 +
p
3
+
1
p
3 +
p
5
+ · · · +
1
p
2n − 1 +
p
2n + 1
Chı’ dˆa˜n. Tru.
c c˘an th´u.c o. ’ mˆa˜u sˆo´ c´ac biˆe’u th´u.c trong dˆa´u ngo˘a.
c.
(DS.
1
p
2
)
23. an =
1
1 · 2 · 3
+
1
2 · 3 · 4
+ · · · +
1
n(n + 1)(n + 2)
Chı’ dˆa˜n. Tru.´o.c hˆe´t ta ch´u.ng minh r˘a`ng
1
n(n + 1)(n + 2)
=
1
2
h 1
n(n + 1)
−
1
(n + 1)(n + 2)
i
(DS.
1
4
)
24. an =
1
a1a2
+
1
a2a3
+ · · · +
1
anan+1
. (DS.
1
a1d
)
ng v´o.i cˆong sai d6= 0, an6= 0.
trong d´o {an} l`a cˆa´p sˆo´ cˆo.
25. an = (1 − 1/4)(1 − 1/9) · · · (1 − 1/(n + 1)2). (DS.
1
2
)
Chı’ dˆa˜n. B˘a`ng quy na.p to´an ho.c ch´u.ng to’ r˘a`ng an =
n + 2
2n + 2
.
7.1.3 Ch´u.ng minh su.
. hˆo.
i tu.
cu’a d˜ay sˆo´ du.
.a trˆen
n du’ dˆe’ d˜ay hˆo.i tu.
diˆe`u kiˆe.
(nguyˆen l´y
Bolzano-Weierstrass)
D˜ay sˆo´ an du.o.
.c go.
i l`a:
i) D˜ay t˘ang nˆe´u an+1 an 8 n
ii) D˜ay gia’m nˆe´u an+1 an 8 n
C´ac d˜ay t˘ang ho˘a.c gia’m c`on du.o.
.c go.
i l`a d˜ay do.n diˆe.
u. Ta lu.u ´y
r˘a`ng d˜ay do.n diˆe.
u bao gi`o. c˜ung bi. ch˘a.
t ph´ıa. Nˆe´u d˜ay
n ´ıt nhˆa´t l`a mˆo.
43. 18 Chu.o.ng 7. Gi´o.i ha.
n v`a liˆen tu.c cu’ a h`am sˆo´
do.n diˆe.
n du .
u t˘ang th`ı n´o bi. ch˘a.
´o.i bo .
’ i sˆo´ ha.ng dˆa ` u tiˆen cu’ a n´o, d˜ay
do.n diˆe.
n trˆen bo. ’ i sˆo´ ha.ng dˆa ` u. Ta c´o di.
u gia’m th`ı bi. ch˘a.
nh l´y sau dˆay
thu.`o.ng du.o.
.c su .
’ du.ng dˆe’ t´ınh gi´o.i ha.
n cu’ a d˜ay do.n diˆe.
u.
D-
nh l´y Bolzano-Weierstrass. D˜ay do.n diˆe.
i.
u v`a bi. ch˘a.
n th`ı hˆo.
i tu.
.
Di.
nh l´y n`ay kh˘a’
ng di.
nh vˆe` su.
. tˆo`n ta.
i cu’ a gi´o.i ha.
n m`a khˆong chı’
ra du.o.
.c phu .
o.ng ph´ap t`ım gi´o.i ha.
y, trong nhiˆe`u tru.`o.ng
n d´o. Tuy vˆa.
ho.
.p khi biˆe´t gi´o.i ha.
i, c´o thˆe’ chı’ ra phu.o.ng ph´ap t´ınh
n cu’ a d˜ay tˆo`n ta.
c t´ınh to´an thu.`o.ng du.
n´o. Viˆe.
.a trˆen d˘a’
ng th´u.c d´ung v´o.i mo.
i d˜ay hˆo.
i
tu. :
lim
n!1
an+1 = lim
n!1
an.
Khi t´ınh gi´o.i ha.
.a trˆen d˘a’
ng th´u.c v`u.a
n du.
.i ho .
n ca’ la` su’.
nˆeu ti ˆe.n lo.
du. ng c´ach cho d˜ay b˘a`ng cˆong th´u.c truy hˆo `i.
C´AC V´I DU.
V´ı du.
1. Ch´u.nh minh r˘a`ng d˜ay:
an =
1
5 + 1
+
1
52 + 1
+ · · · +
1
5n + 1
hˆi .
ut.
o.Gia’ i. D˜ay d˜a cho do.n diˆe.
u t˘ang. Thˆa.
t vˆa.
y v`ı:
an+1 = an +
1
5n+1 + 1
nˆen an+1 an.
D˜ay d˜a cho bi. ch˘a.
n trˆen. Thˆa.
t vˆa.
y:
an =
1
5 + 1
+
1
52 + 1
+
1
53 + 1
+ · · · +
1
5n + 1
1
5
+
1
52 + · · · +
1
5n
=
1
5
−
1
5n+1
1 −
1
5
=
1
4
1 −
1
5n
1
4
·
Nhu. vˆa.
y d˜ay an d˜a cho do.n diˆe.
u t˘ang v`a bi. ch˘a.
n trˆen nˆen n´o hˆo.
i
tu. . N
44. 7.1. Gi´o.i ha.
n cu’ a d˜ay sˆo´ 19
V´ı du.
2. Ch´u.ng minh r˘a`ng d˜ay an =
2n
n!
hˆo.
i tu.
v`a t`ım gi´o.i ha.
n cu’ a
n´o.
Gia’ i. D˜ay d˜a cho c´o da.ng
2
1
,
22
2
, . . . ,
2n
n!
, . . .
D˜ay an do.n diˆe.
u gia’m. Thˆa.
t vˆa.
y
an+1
an
=
2n+1
(n + 1)!
:
2n
n!
=
2
n + 1
1 8n 1.
n trˆen bo. ’ i phˆa ` n tu .
Do d´o an+1 an v`a d˜ay bi. ch˘a.
’ a1. Ngo`ai ra
n du .
an 0, 8 n nˆen d˜ay bi. ch˘a.
´o.i. Do d´o d˜ay do.n diˆe.
u gia’m v`a bi.
ch˘a.
n. N´o hˆo.
i tu.
nh l´y Weierstrass. Gia’ su. ’ a l`a gi´o.i ha.
theo di.
n cu’ a n´o.
Ta c´o:
an+1
an
=
2
n + 1
) an+1 =
2
n + 1
an.
T`u. d´o
lim an+1 = lim
2an
n + 1
= lim
2
n + 1
lim an
v`a nhu. vˆa.
y: a = 0· a ! a = 0. Vˆa.
y: lim
2n
n!
= 0. N
V´ı du.
3. Cho d˜ay an =
p
2, an+1 =
p
2an. Ch´u.ng minh r˘a`ng d˜ay hˆo.
i
tu. v`a t`ım gi´o.i ha.
n cu’ a n´o.
Gia’ i. Hiˆe’n nhiˆen r˘a`ng: a1 a2 a3 · · · . D´o l`a d˜ay do.n diˆe.
u
n du .
t˘ang v`a bi. ch˘a.
´o.i bo .
’ i sˆo´
p
2. Ta ch´u.ng minh r˘a`ng n´o bi. ch˘a.
n trˆen
bo’.i sˆo´ 2.
Thˆa.
t vˆa.
y
a1 =
p
2; a2 =
p
2a1
p
2 · 2 = 2.
Gia’ su. ’ d˜a ch´u.ng minh du.o..c r˘a`ng an 6 2.
Khi d´o:
an+1 =
p
2an 6
p
2 ·2 = 2.
45. 20 Chu.o.ng 7. Gi´o.i ha.
n v`a liˆen tu.c cu’ a h`am sˆo´
e.
Vˆy theo tiˆen d`ˆe quy na.p ta co´ an 6 2 8 n.
Nhu. thˆe´ day ˜an do.n dia.
ˆn nˆen n´o c´o gi´o.i ha.
u t˘ang v`a bi. ch˘a.
n d´o
l`a a.
Ta c´o:
an+1 =
p
2an ) a2
n+1 = 2an.
Do d´o:
lim a2
n+1 = 2 lim an
hay a2 − 2a = 0 v`a thu du.o.
.c a1 = 0, a2 = 2.
V`ı d˜ay do.n diˆe.
u t˘ang 8 n nˆen gi´o.i ha.
n a = 2. N
V´ı du.
4. Ch´u.ng minh t´ınh hˆo.
i tu.
v`a t`ım gi´o.i ha.
n cu’ a d˜ay
x1 =
p
a; x2 =
q
a +
p
a, . . . ,
xn =
r
a +
q
a + · · · +
p
a, a 0, n dˆa´u c˘an.
Gia’ i. i) R˜o r`ang: x1 x2 x3 · · · xn xn+1 .. . ngh˜ıa l`a
d˜ay d˜a cho l`a d˜ay t˘ang.
ii) Ta ch´u.ng minh d˜ay xn l`a d˜ay bi. ch˘a.
n. Thˆa.
t vˆa.
y, ta c´o:
x1 =
p
a
p
a+ 1
x2 =
q
a +
p
a
q
a +
p
a+ 1
q
p
a+ 1 =
a + 2
p
a + 1.
Gia’ su. ’ d˜a ch´u.ng minh du.o..c r˘a`ng: xn
p
a + 1.
Ta cˆa `n ch´u.ng minh xn+1
p
a + 1. Thˆa.
t vˆa.
y, ta c´o:
xn+1 =
p
a + xn
q
a +
p
a + 1
q
p
a + 1 =
a+ 2
p
a + 1.
Do d´o nh`o. ph´ep quy na.p to´an ho.c ta d˜a ch´u.ng minh r˘a`ng d˜ay d˜a
cho bi. ch˘a.n trˆen bo’.i
p
a+ 1.
46. 7.1. Gi´o.i ha.
n cu’ a d˜ay sˆo´ 21
iii) Dˆe’ t`ım gi´o.i ha.
n ta x´et hˆe.
th´u.c xn =
p
a + xn−1 hay
x2
n = a + xn−1.
T`u. d´o:
lim x2
n = lim(a + xn−1) = a + lim xn−1
hay nˆe´u gia’ thiˆe´t limxn = A th`ı: A2 = a + A ! A2 − A − a = 0 v`a
A1 =
1 +
p
1 + 4a
2
, A2 =
1 −
p
1 + 4a
2
·
V`ı A2 0 nˆen gi´a tri. A2 bi. loa.i v`ı xn 0.
Do d´o;
lim xn =
1 +
p
1 + 4a
2
· N
V´ı du.
5. T`ım gi´o.i ha.
n cu’ a d˜ay an du.o.
.c x´ac di.
nh nhu. sau: a1 l`a sˆo´
t`uy ´y m`a
0 a1 1, an+1 = an(2 − an) 8 n 1. (7.10)
Gia’ i. i) Dˆa ` u tiˆen ch´u.ng minh r˘a`ng an bi. ch˘a.
n, m`a cu. thˆe’ l`a b˘a`ng
ph´ep quy na.p to´an ho.c ta ch´u.ng minh r˘a`ng
0 an 1. (7.11)
Ta c´o 0 a1 1. Gia’ su. ’ (7.11) d˜a du.o.
.c ch´u.ng minh v´o.i n v`a ta
s˜e ch´u.ng minh (7.11) d´ung v´o.i n + 1 .
T`u. (7.10) ta c´o; an+1 = 1− (1 − an)2.
T`u. hˆe.
th´u.c n`ay suy ra 0 (1 − an)2 1, v`ı 0 an 1.
.
aa.
T`u. do ´suy ra: 0 an+1 1 8 n.
ii) Bay ˆgi`o. ta chu.´ng minh r˘a`ng an l`a day ˜tang.
˘Thˆt vˆy, v`ı an 1 nˆen 2 − an 1. Chia (7.10) cho an ta thu
du.o.
.c:
an+1
an
= 2− an 1.
47. 22 Chu.o.ng 7. Gi´o.i ha.
n v`a liˆen tu.c cu’ a h`am sˆo´
T`u. d´o an+1 an 8 n. Nhu .
vˆa.
y d˜ay an do.n diˆe.
u t˘ang v`a bi. ch˘a.
n.
nh l´y Weierstrass, limAn tˆo`n ta.
Do d´o theo di.
i v`a ta k´y hiˆe.
u n´o l`a a.
iii) T`u. (7.10) ta c´o:
lim an+1 = lim an · lim(2 − an)
hay a = a(2 − a).
T`u. d´o a = 0 v`a a = 1. V`ı x1 0 v`a d˜ay an t˘ang nˆen
a = 1 = liman. N
V´ı du.
6. Ch´u.ng minh r˘a`ng d˜ay an =
n!
nn hˆo.
i tu.
v`a t`ım gi´o.i ha.
n cu’ a
n´o.
Gia’ i. i) Ta ch´u.ng minh r˘a`ng d˜ay an do.n diˆe.
u gia’m, thˆa.
t vˆa.
y:
an+1 =
(n + 1)!
(n + 1)n+1 =
n!
(n + 1)n =
n!
nn
·
nn
(n + 1)n =
nn
(n + 1)n an
v`ı
nn
(n + 1)n 1 nˆen an+1 an.
n du .
V`ı an 0 nˆen n´o bi. ch˘a.
´o.i v`a do d´o lim an tˆo`n ta.
i, k´y hiˆe.
u
lim an = a v`a r˜o r`ang l`a a = lim an 0.
ii) Ta ch´u.ng minh a = 0. Thˆa.
t vˆa.
y ta c´o:
(n + 1)n
nn =
n + 1
n
n
=
1 +
1
n
n
1 +
n
n
= 2.
Do d´o:
nn
(n + 1)n
1
2
v`a an+1
1
2
an.
Chuyˆe’n qua gi´o.i ha.
.c a 6 a
n ta du.o.
2
) a = 0. N
B`AI TˆA.
P
48. 7.1. Gi´o.i ha.
n cu’ a d˜ay sˆo´ 23
1. Cho c´ac d˜ay sˆo´:
1) an =
5n2
n2 + 3
· 2) bn = (−1)n 2n
n + 1
sin n. 3) cn = n cos n.
H˜ay chı’ ra d˜ay n`ao bi. ch˘a.
n v`a d˜ay n`ao khˆong bi. ch˘a.
n.
(DS. 1) v`a 2) bi. ch˘a.
n; 3) khˆong bi. ch˘a.
n)
2. Ch´u.ng minh r˘a`ng d˜ay:
a1 =
a0
a + a0
, a2 =
a1
a + a1
, a3 =
a2
a + a2
, . . . ,
an =
an−1
a + an−1
, . . . (a 1, a0 0)
hˆo.
i t.
.
u3. Ch´u.ng minh c´ac d˜ay sau dˆay hˆo.
i tu.
1) an =
n2 − 1
n2
2) an = 2+
1
2!
+
1
3!
+ · · · +
1
n!
n du.o.
Chı’ dˆa˜n. T´ınh bi. ch˘a.
.c suy t`u. n! 2n−1 v`a do d´o
an 6 2 +
1
2
+
1
22 + · · · +
1
2n−1 = 3−
1
2n−1 3.
4. Ch´u.ng minh c´ac d˜ay sau dˆay hˆo.
i tu.
v`a t`ım gi´o.i ha.
n a cu’a ch´ung
1) a1 = k p
5, an+1 = k p
5an, k 2 N. (DS. k−1 p
5)
2) an =
2n
(n + 2)!
Chı’ dˆa˜n.
an+1
an
=
2
n + 3
1. (DS. a = 0)
3) an =
E(nx)
n
trong d´o E(nx) l`a phˆa ` n nguyˆen cu’ a nx.
Chı’ dˆa˜n. Su. ’ du.ng hˆe.
th´u.c: nx−1 E(nx) 6 nx. (DS. a = x)
5. Ch´u.ng minh r˘a`ng d˜ay: an = a1/2n hˆo.
i tu.
v`a t`ım gi´o.i ha.
n cu’a n´o
(a 1).
49. 24 Chu.o.ng 7. Gi´o.i ha.
n v`a liˆen tu.c cu’ a h`am sˆo´
(DS. a = 1. Chı’ dˆa˜n. Ch´u.ng minh r˘a`ng an l`a d˜ay do.n diˆe.
u gia’m
v`ı
an+1 = a1/2n+1
= a1/(2n·2) =
p
an, an 1)
6. Ch´u.ng minh r˘a`ng d˜ay
an = 1+
1
22 +
1
32 + · · · +
1
n2
hˆo.
i t.
.
uChı’ dˆa˜n. Ch´u.ng to’ r˘a`ng d˜ay do.n diˆe.
n cu’a n´o
u t˘ang, t´ınh bi. ch˘a.
du.o.
.c x´ac lˆa.
p b˘a`ng c´ach su. ’ du.ng c´ac bˆa´t d˘a’
ng th´u.c:
1
n2
1
n(n − 1)
=
1
n − 1
−
1
n
, n 2.
7. Ch´u.ng minh r˘a`ng d˜ay
an =
1
3 + 1
+
1
32 + 2
+ · · · +
1
3n + n
c´o gi´o.i ha.
n h˜u.u ha.
n.
n cu’ a an du.o.
Chı’ dˆa˜n. T´ınh bi. ch˘a.
.c x´ac lˆa.
p b˘a`ng c´ach so s´anh an
v´o.i tˆo’
ng mˆo.
t cˆa´p sˆo´ nhˆan n`ao d´o.
8. Ch´u.ng minh r˘a`ng d˜ay
1 +
1
n
n+1
do.n diˆe.
u gia’m v`a
lim
n!1
1 +
1
n
n+1
= e.
9. T´ınh lim
n!1
an, nˆe´u
1) an =
1 +
1
n + k
n
, k 2 N. (DS. e)
2) an =
n
n + 1
n
. (DS.
1
e
)
3) an =
1 +
1
2n
n
. (DS.
p
e)
4) an =
2n + 1
2n
2n
. (DS. e)
50. 7.1. Gi´o.i ha.
n cu’ a d˜ay sˆo´ 25
7.1.4 Ch´u.ng minh su.
. hˆo.
i tu.
cu’a d˜ay sˆo´ du.
.a trˆen
n cˆa`n v`a du’ dˆe’ d˜ay hˆo.
diˆe`u kiˆe.
i tu.
(nguyˆen
l´y hˆo.
i tu.
Bolzano-Cauchy)
Trˆen dˆay ta d˜a nˆeu hai phu.o.ng ph´ap ch´u.ng minh su.
. hˆo.
i tu.
cu’ a d˜ay.
Hai phu.o.ng ph´ap n`ay khˆong ´ap du.ng du.o.
.c dˆo´i v´o .
i c´ac d˜ay khˆong do.n
u du.o.
diˆe.
.c cho khˆong b˘a`ng phu.o.ng ph´ap gia’ i t´ıch m`a du.o.
.c cho b˘a`ng
phu.o.ng ph´ap kh´ac (ch˘a’
ng ha.n b˘a`ng phu.o.ng ph´ap truy hˆo `i). M˘a.
t
kh´ac, trong nhiˆe`u tru.`o.ng ho.
.p ngu.`o.i ta chı’ quan tˆam dˆe´n su.
. hˆo.
i tu.
hay phˆan k`y cu’ a d˜ay m`a thˆoi. Sau dˆay ta ph´at biˆe’u mˆo.
t tiˆeu chuˆa’n
c´o t´ınh chˆa´t “nˆo.
i” cho ph´ep kˆe´t luˆa.
i ta.
. hˆo.
n su.
i tu.
cu’ a d˜ay chı’ du.
.a
trˆen gi´a tri. cu’a c´ac sˆo´ ha. ng cu’ a d˜ay:
Nguyˆen l´y hˆo.
. D˜ay (an) c´o gi´o.i ha.
i tu.
n h˜u.u ha.
n khi v`a chı’ khi n´o
tho’a m˜an diˆe`u kiˆe.
n:
8 0, 9N0 = N0() 2 N : 8n N0 v`a 8 p 2 N
) |an − an+p| .
T`u. nguyˆen l´y hˆo.
i tu.
r´ut ra: D˜ay (an) khˆong c´o gi´o.i ha.n khi v`a chı’
khi n´o tho’a m˜an diˆe`u kiˆe.
n:
9 0, 8N 2 N 9 n N 9m N ! |an − am| .
C´AC V´I DU.
V´ı du.
1. Ch´u.ng minh r˘a`ng d˜ay
an =
cos 1
3
+
cos 2
32 + · · · +
cos n
3n , n2 N
hˆo.
i t.
.
u
51. 26 Chu.o.ng 7. Gi´o.i ha.
n v`a liˆen tu.c cu’ a h`am sˆo´
Gia’ i. Ta u.´o.c lu .
.ng hiˆe.
o.
u
|an+p − an| =
63. p
p
n + p
8 n, p 2 N.
D˘a.
t v´o .
c biˆe.
i p = n ta c´o
|an − a2n|
p
n
p
2
1
p
2
8 n. (*)
Ta lˆa´y =
1
p
2
i nh˜u.ng gi´a tri. n N v`a
. Khi d´o 8N 2 N tˆo`n ta.
9 p 2 N sao cho |an − an+p| . Thˆa.
y, theo bˆa´t d˘a’ng th´u.c (*) ta
t vˆa.
64. 7.2. Gi´o.i ha.
n h`am mˆo.
t biˆe´n 27
chı’ cˆa `n lˆa´y sˆo´n N bˆa´t k`y v`a p = n. T`u. d´o theo mˆe.
nh dˆe` phu’ di.
nh
nguyˆen l´y hˆo.
i tu.
ta c´o d˜ay d˜a cho phˆan k`y. N
B`AI TˆA.
P
Su. ’ du.ng tiˆeu chuˆa’n hˆo.
i tu.
dˆe’ ch´u.ng minh su.
. hˆo.
i tu.
cu’ a d˜ay (an)
nˆe´u
1. an =
Pn
k=1
sin n
2n , 2 R.
2. an =
Pn
k=1
akqk, |q| 1, |ak| M 8 k,M 0.
3. an =
Pn
k=1
(−1)k−1
k(k + 1)
·
4. an =
Pn
k=1
(−1)k
k!
·
5. an = 0, |77{.z. . 7}
nch˜u. sˆo´
.
6. an =
Pn
k=1
1
2k + k
·
Ch´u.ng minh r˘a`ng c´ac d˜ay sau dˆay phˆan k`y:
7. an = 1+
1
2
+ · · · +
1
n
, n 2 N.
8. an =
1
ln2
+
1
ln3
+ · · · +
1
lnn
, n = 2, . . .
7.2 Gi´o.i ha.
t biˆe´n
n h`am mˆo.
7.2.1 C´ac kh´ai niˆe.m v`a di.
nh l´y co .
ba’n vˆe` gi´o.i ha.
n
nh ngh˜ıa gi´o.i ha.
Di.
n cu’ a c´ac h`am dˆo´i v´o .
i n˘am tru.`o.ng ho.
.p: x ! a,
x ! a ± 0, x!±1du.o.
.c ph´at biˆe’u nhu .
sau.
65. 28 Chu.o.ng 7. Gi´o.i ha.
n v`a liˆen tu.c cu’ a h`am sˆo´
1) Sˆo´ A du.o.
.c go.
i l`a gi´o.i ha.
i diˆe’m a (khi x ! a)
n cu’ a h`am f(x) ta.
nˆe´u 8 0 b´e bao nhiˆeu t`uy ´y t`ım du.o.
.c sˆo´ = () 0 (9 = ()
0) sao cho 8 x m`a
x 2 Df {x; 0 |x − a| ()}
th`ı
|f(x) − A| .
K´y hiˆe.
u: lim
x!a
f(x) = A.
2) Sˆo´ A du.o.
.c go.
i l`a gi´o.i ha.
.
an bˆen pha’ i (bˆen tr´ai) cu’ a h`am f(x) ti
diˆe’m x = a nˆe´u 8 0, 9 = () 0 sao cho v´o.i mo.
i x tho’a m˜an
diˆe`u kiˆe.
n
x 2 Df {x : a x a+ } (x 2 Df {x : a − x a})
th`ı
|f(x) − A| .
K´y hiˆe.
u:
lim
x!a+0
f(x) = f(a + 0)
lim
x!a−0
f(x) = f(a − 0)
.
Tu.o.ng tu.
.:
3) lim
x!+1
f(x) = A, 8 0 9 0 : 8 x 2 Df {x : x }
) |f(x) − A| .
nh ngh˜ıa gi´o.i ha.n khi x!−1du.o.
Di.
.c ph´at biˆe’u tu .
o.ng tu.
..
4) Nˆe´u lim
x!+1
f(x) = lim
x!−1
f(x) = A th`ı ngu.`o.i ta viˆe´t
lim
x!1
f(x) = A.
66. 7.2. Gi´o.i ha.
n h`am mˆo.
t biˆe´n 29
Tru.`o.ng ho.
.p d˘a.
t nˆe´u A = 0 th`ı h`am f(x) du.o.
c biˆe.
.c go.
i l`a h`am vˆo
c`ung b´e khi x ! a (x ! a ± 0, x!±1).
m h`am vˆo c`ung l´o.n ta.
Kh´ai niˆe.
i diˆe’m a c˜ung du.o.
.c ph´at biˆe’u dˆo´i
v´o.i ca’ n˘am tru.`o.ng ho.
.p.
Ch˘a’
ng ha.n, h`am f(x) du.o.
.c go.
i l`a h`am vˆo c`ung l´o.n ta.
i diˆe’m a nˆe´u
8M 0 9 = (M) 0 : 8 x 2 Df {x : 0 |x − a| }
) |f(x)|M.
Ngo`ai ra, nˆe´u f(x) 0 (f(x) 0) 8 x 2 Df {x : 0 |x−a| }
th`ı ta viˆe´t
lim
x!a
f(x) = +1
lim
x!a
f(x) = −1
.
Ta lu.u ´y r˘a`ng c´ac k´y hiˆe.
u v`u.a nˆeu chı’ ch´u.ng to’ f(x) l`a vˆo c`ung
l´o.n ch´u. ho`an to`an khˆong c´o ngh˜ıa r˘a`ng f c´o gi´o.i ha.
n.
Khi t´ınh gi´o.i ha.
n ta thu .
`o.ng su. ’ du.ng c´ac diˆe`u kh˘a’
ng di.
nh sau dˆay.
D-
nh l´y 7.2.1. Nˆe´u c´ac gi´o.i ha.
i.
n lim
x!a
f1(x), lim
x!a
i h˜u .
f2(x) tˆo`n ta.
u ha.
n
th`ı
1) lim
[f1(x) + f2(x)] = lim
x!a
x!a
f1(x) + lim
x!a
f2(x)
2) lim
[f1(x) · f2(x)] = lim
x!a
x!a
f1(x) · lim
x!a
f2(x)
3) Nˆe´u lim
x!a
f2(x)6= 0 th`ı lim
x!a
f1(x)
f2(x)
=
lim
x!a
f1(x)
lim
x!a
f2(x)
4) Nˆe´u trong lˆan cˆa.
n U(a; ) = {x : 0 |x − a| } ta c´o
f1(x) 6 f(x) 6 f2(x) v`a lim
x!a
f1(x) = lim
x!a
f2(x) = A th`ı lim
x!a
f(x) = A
(nguyˆen l´y bi. ch˘a.
n hai phi´a).
nh ngh˜ıa gi´o.i ha.
Di.
n h`am sˆo´ c´o thˆe’ ph´at biˆe’u du .
´o.i da.
ng ngˆon ng˜u.
d˜ay nhu. sau.
D-
nh l´y 7.2.2. Gia’ su. ’ D R, a 2 R l`a diˆe’m tu.
i.
cu’a n´o; A 2 R,
f : D ! R. Khi d´o
lim
x!a
f(x) = A
67. 30 Chu.o.ng 7. Gi´o.i ha.
n v`a liˆen tu.c cu’ a h`am sˆo´
khi v`a chı’ khi 8(an), an 2 D {a}, an ! a
f(an) ! A
T`u. d´o dˆe’ ch´u.ng minh mˆo.
t h`am n`ao d´o khˆong c´o gi´o.i ha.
n khi x ! a,
ta chı’ cˆa `n ch´u.ng minh r˘a`ng 9(an), 9(a0
n) dˆe`u hˆo.
i tu.
dˆe´n a nhu.ng
lim
x!a
f(an)6= lim
x!a
f0(an).
nh l´y co .
C´ac di.
ba’n vˆe` gi´o.i ha.
n d˜a ph´at biˆe’u trˆen dˆay khˆong ´ap
du. ng du.o.
.c dˆo´i v´o .
i c´ac gi´o.i ha.
n sau dˆay khi x ! a, a 2 R.
1) lim
x!a
[f(x)+g(x)]; f, g l`a c´ac vˆo c`ung l´o.n (vˆo di.
nh da.ng “1±1”).
2) lim
x!a
f(x)
g(x)
; f, g ho˘a.
c dˆo`ng th`o.i l`a hai vˆo c`ung b´e, ho˘a.
c dˆo`ng th`o.i
l`a hai vˆo c`ung l´o.n (vˆo di.
nh da.ng “0/0” ho˘a.
c “1/1”).
3) lim
x!a
f(x) ·g(x); f l`a vˆo c`ung b´e, c`on g l`a vˆo c`ung l´o.n ho˘a.
c ngu.o.
.c
la.i (vˆo di.
nh da.ng “0 · 1”).
4) lim
x!a
f(x)
g(x)
:
nh da.ng “11”)
a) khi f(x) ! 1, g(x)!1(vˆo di.
b) khi f(x) ! 0, g(x) ! 0 (vˆo di.
nh da.ng “00”)
c) khi f(x)!1, g(x) ! 0 (vˆo di.
nh da.ng “10”)
c t´ınh gi´o.i ha.
Viˆe.
n trong c´ac tru.`o.ng ho.
.p n`ay thu.`o.ng du.o.
.c gi
.
ola` khu’. da. ng voˆ d.inh. Trong nhi`ˆeu tru.o`.ng ho.
.p khi t´ınh gi´o.i ha.
n ta
thu.`o.ng su. ’ du.ng c´ac gi´o.i ha.
n quan tro.ng sau dˆay:
lim
x!0
sin x
x
= 1, (7.12)
lim
x!0
(1 + x)
1
x = e (7.13)
68. 7.2. Gi´o.i ha.
n h`am mˆo.
t biˆe´n 31
v`a c´ac hˆe.
qua’ cu’ a (7.13)
lim
x!1
1 +
1
x
x
= e, (7.14)
lim
x!0
loga(1 + x)
x
=
1
lna
, 0 a6= 1, (7.15)
lim
x!0
ax − 1
x
= lna, 0 a6= 1. (7.16)
C´AC V´I DU.
V´ı du.
1. Su. ’ du.ng ( − ) - di.
nh ngh˜ıa gi´o.i ha.
n dˆe’ ch´u.ng minh r˘a`ng
lim
x!−3
x2 = 9.
Gia’ i. Ta cˆa `n ch´u.ng minh r˘a`ng 8 0, 9 0 sao cho v´o.i
|x+ 3| th`ı ta c´o |x2 − 9| .
Ta cˆa `n u.´o.c lu .
.ng hiˆe.
o.
u |x2 − 9|. ta c´o
|x2 − 9| = |x − 3||x + 3|.
Do th`u.a sˆo´ |x − 3| khˆong bi. ch˘a.
n trˆen to`an tru.c sˆo´ nˆen dˆe’ u.´o.c lu .
.ng
o.
t´ıch do.n gia’n ho .
n cu’a diˆe’m a = −3 t´u.c l`a
n ta tr´ıch ra 1 - lˆan cˆa.
khoa’ng (−4;−2). V´o.i mo.
i x 2 (−4;−2) ta c´o |x − 3| 7 v`a do d´o
|x2 − 9| 7|x + 3|.
n diˆe’m a = −3 [t´u.c l`a khoa’ng (−3 − ;−3 + )] khˆong
V`ı -lˆan cˆa.
du.o.
.c vu .
.t ra kho’i ranh gi´o.i cu’ a 1-lˆan cˆa.n nˆen ta lˆa´y = min
o.
1,
7
.
Khi d´o v´o .
i 0 |x + 3| ) |x2 − 9| . Do vˆa.
y lim
x!−3
x2 = 9. N
V´ı du.
2. Ch´u.ng minh r˘a`ng lim
x!2
p
11 − x = 3.
Gia’ i. Gia’ su’. 0 la` sˆo´ du.o.ng cho tru.o´.c b´e bao nhiˆeu tu`y y´. Ta
x´et bˆa´t phu .
o.ng tr`ınh
p
11 − x − 3| . (7.17)
|
69. 32 Chu.o.ng 7. Gi´o.i ha.
n v`a liˆen tu.c cu’ a h`am sˆo´
Ta c´o
(7.17),−
p
11 − x − 3 ,
8
p
11 − x − 3 −
p
11 − x − 3
:
,
8
:
x − 11 −(3 − )2
x − 11 −(3 + )2
,
8
:
x − 2 6 − 3
x − 2 −(6 + 2).
V`ı 6 − 2 | − (6 + )2| = 6 + 2 nˆen ta c´o thˆe’ lˆa´y () l`a sˆo´
6 6 − 2. V´o .
i sˆo´ d´o ta thˆa´y r˘a`ng khi x tho’a m˜an bˆa´t d˘a’
ng th´u.c
p
11 − x − 3| v`a
0 |x − 2| th`ı |
lim
x!2
p
11 − x = 3. N
V´ı du.
3. T´ınh c´ac gi´o.i ha.
n
1) lim
x!2
2x − x2
x − 2
(vˆo di.
nh da.ng
0
0
);
2) lim
x!
4
cotg2x · cotg
4
− x
(vˆo di.
nh da.ng 0 · 1);
3) lim
x!1
e
1
x +
1
x
x
nh da.ng 11).
(vˆo di.
Gia’ i
1) Ta c´o
2x − x2
x − 2
=
2x − 22 − (x2 − 22)
x − 2
= 4 ·
2x−2 − 1
x − 2
−
x2 − 4
x − 2
·
T`u. d´o suy r˘a`ng
lim
x!2
2x − x2
x − 2
= 4 lim
x!2
2x−2 − 1
x − 2
− lim
x!2
x2 − 4
x − 2
= 4ln2 − 4.
2) D˘a.
t y =
4
− x. Khi d´o
lim
x!
4
cotg2x · cotg
4
− x
= lim
y!0
cotg
2
− 2y
cotgy
= lim
y!0
sin 2y
sin y
·
cos y
cos 2y
= 2.
70. 7.2. Gi´o.i ha.
n h`am mˆo.
t biˆe´n 33
3) D˘a. t y =
1
x
. Khi d´o
lim
x!1
e
1
x +
1
x
x
= lim
y!0
(ey + y)
lim
y!0
1
y = e
ln(ey+y)
y ;
lim
y!0
ln(ey + y)
y
= lim
y!0
ln[1 + (ey + y − 1)]
ey + y − 1
·
ey + y − 1
y
= lim
t!0
ln(1 + t)
t
· lim
y!0
1 +
ey − 1
y
= 2.
T`u. d´o suy r˘a`ng
lim
y!0
ey + y
1
y = e2. N
V´ı du.
4. Ch´u.ng to’ r˘a`ng h`am f(x) = sin
1
x
khˆong c´o gi´o.i ha.
n khi
x ! 0.
Gia’ i. Ta lu.u ´y mˆe.
nh dˆo´i v´o .
nh dˆe` phu’ di.
nh ngh˜ıa gi´o.i ha.
i di.
n:
lim
x!a
f(x)6= A, 90 0 8 0 9 x (0 |x − a| )
! |f(x0) − A| 0.
Nˆe´u A = 0 ta lˆa´y 0 =
1
2
v`a xk =
2
2
+ 2k
. Khi d´o 8 0,
9 k 2 N : 0 xk v`a
|f(xk) − 0| = |f(xk)| = 1 0
v`a nhu. vˆa.
y A = 0 khˆong pha’ i l`a gi´o.i ha.
n cu’a h`am d˜a cho khi x ! 0.
Nˆe´u A6= 0 th`ı ta lˆa´y 0 =
|A|
2
v`a xk =
1
2k
. Khi d´o 8 0,
9 k 2 N : 0 xk th`ı |f(xk) − A| = |A| . Nhu .
vˆa.
i sˆo´
y mo.
A6= 0 dˆe`u khˆong l`a gi´o.i ha.
n cu’ a h`am sin
1
x
khi x ! 0. N
V´ı du.
5. H`am Dirichlet D(x):
D(x) =
8
:
1 nˆe´u x 2 Q,
0 nˆe´u x 2 R Q
71. 34 Chu.o.ng 7. Gi´o.i ha.
n v`a liˆen tu.c cu’ a h`am sˆo´
khˆong c´o gi´o.i ha.
i 8 a 2 R.
n ta.
Gia’ i. Ta ch´u.ng minh r˘a`ng ta.i mo.
i diˆe’m a 2 R h`am D(x) khˆong
nh l´y 2. Dˆe’ l`am viˆe.
tho’a m˜an Di.
c d´o, ta chı’ cˆa `n chı’ ra hai d˜ay (an) v`a
(a0
n) c`ung hˆo.
i tu.
dˆe´n a sao cho lim
n!1
D(an)6= lim
n!1
D(a0
n).
Dˆa ` u tiˆen ta x´et d˜ay c´ac diˆe’m h˜u.u ty’ (an) hˆo.
i tu.
dˆe´n a. Ta c´o
D(an) = 1 8 n v`a do d´o lim
n!1
D(an) = 1. Bˆay gi`o. ta x´et d˜ay (a0
n) -
d˜ay c´ac diˆe’m vˆo ty’ hˆo.
dˆe´n a. Ta c´o D(a0
i tu.
n) = 0 8 n v`a do vˆa.
y
lim
n!1
D(a0
n) = 0.
Nhu. vˆa.
y lim
n!1
D(an)6= lim
n!1
n). T`u. d´o suy ra r˘a`ng ta.i diˆe’m a
D(a0
h`am D(x) khˆong c´o gi´o.i ha.
n . N
V´ı du.
6. Gia’ su’. lim
x!a
f(x) = b, lim
x!a
g(x) = +1. Ch´u.ng minh r˘a`ng
lim
x!a
[f(x) + g(x)] = +1.
Gia’ i. Ta cˆa `n ch´u.ng minh r˘a`ng 8M 0, 9 0 sao cho 8 x : 0
|x − a| th`ı f(x) + g(x)M.
V`ı lim
x!a
f(x) = b nˆen tˆo`n ta.
n U(a, 1) cu’a diˆe’m a sao cho
i 1-lˆan cˆa.
|f(x)| C, x6= a (7.18)
trong d´o C l`a h˘a`ng sˆo´ du.o.ng n`ao d´o.
Gia’ su’. M 0 la` sˆo´ cho tru.o´.c tu`y y´. V`ı lim
x!a
g(x) = +1 nˆen dˆo´i
v´o.i sˆo´ M + C, 9 0 ( 6 1) sao cho 8 x : 0 |x − a| th`ı
g(x) M + C (7.19)
T`u. c´ac bˆa´t d˘a’
ng th´u.c (7.18) v`a(7.19) ta thu du.o.
.c l`a: v´o.i x tho’a
n 0 |x − a| 6 1 th`ı
m˜an diˆe`u kiˆe.
f(x) + g(x) g(x) − |f(x)|M + C − C = M. N
B`AI TˆA.
P
72. 7.2. Gi´o.i ha.
n h`am mˆo.
t biˆe´n 35
1. Su. ’ du.ng di.
nh ngh˜ıa gi´o.i ha.
n h`am sˆo´ dˆe’ ch´u.ng minh c´ac d˘a’
ng th´u.c
sau dˆay:
1) lim
x!
6
sin x =
1
2
; 2) lim
x!
2
sin x = 1;
3) lim
x!0
x sin
1
x
= 0; 4) lim
x!+1
arctgx =
2
.
Chı’ dˆa˜n. D`ung hˆe.
th´u.c
2
− arctgx tg
2
− arctgx
=
1
x
)
5) lim
x!1
x − 1
3x + 2
=
1
3
; 6) lim
x!+1
logax = +1;
7) lim
x!+1
p
x2 + 1 − x
= 0; 8) lim
x!−5
x2 + 2x − 15
x + 5
= −8;
9) lim
x!1
(5x2 − 7x + 6) = 4; 10) lim
x!2
x2 − 3x + 2
x2 + x − 6
=
1
5
;
11) lim
x!+1
x sin x
x2 − 100x + 3000
= 0.
2. Ch´u.ng minh c´ac gi´o.i ha.
n sau dˆay khˆong tˆo`n ta.
i:
1) lim
x!1
sin
1
x − 1
; 2) lim
x!1
sin x; 3) lim
x!o
2
1
x ;
4) lim
x!0
e
1
x ; 5) lim
x!1
cos x.
Nˆe´u tu .
’ sˆo´ v`a mˆa˜u sˆo´ cu’ a phˆan th´u.c h˜u.u ty’ dˆe`u triˆe.
t tiˆeu ta.i diˆe’m
x = a th`ı c´o thˆe’ gia’n u.´o.c phˆan th´u.c cho x − a (6= 0) mˆo.
t ho˘a.
c mˆo.
t
sˆo´ lˆa ` n.
Su. ’ du. ng phu.o.ng ph´ap gia’n u.´o.c d´o, h˜ay t´ınh c´ac gi´o.i ha.
n sau dˆay
(3-10).
3. lim
x!7
2x2 − 11x − 21
x2 − 9x + 14
(DS.
17
5
)
4. lim
x!1
x4 − x3 + x2 − 3x + 2
x3 − x2 − x + 1
(DS. 2)
5. lim
x!1
x4 + 2x2 − 3
x2 − 3x + 2
(DS. −8)
6. lim
x!1
xm − 1
xn − 1
; m, n 2 Z (DS.
m
n
)
73. 36 Chu.o.ng 7. Gi´o.i ha.
n v`a liˆen tu.c cu’ a h`am sˆo´
7. lim
x!1
1
1 − x
−
3
1 − x3
(DS. −1)
8. lim
x!1
a
1 − xa
−
b
1 − xb
; a, b 2 N (DS.
a − b
2
)
9. lim
x!1
(xn − 1)(xn−1 − 1) · · · (xn−k+1 − 1)
(x − 1)(x2 − 1) · · · (xk − 1)
(DS. Ck
n)
10. lim
x!a
(xn − an) − nan−1(x − a)
(x − a)2 , n 2 N (DS.
n(n − 1)
2
an−1)
Chı’ dˆa˜n. Dˆo’i biˆe´n x − a = t.
C´ac b`ai to´an sau dˆay c´o thˆe’ du.a vˆe` da.ng trˆen nh`o. ph´ep dˆo’i biˆe´n
(11-14)
11. lim
x!1
p
x
q − 1
x
r
s − 1
(DS.
ps
qr
)
12. lim
x!−1
1 + 3 p
x
1 + 5 p
x
(DS.
5
3
)
13. lim
x!0
3 3 p
1 + x − 4 4 p
1 + x + 1
2 − 2
p
1 + x + x
(DS.
1
6
)
14. lim
x!0
n p
1 + x − 1
x
(DS.
1
n
)
Mˆo.
t trong c´ac phu.o.ng ph´ap t´ınh gi´o.i ha.
n cu’ a c´ac biˆe’u th´u.c vˆo ty’
l`a chuyˆe’n vˆo ty’ t`u. mˆa˜u sˆo´ lˆen tu. ’ sˆo´ ho˘a.
c ngu.o.
.c la.
i (15-26)
15. lim
x!0
p
1 + x + x2 − 1
x
(DS.
1
2
)
16. lim
x!2
p
3 + x + x2 −
p
9 − 2x + x2
x2 − 3x + 2
(DS.
1
2
)
17. lim
x!0
5x
3 p
1 + x − 3 p 1 − x
(DS.
15
2
)
18. lim
x!0
3 p
1 + 3x − 3 p
1 − 2x
x + x2 (DS. 2)
19. lim
x!1
p
x2 + 1 −
p
x2 − 1
(DS. 0)
74. 7.2. Gi´o.i ha.
n h`am mˆo.
t biˆe´n 37
20. lim
x!1
3 p
1 − x3 + x
(DS. 0)
21. lim
x!+1
p
x2 + 5x + x
(DS. +1)
22. lim
x!−1
p
x2 + 5x + x
(DS. −
5
2
)
23. lim
x!+1
p
x2 + 2x − x
(DS. 1)
24. lim
x!−1
p
x2 + 2x − x
. (DS. +1)
25. lim
x!1
h
(x + 1)
2
3 − (x − 1)
2
3
i
(DS. 0)
26. lim
x!+1
n p
(x + a1)(x + a2) · · · (x + an) − x
(DS.
a1 + a2 + · · · + an
n
)
Khi gia’ i c´ac b`ai to´an sau dˆay ta thu.`o.ng su. ’ du.ng hˆe.
th´u.c
lim
t!0
(1 + t) − 1
t
= (27-34)
27. lim
x!0
5 p
1 + 3x4 −
p
1 − 2x
3 p
1 + x −
p
1 + x
(DS. −6)
28. lim
x!0
n p
a + x − n p
a − x
x
, n 2 N (DS.
2
n
a
1
n
−1)
29. lim
x!0
p
1 + 3x + 3 p
1 + x − 5 p
1 + x − 7 p
1 + x
4 p
1 + 2x + x − 6 p
1 + x
(DS.
313
280
)
30. lim
x!0
3 p
a2 + ax + x2 − 3 p
a2 − ax + x2
p
a + x −
p
a − x
(DS.
3
2
a
1
6 )
31. lim
x!0
p
1 + x2 + x
n
−
p
1 + x2 − x
n
x
(DS. 2n)
32. lim
x!0
n p
a + x − n p
a − x
x
, n 2 N, a 0 (DS.
2 n p
a
na
)
33. lim
x!0
1 + ax − k p
1 + bx
n p
x
, n 2 N, a 0 (DS.
ak − bn
nk
)
34. lim
x!1
n p
(1 + x2)(2 + x2) · · · (n + x2) − x2
(DS.
n + 1
2
)
75. 38 Chu.o.ng 7. Gi´o.i ha.
n v`a liˆen tu.c cu’ a h`am sˆo´
Khi t´ınh gi´o.i ha.
n c´ac biˆe’u th´u.c lu .
.ng gia´c ta thu.o`.ng su’. du.ng cˆong
o.
th´u.c co .
ba’n
lim
x!0
sin x
x
= 1
c`ung v´o.i su.
. kˆe´t ho.
.p c´ac phu.o.ng ph´ap t`ım gi´o.i ha.
n da˜ nˆeu o’. trˆen
(35-56).
35. lim
x!1
x
2
x
sin
(DS. 0)
36. lim
x!1
arctgx
2x
(DS. 0)
37. lim
x!−2
x2 − 4
arctg(x + 2)
(DS. −4)
38. lim
x!0
tgx − sin x
x3 (DS.
1
2
)
39. lim
x!0
xcotg5x (DS.
1
5
)
40. lim
x!1
(1 − x)tg
x
2
(DS.
2
)
41. lim
x!1
1 − x2
sin x
(DS.
2
)
42. lim
x!
sin x
2 − x2 (DS.
1
2
)
43. lim
x!0
cos mx − cos nx
x2 (DS.
1
2
(n2 − m2))
44. lim
x!1
x2
h
cos
1
x
− cos
3
x
i
(DS. 4)
45. lim
x!0
sin(a + x) + sin(a − x) − 2 sin a
x2 (DS. −sin a)
46. lim
x!0
cos(a + x) + cos(a − x) − 2 cos a
1 − cos x
(DS. −2 cos a)
47. lim
x!1
sin
p
x2 + 1 − sin
p
x2 − 1
(DS. 0)
76. 7.2. Gi´o.i ha.
n h`am mˆo.
t biˆe´n 39
48. lim
x!0
p
cos x − 1
x2 (DS. −
1
4
)
49. lim
x!
2
cos
x
2
− sin
x
2
cos x
(DS.
1
p
2
)
50. lim
x!
3
sin
x −
3
1 − 2 cos x
(DS.
1
p
3
)
51. lim
x!
4
p
2 cos x − 1
1 − tg2x
(DS.
1
4
)
52. lim
x!0
p
1 + tgx −
p
1 − tgx
sin x
(DS. 1)
53. lim
x!0
p
m
cos x − m
p
cos
78. 2 − 2
2m
)
54. lim
x!0
cos x − 3 p
cos x
sin2 x
(DS. −
1
3
)
55. lim
x!0
1 − cos x
p
cos 2x
tgx2 (DS.
3
2
)
56. lim
x!0
p
1 + x sin x − cos x
sin2 x
2
(DS. 4)
Dˆe’ t´ınh gi´o.i ha.
n lim
x!a
[f(x)]'(x), trong d´o
f(x) ! 1, '(x) ! 1 khi x ! a ta c´o thˆe’ biˆe´n dˆo’i biˆe’u th´u.c
[f(x)]'(x) nhu. sau:
lim
x!a
[f(x)]'(x) = lim
x!a
n
[1 + (f(x) − 1)]
1
f(x)−1
o'(x)[f(x)−1]
= e lim
x!a
'(x)[f(x)−1]
o’. daˆy lim
x!a
'(x)[f(x)−1] du.o.
.c t´ınh theo c´ac phu.o.ng ph´ap d˜a nˆeu trˆen
dˆay. Nˆe´u lim
x!a
'(x)[f(x) − 1] = A th`ı
lim
x!a
[f(x)]'(x) = eA (57-68).
79. 40 Chu.o.ng 7. Gi´o.i ha.
n v`a liˆen tu.c cu’ a h`am sˆo´
57. lim
x!1
2x + 3
2x + 1
x+1
(DS. e)
58. lim
x!1
x2 − 1
x2
x4
(DS. 0)
59. lim
x!0
(1 + tgx)cotgx (DS. e)
60. lim
x!0
(1 + 3tg2x)cotg2x (DS. e3)
61. lim
x!0
cos x
cos 2x
1
x2
(DS. e
3
2 )
62. lim
x!
2
(sin x)
1
cotgx (DS. −1)
63. lim
x!
2
(tgx)tg2x (DS. e−1)
64. lim
x!0
h
tg
4
+ x
icotg2x
(DS. e)
65. lim
x!0
cos x
1
x2 (DS. e−1
2 )
66. lim
x!0
cos 3x
1
sin2 x (DS. e−9
2 )
67. lim
x!0
1 + tgx
1 + sin x
1
sin x (DS. 1)
68. lim
x!
4
sin 2x
tg22x (DS. e−1
2 )
Khi t´ınh gi´o.i ha.
n c´ac biˆe’u th´u.c c´o ch´u.a h`am lˆodarit v`a h`am m˜u ta
o .
.
athu.o`.ng su’. du.ng ca´c coˆng thu´.c (7.15) va` (7.16) va` ca´c phu.o.ng pha´p
t´ınh gio.´i hn da ˜nˆeu ’ trˆen (69-76).
69. lim
x!e
lnx − 1
x − e
(DS. e−1)
70. lim
x!10
lgx − 1
x − 10
(DS.
1
10ln10
)
71. lim
x!0
ex2
− 1
p
1 + sin2 x − 1
(DS. 2)
72. lim
x!0
ex2
− cos x
sin2 x
(DS.
3
2
)
93. 6 |xn|.
V`ı |xn| ! 0 khi n!1 nˆen lim
n!1
f(xn) = 0.
ii) H`am d˜a cho khˆong c´o gi´o.i ha.
i diˆe’m x0 = 0. Dˆe’
n bˆen pha’i ta.
ch´u.ng minh diˆe`u d´o t a x´et hai d˜ay hˆo.
i tu.
p nˆen t`u. c´ac d˜ay
dˆe´n 0 lˆa.
sˆo´ du.o.ng xn =
1
2
+ n
v`a x0
n =
1
2n
. Nˆe´u nhu .
h`am f c´o gi´o.i ha.
n
i diˆe’m x0 = 0 th`ı hai d˜ay f(xn) v`a f(x0
bˆen pha’i ta.
n) pha’i hˆo.
i tu.
dˆe´n
t gi´o.i ha.
c`ung mˆo.
n. Thˆe´ nhu.ng f(x0
n) = cos2n = 1 hˆo.
i tu.
dˆe´n 1, c`on
f(xn) = cos
2
+ n
= 0 hˆo.
i tu.
dˆe´n 0.
T`u. d´o suy r˘a`ng h`am c´o gi´an doa.n kiˆe’
i diˆe’m x0 = 0. N
u II ta.
V´ı du.
7. T`ım v`a phˆan loa.i c´ac diˆe’m gi´an doa.n cu’ a c´ac h`am:
1) y = (signx)2; 2) y = [x]
Gia’ i
1) T`u. di.
nh ngh˜ıa h`am signx suy r˘a`ng
(signx)2 =
8
:
1, x6= 0
0, x= 0.
T`u. d´o suy r˘a`ng h`am y = (signx)2 liˆen tu. c 8 x6= 0 (h˜ay du.
.ng d`oˆ
o.
.
thi. cu’ a u h`am) v`a ta.i diˆe’m x0 = 0 ta co ´y(0 − 0) = y(0 + 0)6= y(0).
Diˆe`u do ´co ´ngh˜ıa r˘a`ng x0 = 0 l`a diˆe’m gian ´doa.n kh’ du..c.
2) Gia’ su’. n 2 Z. Nˆe´u n − 1 6 x n th`ı [x] = n − 1, nˆe´u
n 6 x n + 1 th`ı [x] = n (h˜ay du.
.ng dˆo` thi. cu’ a h`am phˆa ` n nguyˆen
[x]). Nˆe´u x062 Z th`ı tˆo`n ta.
n cu’a diˆe’m x0 (khˆong ch´u.a c´ac sˆo´
i lˆan cˆa.
94. 7.3. H`am liˆen tu. c 47
nguyˆen) sao cho ta.i d´o h`am b˘a`ng h˘a`ng sˆo´. Do vˆa.
y n´o liˆen tu.c ta.
i x0.
Nˆe´u x0 = n l`a sˆo´ nguyˆen th`ı [n − 0] = n − 1, [n + 0] = n. T`u. d´o suy
r˘a`ng x0 = n l`a diˆe’m gi´an doa.n kiˆe’
u I. N
V´ı du.
8. Kha’o s´at su.
. liˆen tu. c v`a phˆan loa.i diˆe’m gi´an doa.n cu’ a c´ac
h`am
1) f(x) =
x2
x
, 2) f(x) = e−1
x , 3) f(x) =
8
x nˆe´u x 6 1
lnx nˆe´u x 1.
:
Gia’ i
1) H`am f(x) = x nˆe´u x6= 0 v`a khˆong x´ac di.
nh khi x = 0. V`ı 8 a
ta c´o lim
x!a
x = a nˆen khi a6= 0:
lim
x!a
f(x) = a = f(a)
v`a do vˆa.
y h`am f(x) liˆen tu. c 8 x6= 0. Ta.i diˆe’m x = 0 ta c´o gi´an doa.n
khu. ’ du.o.
.
.c av`ı tˆo`n ti
lim
x!0
f(x) = lim
x!0
x = 0.
x l`a h`am so. cˆa´p v`ı n´o l`a ho.
2) H`am f(x) = e−1
.p cu’ a c´ac h`am
y = −x−1 v`a f = ey. Hiˆe’
n nhiˆen l`a h`am f(x) x´ac di.nh 8 x6= 0 v`a
do d´o n´o liˆen tu. c 8 x6= 0. V`ı h`am f(x) x´ac di.
n diˆe’m
nh trong lˆan cˆa.
nh ta.i ch´ınh diˆe’m x = 0 nˆen diˆe’m x = 0 l`a diˆe’m
x = 0 v`a khˆong x´ac di.
gi´an doa. n. Ta t´ınh f(0 + 0) v`a f(0 − 0).
Ta x´et d˜ay vˆo c`ung b´e t`uy ´y (xn) sao cho xn 0 8 n. V`ı
lim
x!1
−
1
xn
= −1nˆen lim
x!1
e− 1
xn = 0. T`u. d´o suy r˘a`ng lim
x!0+0
e−1
x = 0.
Bˆay gi`o. ta x´et d˜ay vˆo c`ung b´e bˆa´t k`y (x0
n) sao cho x0
0 0 8 n. V`ı
lim
n!1
−
1
x0
n
= +1 nˆen lim
x!0
− 1
e
x0
n = +1. Do d´o lim
x!0−0
e−1
x = +1
t´u.c l`a f(0 − 0) = +1.
Nhu. vˆa.
y gi´o.i ha.
i diˆe’m x = 0 khˆong tˆo`n
n bˆen tr´ai cu’ a h`am f(x) ta.
ta.i do d´o diˆe’m x = 0 l`a diˆe’m gi´an doa.n kiˆe’
u II.
95. 48 Chu.o.ng 7. Gi´o.i ha.
n v`a liˆen tu.c cu’ a h`am sˆo´
3) Ta ch´u.ng minh r˘a`ng f(x) liˆen tu.c ta.
i diˆe’m x = a6= 1. Ta lˆa´y
n cu’a diˆe’m x = a khˆong ch´u.a diˆe’m
|a − 1|, 0. Khi d´o -lˆan cˆa.
x = 1 nˆe´u |a − 1|. Trong -lˆan cˆa.
n n`ay h`am f(x) ho˘a.
c tr`ung v´o.i
h`am '(x) = x nˆe´u a 1 ho˘a.
c tr`ung v´o.i h`am '(x) = lnx nˆe´u a 1.
V`ı c´ac h`am so. cˆa´p co .
ba’n n`ay liˆen tu.c ta.
i diˆe’m x = a nˆen h`am f(x)
i diˆe’m x = a6= 1.
liˆen tu.c ta.
i diˆe’m x = a = 1. Dˆe’ l`am
Ta kha’o s´at t´ınh liˆen tu.c cu’ a h`am f(x) ta.
c d´o ta cˆa ` n t´ınh c´ac gi´o.i ha.
viˆe.
t ph´ıa cu’ a f(x) ta.
n mˆo.
i diˆe’m x = a = 1.
Ta c´o
f(1 + 0) = lim
x!1+0
f(x) = lim
x!1+0
lnx = 0,
f(1 − 0) = lim
x!1−0
f(x) = lim
x!1−0
x = lim
x!1
x = 1.
Nhu. vˆa.
y f(1 +0)6= f(1−0) v`a do d´o h`am f(x) c´o gi´an doa.n kiˆe’u
I ta.
i x = a = 1.
B`AI TˆA.
P
Kha’o s´at t´ınh liˆen tu. c v`a phˆan loa.i diˆe’m gi´an doa.n cu’ a h`am
1. f(x) =
|2x − 3|
2x − 3
(DS. H`am x´ac di.
nh v`a liˆen tu. c 8 x6=
3
2
; ti
.
ax0 =
3
2
h`am c´o gi´an doa.n kiˆe’
u I)
2. f(x) =
8
:
1
x
nˆe´u x6= 0
1 nˆe´u x = 0.
(DS. H`am liˆen tu. c 8 x 2 R)
i hay khˆong gi´a tri. a dˆe’ h`am f(x) liˆen tu.c ta.
3. C´o tˆo`n ta.
i x0 nˆe´u:
1) f(x) =
8 :
4 · 3x nˆe´u x 0
2a + x khi x 0.
(DS. H`am f liˆen tu. c 8 x 2 R nˆe´u a = 2)
96. 7.3. H`am liˆen tu. c 49
2) f(x) =
8
:
x sin
1
x
, x6= 0;
a, x = 0, x0 = 0.
.
(DS. a = 0)
3) f(x) =
8
:
1 + x
1 + x3, x6= −1
a, x = −1, x0 = −1.
(DS. a =
1
3
)
4) f(x) =
8
:
cos x, x 6 0;
a(x − 1), x 0; x0 = 0.
(DS. a = −1)
4. f(x) =
| sin x|
sin x
i x = k, k 2 Z v`ı:
(DS. H`am c´o gi´an doa.n ta.
f(x) =
8
:
1 nˆe´u sinx 0
−1 nˆe´u sinx 0)
5. f(x) = E(x) − E(−x)
(DS. H`am c´o gi´an doa.n khu .
’ du.o.
.c ta.
i x = n, x 2 Z v`ı:
f(x) =
8
:
−1 nˆe´u x = n
0 nˆe´u x6= n.)
6. f(x) =
8
:
e1/x khi x6= 0
0 khi x = 0.
(DS. Ta.i diˆe’m x = 0 h`am c´o gi´an doa.n kiˆe’u II; f(−0) = 0, f(+0) =
1)
T`ım diˆe’m gi´an doa.n v`a t´ınh bu.´o.c nha’y cu’ a c´ac h`am:
7. f(x) = x +
x + 2
|x + 2|
(DS. x = −2 l`a diˆe’m gi´an doa.n kiˆe’
u I, (−2) = 2)
97. 50 Chu.o.ng 7. Gi´o.i ha.
n v`a liˆen tu.c cu’ a h`am sˆo´
8. f(x) =
2|x − 1|
x2 − x3
(DS. x = 0 l`a diˆe’m gi´an doa.n kiˆe’
u II, x = 1 l`a diˆe’m gi´an doa.n kiˆe’
u
I, (1) = −4)
H˜ay bˆo’ sung c´ac h`am sau daˆy ta. i diˆe’m x = 0 dˆe’ chu´ng tro’. tha`nh
liˆen tu. c
9. f(x) =
tgx
x
(DS. f(0) = 1)
10. f(x) =
p
1 + x − 1
x
(DS. f(0) =
1
2
)
11. f(x) =
sin2 x
1 − cos x
(DS. f(0) = 2)
u cu’ a c´ac gi´o.i ha.
12. Hiˆe.
t ph´ıa cu’ a h`am f(x):
n mˆo.
d = lim
x!x0+0
f(x) − lim
x!x0−0
f(x)
du.o.
.c go.
i l`a bu.´o.c nha’y cu’ a h`am f(x) ta.
i diˆe’m x0. T`ım diˆe’m gi´an doa. n
v`a bu.´o.c nha’y cu’ a h`am f(x) nˆe´u:
1) f(x) =
8
:
−
1
2
x2 nˆe´u x 6 2,
x nˆe´u x 2.
(DS. x0 = 2 l`a diˆe’m gi´an doa.n kiˆe’
u I; d = 4)
2) f(x) =
8
:
p
2
x nˆe´u 0 6 x 6 1;
4 − 2x nˆe´u 1 x 6 2, 5;
2x −7 nˆe´u 2, 5 6 x +1.
(DS. x0 = 2,5 l`a diˆe’m gi´an doa.n kiˆe’
u I; d = −1)
3) f(x) =
8
:
2x + 5 nˆe´u −1 x −1,
1
x
nˆe´u − 1 6 x +1.
(DS. x0 = 0 l`a diˆe’m gi´an doa.n kiˆe’
u II; diˆe’m x0 = −1 l`a diˆe’m gi´an
doa.n kiˆe’u I, d = −4)
98. 7.4. Gi´o.i ha.
n v`a liˆen tu.c cu’ a h`am nhiˆe`u biˆe´n 51
7.4 Gi´o.i ha.
n v`a liˆen tu.c cu’a h`am nhiˆe`u
biˆe´n
1. Gia’ su. ’ u = f(M) = f(x, y) x´ac di.
nh trˆen tˆa.
.p D. Gia’ su’.
p ho.
M0(x0, y0) l`a diˆe’m cˆo´ di.
nh n`ao d´o cu’a m˘a.
t ph˘a’
ng v`a x ! x0, y ! y0,
khi d.
u o ´diˆe’m M(x, y) ! M0(x0, y0). Diˆe`u n`ay tu.o.ng du.o.ng vo.´i khoa’ng
cach ´(M,M0) giu.˜a hai diˆe’m M v`a M0 d`a ˆn dˆe´n 0. Ta lu y ´r˘a`ng
(M,M0) = [(x − x0)2 + (y − y0)2]1/2.
Ta c´o c´ac di.
nh ngh˜ıa sau dˆay:
nh ngh˜ıa gi´o.i ha.
i) Di.
n (theo Cauchy)
Sˆo´ b du.o.
.c go.
i l`a gi´o.i ha.
n cu’ a h`am f(M) khi M ! M0 (hay ta. i
diˆe’m M0) nˆe´u
8 0, 9 = () 0 : 8M 2 {D : 0 (M,M0) ()}
) |f(M) − b| .
nh ngh˜ıa gi´o.i ha.
ii) Di.
n (theo Heine)
Sˆo´ b du.o.
.c go.
i l`a gi´o.i ha.
i diˆe’m M0 nˆe´u dˆo´i v´o .
n cu’ a h`am f(M) ta.
i
d˜ay diˆe’m {Mn} bˆa´t k`y hˆo.
i tu.
dˆe´n M0 sao cho Mn 2 D, Mn6= M0
8 n 2 N th`ı d˜ay c´ac gi´a tri. tu.o.ng ´u.ng cu’ a h`am {f(Mn)} hˆo.
i tu.
dˆe´n b.
K´y hiˆe.
u:
i) lim
M!M0
f(M) = b, ho˘a.
c
ii) lim
x ! x0
y ! y0
f(x, y) = b
Hai di.
nh ngh˜ıa gi´o.i ha.
n trˆen dˆay tu.o.ng du.o.ng v´o.i nhau.
nh r˘a`ng theo di.
Ch´u ´y. Ta nhˆa´n ma.
nh ngh˜ıa, gi´o.i ha.
n cu’ a h`am khˆong
phu. thuˆo.
c v`ao phu.o.ng M dˆa `n t´o .
i M0. Do d´o nˆe´u M ! M0 theo
.
acac ´hu.o.´ng khac ´nhau m`a f(M) da`ˆn dˆe´n cac ´gia ´tri. khac ´nhau th`ı khi
M ! M0 h`am f(M) khong ˆco ´gio.´i hn.
99. 52 Chu.o.ng 7. Gi´o.i ha.
n v`a liˆen tu.c cu’ a h`am sˆo´
iii) Sˆo´ b du.o.
.c go.
i l`a gi´o.i ha.
n cu’ a h`am f(M) khi M !1nˆe´u
8 0, 9R 0 : 8M 2 {D : (M,0) R} ) |f(M) − b| .
Dˆo´i v´o .
i h`am nhiˆe`u biˆe´n, c`ung v´o.i gi´o.i ha.
.
n tho ˆong thu.`o.ng d˜a nˆeu ’
trˆen (gi´o.i ha.
n k´ep !), ngu.`o.i ta c`on x´et gi´o.i ha.
n l˘a.
p. Ta s˜e x´et kh´ai
m n`ay cho h`am hai biˆe´n u = f(M) = f(x, y).
Gia’ su. ’ u = f(x, y) x´ac di.
niˆe.
nh trong h`ınh ch˜u. nhˆa.
t
Q = {(x, y) : |x − x0| d1, |y − y0| d2}
c´o thˆe’ tr`u. ra ch´ınh c´ac diˆe’m x = x0, y = y0. Khi cˆo´ di.
nh mˆo.
t gi´a tri.
y th`ı h`am f(x, y) tro. ’ th`anh h`am mˆo.
t biˆe´n. Gia’ su. ’ dˆo´i v´o .
i gi´a tri. cˆo´
nh y bˆa´t k`y tho’a m˜an diˆe`u kiˆe.
di.
i gi´o.i ha.
n 0 |y − y0| d2 tˆo`n ta.
n
lim
x!x0
y cˆo´ di.
nh
f(x, y) = '(y).
Tiˆe´p theo, gia’ su’. lim
y!y0
i. Khi d´o ngu.`o.i ta n´oi r˘a`ng
'(y) = b tˆo`n ta.
i gi´o.i ha.
tˆo`n ta.
p cu’ a h`am f(x, y) ta.
n l˘a.
i diˆe’m M0(x0, y0) v`a viˆe´t
lim
y!y0
lim
x!x0
f(x, y) = b,
trong d´o gi´o.i ha.
n lim
x!x0
y cˆo´ di.
nh
0|y−y0|d2
i l`a gi´o.i ha.
f(x, y) go.
n trong. Tu.o.ng tu.
., ta
nh ngh˜ıa gi´o.i ha.n l˘a.
c´o thˆe’ ph´at biˆe’u di.
p kh´ac lim
x!x0
lim
y!y0
f(x, y) trong
d´o gi ´o.i ha.
n
lim
y!y0
x cˆo´ di.
nh
0|x−x0|d1
f(x, y)
l`a gi´o.i ha.
n trong.
Mˆo´i quan hˆe.
gi˜u.a gi´o.i ha.
n k´ep v`a c´ac gi´o.i ha.
p du.o.
n l˘a.
.c thˆe’
hiˆe.
n
trong di.
nh l´y sau dˆay:
100. 7.4. Gi´o.i ha.
n v`a liˆen tu.c cu’ a h`am nhiˆe`u biˆe´n 53
Gia’ su. ’ ta. i diˆe’m M0(x0, y0) gi´o.i ha.
n k´ep v`a c´ac gi´o.i ha.
n trong cu’a
c´ac gi´o.i ha.
p cu’a h`am tˆo`n ta.
n l˘a.
i. Khi d´o c´ac gi´o.i ha.
p tˆo`n ta.
n l˘a.
i v`a
lim
x!x0
lim
y!y0
f(x, y) = lim
y!y0
lim
x!x0
= lim
x!x0
y!y0
f(x, y).
T`u. di.
c thay dˆo’i th´u. tu.
nh l´y n`ay ta thˆa´y r˘a`ng viˆe.
. trong c´ac gi´o.i
ha.n khˆong pha’ i bao gi`o. c˜ung du.o.
.c ph´ep.
Dˆo´i v´o .
i h`am nhiˆe`u biˆe´n ta c˜ung c´o nh˜u.ng di.
nh l´y vˆe` c´ac t´ınh chˆa´t
sˆo´ ho.c cu’ a gi´o.i ha.
n tu .
o.ng tu.
. c´ac di.
nh l´y vˆe` gi´o.i ha.
n cu’ a h`am mˆo.
t
biˆe´n.
2. T`u. kh´ai niˆe.
m gi´o.i ha.
m vˆe` t´ınh liˆen tu. c
n ta s˜e tr`ınh b`ay kh´ai niˆe.
cu’ a h`am nhiˆe`u biˆe´n.
H`am u = f(M) du.o.
.c go.
i l`a liˆen tu. c ta.i diˆe’m M0 nˆe´u:
nh ta.i ch´ınh diˆe’m M0 c˜ung nhu. trong mˆo.
i) f(M) x´ac di.
t lˆan cˆa.
n
n`ao d´o cu’a diˆe’m M0.
ii) Gi´o.i ha.
n lim
M!M0
f(M) tˆo`n ta.
i.
iii) lim
M!M0
f(M) = f(M0).
. liˆen tu.c v`u.a du.o.
Su.
.c di.
. liˆen tu. c theo tˆa.p ho.
nh ngh˜ıa go. i l`a su.
.p
biˆe´n sˆo´.
i diˆe’m cu’ a
H`am f(M) liˆen tu. c trong miˆe`n D nˆe´u n´o liˆen tu.c ta.i mo.
miˆe`n d´o.
Diˆe’m M0 du.o.
.c go.
.
o i l`a diˆe’m gian ´doa. n cu’ a h`am f(M) nˆe´u doˆ´i v´i
diˆe’m M0 c´o ´ıt nhˆa´t mˆo.
t trong ba diˆe`u kiˆe.
n trong di.
nh ngh˜ıa liˆen tu. c
l`a nh˜u.ng
khˆong tho’a m˜an. Diˆe’m gi´an doa.n cu’ a h`am nhiˆe `u biˆe´n c´o thˆe’
diˆe’m cˆo lˆa.
t du.`o.ng (du.`o.ng gi´an doa. n).
p, v`a c˜ung c´o thˆe’ l`a ca’ mˆo.
i diˆe’m M0(x0, y0) theo tˆa.
Nˆe´u h`am f(x, y) liˆen tu.c ta.
.p biˆe´n sˆo´
p ho.
th`ı n´o liˆen tu. c theo t`u.ng biˆe´n sˆo´. Diˆe`u kh˘a’
ng di.
nh ngu.o.
.c la.
i l`a khˆong
d´ung.
C˜ung nhu. dˆo´i v´o .
t biˆe´n, tˆo’ng, hiˆe.
i h`am mˆo.
u v`a t´ıch c´ac h`am liˆen
i diˆe’m M0 l`a h`am liˆen tu.c ta.
tu. c hai biˆe´n ta.
i diˆe’m d´o; thu.o.ng cu’ a hai
h`am liˆen tu.c ta.
i M0 nˆe´u ta.
i M0 c˜ung l`a h`am liˆen tu.c ta.
i diˆe’m M0 h`am
101. 54 Chu.o.ng 7. Gi´o.i ha.
n v`a liˆen tu.c cu’ a h`am sˆo´
mˆa˜u sˆo´ kh´ac 0. Ngo`ai ra, di.
.p vˆa˜n
nh l´y vˆe ` t´ınh liˆen tu.c cu’ a h`am ho.
d´ung trong tru.`o.ng ho.
.p n`ay.
n x´et. Tu.o.ng tu.
Nhˆa.
. nhu. trˆen ta c´o thˆe’ tr`ınh b`ay c´ac kh´ai niˆe.
m co .
ba’n liˆen quan dˆe´n gi´o.i ha.
n v`a liˆen tu.c cu’ a h`am ba biˆe´n,...
C´AC V´I DU.
V´ı du.
1. Ch´u.ng minh r˘a`ng h`am
f(x, y) = (x + y) sin
1
x
sin
1
y
l`a vˆo c`ung b´e ta.i diˆe’m O(0, 0).
nh ngh˜ıa vˆo c`ung b´e (tu.o.ng tu.
Gia’ i. Theo di.
. nhu. dˆo´i v´o .
i h`am mˆo.
t
biˆe´n) ta cˆa `n ch´u.ng minh r˘a`ng
lim
x!0
y!0
f(x, y) = 0.
Ta ´ap du.
nh ngh˜ıa gi´o.i ha.
ng di.
n theo Cauchy. Ta cho sˆo´ 0 t`uy
´y v`a d˘a.
t =
2
. Khi d´o nˆe´u
M(x, y),O(0, 0)
=
p
x2 + y2 th`ı |x| , |y| .
Do d´o
|f(x, y) − 0| =
108. 7.4. Gi´o.i ha.
n v`a liˆen tu.c cu’ a h`am nhiˆe`u biˆe´n 55
Gia’ i. 1) Ta biˆe’u diˆe˜n h`am du.´o.i dˆa´u gi´o.i ha.
n du .
´o.i da.
ng
h
1 + xy
1
xy
i 2y
x + y .
V`ı t = xy ! 0 khi
x ! 0
y ! 0
!
nˆen
lim
x!0
y!2
1 + xy
1
xy = lim
t!0
1 + t
1
t = e.
Tiˆe´p theo v`ı lim
x!0
y!2
2
x + y
nh l´y thˆong thu.`o.ng vˆe ` gi´o.i ha.
= 2 (theo di.
n
cu’a thu .
o.ng), do d´o gi´o.i ha.
n cˆa` n t`ım b˘a`ng e2.
2) Ta t`ım gi´o.i ha.
n v´o .
i diˆe`u kiˆe.
n M(x, y) ! M0(0, 2). Khoa’ng c´ach
gi˜u.a hai diˆe’m M v`a M0 b˘a`ng
=
p
x2 + (y − 2)2 .
Do d´o
lim
x!0
y!2
f(x, y) = lim
!0
p
2 + 1 − 1
2 = lim
!0
(2 p
+ 1) − 1
2(
2 + 1 + 1)
= lim
!0
1 p
2 + 1 + 1
=
1
2
·
.c ta c´o x = cos ', y = sin '. Ta c´o
3) Chuyˆe’n sang to.a dˆo.
cu.
x4 + y4
x2 + y2 =
4(cos4 ' + sin4 ')
2(cos2 ' + sin2 ')
= 2(cos4 ' + sin4 ').
V`ı cos4 ' + sin4 ' 6 2 nˆen
lim
x!0
y!0
x4 + y4
x2 + y2 = lim
!0
2(cos4 ' + sin4 ') = 0.
109. 56 Chu.o.ng 7. Gi´o.i ha.
n v`a liˆen tu.c cu’ a h`am sˆo´
V´ı du.
3. 1) Ch´u.ng minh r˘a`ng h`am
f1(x, y) =
x − y
x + y
khˆong c´o gi´o.i ha.
i diˆe’m (0, 0).
n ta.
2) H`am
f2(x, y) =
xy
x2 + y2
c´o gi´o.i ha.
i diˆe’m (0, 0) hay khˆong ?
n ta.
nh kh˘a´p no .
Gia’ i. 1) H`am f1(x, y) x´ac di.
i ngoa.i tr`u. du.`o.ng th˘a’
ng
x + y = 0. Ta ch´u.ng minh r˘a`ng h`am khˆong c´o gi´o.i ha.
n ta.
i (0, 0). Ta
lˆa´y hai d˜ay diˆe’m hˆo.
i tu.
dˆe´n diˆe’m (0, 0):
Mn =
1
n
, 0
! (0, 0), n!1,
M0
n =
0,
1
n
! (0, 0), n!1.
Khi d´o thu du.o.
.c
lim
n!1
f1(Mn) = lim
n!1
1
n
− 0
1
n
+ 0
= 1;
lim
n!1
f1(M0
n) = lim
n!1
0 −
1
n
0 +
1
n
= −1.
Nhu. vˆa.
y hai d˜ay diˆe’m kh´ac nhau c`ung hˆo.
i tu.
dˆe´n diˆe’m (0, 0) nhu.ng
hai d˜ay gi´a tri. tu.o.ng ´u.ng cu’ a h`am khˆong c´o c`ung gi´o.i ha.
n. Do d´o
nh ngh˜ıa h`am khˆong c´o gi´o.i ha.
theo di.
n ta.
i (0, 0).
2) Gia’ su. ’ diˆe’m M(x, y) dˆa ` n dˆe´n diˆe’m (0, 0) theo du.`o.ng th˘a’
ng
y = kx qua gˆo´c to.
a dˆo.
. Khi d´o ta c´o
lim
x!0
y!0
(y=kx)
xy
x2 + y2 = lim
x!0
kx2
x2 + k2x2 =
k
1 + k2
·
110. 7.4. Gi´o.i ha.
n v`a liˆen tu.c cu’ a h`am nhiˆe`u biˆe´n 57
Nhu. vˆa.
y khi dˆa `n dˆe´n diˆe’m (0, 0) theo c´ac du.`o.ng th˘a’
ng kh´ac nhau
(tu.o.ng ´u.ng v´o.i c´ac gi´a tri. k kh´ac nhau) ta thu du.o.
.c c´ac gi´a tri. gi´o.i
ha.n kh´ac nhau, t´u.c l`a h`am d˜a cho khˆong c´o gi´o.i ha.
i (0, 0). N
n ta.
V´ı du.
4. Kha’o s´at t´ınh liˆen tu.c cu’ a c´ac h`am
1) f(x, y) =
x2 + 2xy + 5
y2 − 2x + 1
2) f(x, y) =
1
x2 + y2 − z
3) f(x, y) =
x + y
x3 + y3
Gia’ i. 1) Diˆe`u kiˆe.
i nh˜u.ng
n liˆen tu.c cu’a h`am d˜a cho bi. vi pha.m ta.
t ph˘a’
ng R2 m`a to.a dˆo. cu’a ch´ung tho’a m˜an phu.o.ng tr`ınh
diˆe’m cu’a m˘a.
y2−2x+1 = 0. D´o l`a phu .
o.ng tr`ınh du.`o.ng parabon ˆvo. ´i d’ınh ta. i diˆe’m 1
, 0
. Nhva.
ˆy cac ´diˆe’m cu’ a parabon ˆn`ay l`a nhu.˜ng diˆe’m gian ´doa.n
2
.
u - d´o l`a du.`o.ng gi´an doa.n cu’ a h`am. Nh˜u.ng diˆe’m cu’a m˘a.
t ph˘a’
ng R2
c parabˆon d´o l`a nh˜u.ng diˆe’m liˆen tu. c.
khˆong thuˆo.
2) H`am d˜a cho liˆen tu.c ta.
i diˆe’m cu’ a khˆong gian R3 m`a to.a dˆo.
i mo.
n x2 + y2 − z6= 0. D´o l`a phu.o.ng tr`ınh
cu’a ch´ung tho’a m˜an diˆe`u kiˆe.
t paraboloit tr`on xoay. Trong tru.`o.ng ho.
m˘a.
.p n`ay m˘a.
t paraboloit l`a
t gi´an doa.n cu’ a h`am.
3) V`ı tu’. soˆ´ va` maˆ˜u sˆo´ la` nhu˜.ng ha`m liˆen tu. c nˆen thu.o.ng la` ha`m
m˘a.
i nh˜u.ng diˆe’m m`a mˆa˜
u sˆo´ x3+y36= 0. H`am c´o gi´an doa.n ta.
liˆen tu.c ta.
i
a’
nhu.˜ng diˆe’m m`a x3 + y3 = 0 hay y = −x. Ngh˜ıa l`a h`am co ´gian ´doa.n
trˆen du.`o.ng th˘ng y = −x.
Gia’ su’. x06= 0, y06= 0. Khi do´
lim
x!x0
y!y0
x + y
x3 + y3 = lim
x!x0
y!y0
1
x2 − xy + y2 =
1
x20
− x0y0 + y2
0
·
T`u. d´o suy ra r˘a`ng c´ac diˆe’m cu’a du.`o.ng th˘a’
ng y = x (x6= 0) l`a
111. 58 Chu.o.ng 7. Gi´o.i ha.
n v`a liˆen tu.c cu’ a h`am sˆo´
.ng diˆe’m gi´an doa.n khu .
nhu.
’ du.o.
.c. V`ı
lim
x!0
y!0
x + y
x3 + y3 = lim
x!0
y!0
1
x2 − xy + y2 = +1
nˆen diˆe’m O(0, 0) l`a diˆe’m gi´an doa.n vˆo c`ung.
B`AI TˆA.
P
Trong c´ac b`ai to´an sau dˆay (1-10) h˜ay t`ım miˆe `n x´ac di.
nh cu’ a c´ac
h`am nˆe´u:
1. w =
p
x2 − y2. (DS. |y| 6 |x|)
2. w =
p
xy. (DS. x 0, y 0 ho˘a.
c x 6 0, y 6 0)
3. w =
p
a2 − x2 − y2. (DS. x2 + y2 6 a2)
4. w =
1 p
x2 + y2 − a2
. (DS. x2 + y2 a2)
5. w =
r
1 −
x2
a2
−
y2
b2. (DS.
x2
a2 +
y2
b2 6 1)
6. w = ln(z2 − x2 − y2 − 1). (DS. x2 + y2 − z2 −1)
7. w = arcsin
x
2
+
p
xy. (DS. Hai nu. ’ a b˘ang vˆo ha.n th˘a’
ng d´u.ng
{0 6 x 6 2, 0 6 y +1} v`a {−2 6 x 6 0,−1 y 6 0})
8. w =
p
x2 + y2 − 1 + ln(4 − x2 − y2).
(DS. V`anh tr`on 1 6 x2 + y2 4)
9. w =
p
sin (x2 + y2). (DS. Tˆa.
.p c´ac v`anh dˆo`ng tˆam
p ho.
0 6 x2 + y2 6 1; 2 6 x2 + y2 6 3; . . . )
10. w =
p
ln(1 + z − x2 − y2).
t paraboloid z = x2 + y2 − 1).
(DS. Phˆa ` n trong cu’a mˆa.
Trong c´ac b`ai to´an sau dˆay (11-18) h˜ay t´ınh c´ac gi´o.i ha.
n cu’ a h`am
112. 7.4. Gi´o.i ha.
n v`a liˆen tu.c cu’ a h`am nhiˆe`u biˆe´n 59
11. lim
x!0
y!0
sin xy
xy
. (DS. 1)
12. lim
x!0
y!0
sin xy
x
. (DS. 0)
13. lim
x!0
y!0
xy
p
xy + 1 − 1
. (DS. 2)
14. lim
x!0
y!0
x2 + y2
p
x2 + y2 + 1 − 1
. (DS. 2)
Ch’ı daˆ˜n. Su’. du.ng khoa’ng ca´ch =
p
x2 + y2 ho˘a.
c nhˆan - chia
v´o.i da.
i lu .
.ng liˆen ho.
o.
.p v´o .
i mˆa˜
u sˆo´.
15. lim
x!0
y!3
1 + xy2
y
x2y + xy2. (DS. e3)
16. lim
x!0
y!0
x2y
x2 + y2. (DS. 0)
17. lim
x!0
y!5
p
(x2 + (y − 5)2 + 1 − 1
x2 + (y − 5)2 . (DS.
1
2
)
18. lim
x!1
y!0
tg(2xy)
x2y
. (DS. 2).
114. 8.1. D-a.
o h`am 61
8.1 D-a.
o h`am
8.1.1 D-a.
o h`am cˆa´p 1
Gia’ su. ’ h`am y = f(x) x´ac di.
n cu’a diˆe’m x0 (U(x0; ) =
nh trong -lˆan cˆa.
{x 2 R : |x − x0| ) v`a f(x0) = f(x0 +x) − f(x0) l`a sˆo´ gia cu’ a
n´o ta.i diˆe’m x0 tu.o.ng ´u.ng v´o.i sˆo´ gia x = x − x0 cu’a dˆo´i sˆo´.
nh ngh˜ıa: Nˆe´u tˆo`n ta.
Theo di.
i gi´o.i ha.
n h˜u.u ha.
n
lim
x!0
f(x0+x) − f(x0)
x
khi x ! 0 th`ı gi´o.i ha.
n d´o du.o.
.c go.
i l`a da.
.
o h`aam cu’ a h`am f(x) ti
diˆe’m x0 v`a du.o.
.c chı’ bo. ’ i mˆo.
t trong c´ac k´y hiˆe.
u:
lim
x!0
f(x0+x) − f(x0)
x
dy
dx
d
dx
f(x) f0(x) y0.
i lu .
Da.
.ng
o.
f0
+(x0) = f0(x0 + 0) = lim
x!0
x0
y
x
= lim
x!0+0
y
x
v`a
f0
−(x0) = f0(x0 − 0) = lim
x!0
x0
y
x
= lim
x!0−0
y
x
du.o.
.c go.
o h`am bˆen pha’ i v`a da.
i l`a da.
o h`am bˆen tr´ai cu’ a h`am y = f(x)
ta.i diˆe’m x0 nˆe´u c´ac gi´o.i ha.
n d˜a nˆeu tˆo`n ta.
i.
Su. ’ du.ng kh´ai niˆe.
m gi´o.i ha.
n mˆo.
t ph´ıa ta c´o:
D-
i.
nh l´y 8.1.1. H`am y = f(x) c´o da.
o h`am ta.i diˆe’m x khi v`a chı’ khi
c´ac da.
o h`am mˆo.
i v`a b˘a`ng nhau:
t ph´ıa tˆo`n ta.
f0(x+ 0) = f0(x − 0) = f0(x).
H`am f(x) kha’ vi nˆe´u n´o c´o da.
o h`am f0(x) h˜u.u ha.
n. H`am f(x) kha’
vi liˆen tu. c nˆe´u da.
i v`a liˆen tu. c. Nˆe´u h`am f(x) kha’
o h`am f0(x) tˆo`n ta.
vi th`ı n´o liˆen tu. c. Diˆe`u kh˘a’
nh ngu.o.
ng di.
.c la.
i l`a khˆong d´ung.
115. 62 Chu.o.ng 8. Ph´ep t´ınh vi phˆan h`am mˆo.
t biˆe´n
8.1.2 D-a.
o h`am cˆa´p cao
Da.
o h`am f0(x) du.o.
.c go.
i l`a da.
o h`am cˆa´p 1 (hay da.
c nhˆa´t).
o h`am bˆa.
Da.
o h`am cu’ a f0(x) du.o.
.c go.
i l`a da.
o h`am cˆa´p hai (hay da.
o h`am th´u.
hai) cu’ a h`am f(x) v`a du.o.
.c k´y hiˆe.
u l`a y00 hay f00(x). Da.
o h`am cu’ a
f00(x) du.o.
.c go.
i l`a da.
o h`am cˆa´p 3 (hay da.
o h`am th´u. ba) cu’ a h`am f(x)
v`a du.o.
.c k´y hiˆe.
u y000 hay f000(x) (hay y(3), f(3)(x) v.v...
Ta c´o ba’ng da.
o h`am cu’ a c´ac h`am so. cˆa´p co .
ba’n
f(x) f0(x) f(n)(x)
xa axa−1 a(a − 1)(a − 2) · · · (a − n + 1)xa−n,
x 0
ex ex ex
ax axlna ax(lna)n
lnx
1
x
(−1)n−1(n − 1)!
1
xn , x 0
logax
1
xlna
(−1)n−1(n − 1)!
1
xnlna
, x 0
sin x cos x sin
x +
n
2
116. 8.1. D-a.
o h`am 63
f(x) f0(x) f(n)(x)
cos x −sin x cos
x +
n
2
tgx
1
cos2 x
cotgx −
1
sin2 x
arc sin x
1
p
1 − x2
, |x| 1
arccosx −
1
p
1 − x2
, |x| 1
arctgx
1
1 + x2
arccotgx −
1
1 + x2
Viˆe.
c t´ınh da.
o h`am du.o.
.c du.
.a trˆen c´ac quy t˘a´c sau dˆay.
1+ d
dx
[u + v] =
d
dx
u +
d
dx
v.
2+ d
dx
(u) =
du
dx
, 2 R.
3+ d
dx
(uv) = v
du
dx
+ u
dv
dx
.
4+ d
dx
u
v
=
1
v2
v
du
dx
− u
dv
dx
, v6= 0.
5+ d
dx
f[u(x)] =
df
du
·
du
dx
(da.
.p).
o h`am cu’ a h`am ho.
6+ Nˆe´u h`am y = y(x) c´o h`am ngu.o.
.c x = x(y) v`a
dy
dx
y0
x6= 0 th`ı
dx
dy
x0
y =
1
y0x
·
117. 64 Chu.o.ng 8. Ph´ep t´ınh vi phˆan h`am mˆo.
t biˆe´n
7+ Nˆe´u h`am y = y(x) du.o.
.c cho du.´o.i da.
ng ˆa’n bo .
th´u.c kha’ vi
’i hˆe.
F(x, y) = 0 v`a F0
y6= 0 th`ı
dy
dx
= −
F0
x
F0
y
trong d´o F0
x v`a F0
y l`a da.
o h`am theo biˆe´n tu .
o.ng ´u.ng cu’ a h`am F(x, y)
khi xem biˆe´n kia khˆong dˆo’i.
8+ Nˆe´u h`am y = y(x) du.o.
.c cho du .
´o.i da.
ng tham sˆo´ x = x(t),
y = y(t) (x0(t)6= 0) th`ı
dy
dx
=
y0(t)
x0(t)
·
9+ dn
dxn (u +
119. dnv
dxn ;
dn
dxn uv =
Xn
k=0
Ck
n
dn−k
dxn−k u
dk
dxk v (quy t˘a´c Leibniz).
n x´et. 1) Khi t´ınh da.
Nhˆa.
t biˆe’u th´u.c d˜a cho ta c´o thˆe’
o h`am cu’a mˆo.
biˆe´n dˆo’i so .
bˆo.
biˆe’u th´u.c d´o sao cho qu´a tr`ınh t´ınh da.o h`am do.n gia’n
ho.n. Ch˘ng ha.n na.
ˆe´u biˆe’u thu´.c do´ la` logarit th`ı co´ thˆe’ su’. du.ng ca´c
t´ınh chaˆ´t cu’ a logarit dˆe’ biˆe´n do’i... ˆro`ˆi t´ınh do h`am. Trong nhiˆe`u
a’
tru.`o.ng ho.
.p khi t´ınh da.
o h`am ta nˆen lˆa´y logarit h`am d˜a cho rˆo `i ´ap
du. ng cˆong th´u.c da.
o h`am loga
d
dx
lny(x) =
y0(x)
y(x)
·
2) Nˆe´u h`am kha’ vi trˆen mˆo.
t khoa’ng du.o.
.c cho bo .
’i phu .
o.ng tr`ınh
F(x, y) = 0 th`ı da.
o h`am y0(x) c´o thˆe’ t`ım t`u. phu.o.ng tr`ınh
d
dx
F(x, y) = 0.
C´AC V´I DU.
120. 8.1. D-a.
o h`am 65
V´ı du.
o h`am y0 nˆe´u:
1. T´ınh da.
r
1) y = ln 3
ex
1 + cos x
; x6= (2n + 1), n 2 N
2) y =
1 + x2
3 p
x4 sin7 x
, x6= n, n 2 N.
Gia’ i. 1) Tru.´o.c hˆe´t ta do.n gia’n biˆe’u th´u.c cu’ a h`am y b˘a`ng c´ach
.a v`ao c´ac t´ınh chˆa´t cu’ a logarit. Ta c´o
du.
y =
1
3
lnex −
1
3
ln(1 + cos x) =
x
3
−
1
3
ln(1 + cos x).
Do d´o
y0 =
1
3
−
1
3
(cos x)0
1 + cos x
=
1
3
+
1
3
sin x
1 + cosx
=
1 + tg
x
2
3
·
’. dˆay tiˆe.
2) O
.i ho .
n ca’ l`a x´et h`am z = ln|y|. Ta c´o
n lo.
dz
dx
=
dz
dy
·
dy
dx
=
1
y
dy
dx
)
dy
dx
= y
dz
dx
· (*)
Viˆe´t h`am z du.´o.i da.
ng
x = ln|y| = ln(1 + x2) −
4
3
ln|x| − 7ln| sin x|
)
dz
dx
=
2x
1 + x2
−
4
3x
− 7
cos x
sin x
·
Thˆe´ biˆe’u th´u.c v`u.a thu du.o.
.c v`ao () ta c´o
dy
dx
=
1 + x2
3 p
x4 sin7 x
2x
1 + x2
−
4
3x
− 7
cos x
sin x
. N
V´ı du.
o h`am y0 nˆe´u: 1) y = (2+cos x)x, x 2 R; 2) y = x2x,
2. T´ınh da.
x 0.
Gia’ i. 1) Theo di.
nh ngh˜ıa ta c´o
y = exln(2+cosx).
121. 66 Chu.o.ng 8. Ph´ep t´ınh vi phˆan h`am mˆo.
t biˆe´n
T`u. d´o
y0 = exln(2+cosx)
xln(2 + cos x)
0
= exln(2+cosx)
h
ln(2 + cos x) − x
sin x
2 + cos x
i
, x2 R.
2) V`ı y = e2xlnx nˆen v´o.i x 0 ta c´o
y0 = e2xlnx[2xlnx]0 = e2xlnx
h1
x
i
2x + 2xln2 · lnx
= 2xx2x
1
x
+ ln2 · lnx
. N
V´ı du.
3. T´ınh da.
o h`am cˆa´p 2 cu’ a h`am ngu.o.
.c v´o .
i h`am y = x + x5,
x 2 R.
Gia’ i. H`am d˜a cho liˆen tu.c v`a do.n diˆe.
u kh˘a´p no .
i, da.
o h`am y0 =
t tiˆeu ta.i bˆa´t c´u. diˆe’m n`ao. Do d´o
1 + 5x4 khˆong triˆe.
x0
y =
1
y0x
=
1
1 + 5x4
·
Lˆa´y da.
o h`am d˘a’ng th´u.c n`ay theo y ta thu du.o.
.c
x00
yy =
1
1 + 5x4
0
x
· x0
y =
−20x3
(1 + 5x4)3
· N
V´ı du.
4. Gia’ su. ’ h`am y = f(x) du.o.
.c cho du .
´o.i da.ng tham soˆ´ bo’.i ca´c
coˆng thu´.c x = x(t), y = y(t), t 2 (a; b) va` gia’ su’. x(t), y(t) kha’ vi caˆ´p
2 v`a x0(t)6= 0 t 2 (a, b). T`ım y00
xx.
Gia’ i. Ta c´o
dy
dx
=
dy
dt
dx
dt
=
y0
t
x0
t
) y0
x =
y0
t
x0
t
·
Lˆa´y da.
o h`am hai vˆe´ cu’a d˘a’ng th´u.c n`ay ta c´o
y00
xx =
y0
t
x0
t
0
t
· t0
x =
y0
t
x0
t
0
t
·
1
x0
t
=
x0
ty00
tt − y0
tx00
tt
x0
t
3
· N
122. 8.1. D-a.
o h`am 67
V´ı du.
5. Gia’ su’. y = y(x), |x| a la` ha`m gia´ tri. du.o.ng cho du.o´.i
da.ng ˆa’n bo .
’i phu .
o.ng tr`ınh
x2
a2
−
y2
b2 = 1.
T´ınh y00
xx.
Gia’ i. Dˆe’ t`ım y0 ta ´ap du.ng cˆong th´u.c
d
dx
F(x, y) = 0.
Trong tru.`o.ng ho.
.p n`ay ta c´o
d
dx
x2
a2
−
y2
b2
− 1
= 0.
Lˆa´y da.
o h`am ta c´o
2x
a2
−
2y
b2 y0
x = 0, (8.1)
)y0
x =
b2x
a2y
, |x| 0, y 0. (8.2)
Lˆa´y da.
o h`am (8.1) theo x ta thu du.o.
.c
1
a2
−
1
b2
y0
x
2
−
y
b2y00
xx = 0
v`a t`u. (8.2) ta thu du.o.
.c y00
x:
y00
xx =
1
y
h b2
a2
−
y0
x
2
i
=
1
y
hb2
a2
−
b4
a4
x2
y2
i
= −
b4
a2y3
hx2
a2
−
y2
b2
i
= −
b4
a2y3, y0. N
V´ı du.
6. T´ınh y(n) nˆe´u: 1) y =
1
x2 − 4
; 2) y = x2 cos 2x.
Gia’ i. 1) Biˆe’u diˆe˜n h`am d˜a cho du .
´o.i da.
ng tˆo’ng c´ac phˆan th´u.c co .
ba’n
1
x2 − 4
=
1
4
h 1
x − 2
−
1
x + 2
i
123. 68 Chu.o.ng 8. Ph´ep t´ınh vi phˆan h`am mˆo.
t biˆe´n
v`a khi d´o
1
x2 − 4
(n)
=
1
4
h 1
x − 2
(n)
−
1
x + 2
(n)i
.
Do
1
x ± 2
(n)
= (−1)(−2) · · · (−1 − n + 1)(x ± 2)−1−n
= (−1)nn!
1
(x ± 2)n+1
nˆen
1
x2 − 4
(n)
=
(−1)nn!
4
h 1
(x − 2)n+1
−
1
(x+ 2)n+1
i
.
2) Ta ´ap du.ng cˆong th´u.c Leibniz dˆo´i v´o .
i da.
o h`am cu’a t´ıch
(x2 cos 2x) = C0
nx2(cos 2x)(n) + C1
n(x2)0(cos 2x)n−1
+ C2
n(x2)0(cos 2x)n−2.
Ca´c sˆo´ ha.ng co`n la. i d`ˆeu = 0 v`ı
x2(k) = 0 8k 2.
´Ap du. ng cˆong th´u.c
(cos 2x)(n) = 2n cos
2x +
n
2
ta thu du.o.
.c
(x2 cos 2x)(n) = 2n
x2 −
n(n − 1)
4
cos
2x +
n
2
+ 2nnx sin
2x +
n
2
. N
V´ı du.
7. V´o.i gi´a tri. n`ao cu’ a a v`a b th`ı h`am
f(x) =
8
:
ex, x6 0,
x2 + ax + b, x 0
124. 8.1. D-a.
o h`am 69
o h`am trˆen to`an tru.c sˆo´.
Gia’ i. R˜o r`ang l`a h`am f(x) c´o da.
c´o da.
o h`am 8x 0 v`a 8x 0. Ta chı’
cˆa `n x´et diˆe’m x0 = 0.
i diˆe’m x0 = 0 nˆen
V`ı h`am f(x) pha’ i liˆen tu.c ta.
lim
x!0+0
f(x) = lim
x!0−0
f(x) = lim
x!0
f(x)
t´u.c l`a
lim
x!0+0
(x2 + ax + b) = b = e0 = 1 ) b = 1.
Tiˆe´p d´o, f0
+(0) = (x0 + ax + b)0
128. x0=0 = 1.
i nˆe´u a = 1 v`a b = 1. Nhu. vˆa.
Do d´o f0(0) tˆo`n ta.
y v´o .
i a = 1, b = 1
h`am d˜a cho c´o da.
o h`am 8 x 2 R. N
B`AI TˆA.
P
T´ınh da.
o h`am y0 cu’ a h`am y = f(x) nˆe´u:
1. y = 4 p
x3 +
5
x2
−
3
x3 + 2. (DS.
3
4 4 p
x
−
10
x3 +
9
x4 )
2. y = log2x + 3log3x. (DS.
ln24
xln2 · ln3
)
3. y = 5x + 6x +
1
7
x
. (DS. 5xln5 + 6xln6 − 7−xln7)
4. y = ln(x+ 1+
p
x2 + 2x + 3). (DS.
1
p
x2 + 2x + 3
)
5. y = tg5x. (DS.
10
sin 10x
)
p
x). (DS.
6. y = ln(ln
1
p
x
2xln
)
r
7. y = ln
1 + 2x
1 − 2x
. (DS.
2
1 − 4x2 )
129. 70 Chu.o.ng 8. Ph´ep t´ınh vi phˆan h`am mˆo.
t biˆe´n
8. y = xarctg
p
2x − 1 −
p
2x − 1
2
. (DS. arctg
p
2x − 1)
9. y = sin2 x3. (DS. 3x2 sin 2x3)
10. y = sin4 x + cos4 x. (DS. −sin 4x)
11. y =
p
xe
p
x. (DS.
p
x(1 +
e
p
x)
2
p
x
)
12. y = e
1
cos x. (DS. e
1
cos x
sin x
cos2 x
)
13. y = e
1
lnx. (DS.
−e
1
lnx
xln2x
)
e2x +
14. y = ln
p
e4x + 1. (DS.
2e2x
p
e4x + 1
)
r
15. y = ln
e4x
e4x + 1
. (DS.
2
e4x + 1
)
16. y = log5 cos 7x. (DS. −
7tg7x
ln5
)
17. y = log7 cos
p
1 + x. (DS. −
tg
p
1 + x
2
p
1 + xln7
)
18. y = arccos
e
−
x2
2
. (DS.
−
xe
x2
2
p
1 − e−x2 )
19. y = tg sin cos x. (DS.
−sin cos(cos x)
cos2(sin cos x)
)
20. y = ex2cotg3x. (DS.
xec2cotg3x
sin2 3x
(sin 6x − 3x))
p
1+lnx. (DS.
21. y = e
e
p
1+lnx
2x
p
1 + lnx
)
22. y = x
1
x. (DS. x
1
x
−2(1 − lnx))
23. y = ex. (DS. xx(1 + lnx))
130. 8.1. D-a.
o h`am 71
24. y = xsin x. (DS. xsin x cos x · lnx + xsin x−1 sin x)
25. y = (tgx)sin x. (DS. (tgx)sin x
h
cos xlntgx +
1
cos x
i
)
26. y = xsin x. (DS. xsin x
hsin x
x
+ lnx · cos x
i
)
27. y = xx2. (DS. xx2+1(1 + 2lnx))
28. y = xex. (DS. exxex
1
x + lnx))
29. y = logx7. (DS. −
1
xlnxlog7x
)
30. y =
1
2a
136. . (DS.
1
x2 − a2 )
31. y = sin ln|x|. (DS.
cos ln|x|
x
)
32. y = ln| sin x|. (DS. cotgx)
33. y = ln|x +
p
x2 + 1|. (DS.
1
p
x2 + 1
).
Trong c´ac b`ai to´an sau dˆay (34-40) t´ınh da.
o h`am cu’ a h`am y du.o.
.c
cho du.´o.i da.
ng tham sˆo´.
34. x = a cos t, a sin t, t 2 (0, ). y00
xx? (DS. −
1
a sin3 t
)
35. x = t3, y = t2. y00
xx? (DS. −
2
9t4 )
36. x = 1+eat, y = at + e−at. y00
xx? (DS. 2e−3at − e−2at)
37. x = a cos3 t, y = a sin3 t. y00
xx? (DS.
1
3a sin t cos4 t
)
38. x = et cos t, y = et sin t. y00
xx? (DS.
2
et(cos t − sin t)3 )
39. x = t − sin t, y = 1 − cos t. y00
xx? (DS. −
1
4 sin4 t
2
)
40. x = t2 + 2t, y = ln(1 + t). y00
xx? (DS.
−1
4(1 + t)4 ).
