Một số phương pháp tìm giới hạn của một hàm số

MOÄT SOÁ PHÖÔNG PHAÙP TÌM GIÔÙI HAÏN CUÛA MOÄT HAØM SOÁ
I. Toùm taét lyù thuyeát
1. Giôùi haïn höõu haïn 
Cho khoaûng K chöùa ñieåm x0 vaø haøm soá y=f(x) xaùc ñònh treân K hoaëc treân K\{x0}. limx→x0fx=L khi vaø chæ khi vôùi daõy soá (xn ) baát kyø ,xn ∈\{x0} vaø xn→x0 ,ta coù limf(xn)=L .
Cho haøm soá y=f(x) xaùc ñònh treân khoaûng (xo;b) . limx→x0+fx=L khi vaø chæ khi vôùi daõy soá (xn) baát kyø x0<xn<b vaø xn →x0, ta coù limf(x)=L .
Cho haøm soá y=f(x) xaùc ñònh treân khoaûng (a;x0). limx→x0-fx=L , khi vaø chæ khi vôùi daõy soá (xn) baát kyø , a<xn<x0 vaø xn→x0, ta coù limf(xn)=L .
Cho haøm soá y=f(x) xaùc ñònh treân khoaûng (a;+∞) . limx→+∞fx=L , khi vaø chæ khi vôùi daõy (xn) baát kyø ,xn>a vaø xn→+∞, thì limf(xn)=L 
Cho haøm soá y=f(x) xaùc ñònh treân khoaûng (-∞;a) . limx→-∞fx=L, khi vaø chæ khi vôùi daõy soá (xn) baát kyø ,xn<a vaø xn→-∞ thì limf(xn)=L.
2. Giôùi haïn ôû voâ cöïc 
Cho haøm soá y=f(x) xaùc ñònh treân khoaûng(a;+∞) . limx→+∞fx=-∞ , khi vaø chæ khi vôùi daõy soá (xn) baát kyø , xn>a vaø xn→+∞ ,ta coù limf(xn)=-∞ .
Cho K laø khoaûng chöùa ñieåm x0 vaø haøm soá y=f(x) xaùc ñònh treân K hoaëc treân K\{x0}. limx→x0fx=+∞ .khi vaø chæ khi vôùi moïi daõy soá baát kyø (xn) ,xn thuoäc K\{x0} vaø xn→x0, ta coù limf(xn)=+∞ .
Chuù yù : f(x) coù giôùi haïn +∞ ,khi vaø chæ khi -f(x) coù giôùi haïn -∞
3.Caùc giôùi haïn ñaëc bieät 
a) limx→x0x=x0
b) limx→x0c=c ;limx→±∞c=c ; limx→±∞cx=0
c) limx→+∞xk=+∞ , 
Vôùi k laø moät soá nguyeân döông 
d) limx-∞xk=-∞ , khi k lelimx→-∞xk=+∞ , khi k chan
4. Ñònh lyù veà giôùi haïn höõu haïn 
* Ñònh lyù 1
a) Neáu limx→x0fx=L vaø limx→x0gx=M , thì 
limx→x0fx+gx=L+M
limx→x0fx-gx=L-M
limx→x0fx.gx=L.M
limx→x0f(x)g(x)=LM (M≠0 )
b) Neáu f(x)≥ 0 vaø limx→x0fx=L , thì L ≥ 0 vaø limx→x0f(x)=L
Ñònh lyù 2
limx→x0fx=L khi limx→x0+fx=limx→x0-fx=L
5. Quy taéc veà giôùi haïn voâ cöïc 
 a) Quy taéc tìm giôùi haïn cuûa tích f(x).g(x) .
limx→x0f(x)
limx→x0g(x)
limx→x0fxg(x)
L>0
+∞
+∞
-∞
-∞
L <0
+∞
-∞
-∞
+∞
b) Quy taéc tìm giôùi haïn cuûa thöông f(x)g(x)
limx→x0f(x)
limx→x0g(x)
Daáu cuûa g(x)
limx→x0fxg(x)
L
±∞
Tuyø yù
0
L>0
0
+
+∞
-
-∞
L <0
0
+
-∞
-
+∞
B. Phöông phaùp tìm giôùi haïn cuûa haøm soá
I. Thoâng thöôøng ta aùp duïng caùc quy taéc vaø ñònh lyù veà giôùi haïn cuûa haøm soá laø ta tìm ñöôïc ngay giaù trò cuûa giôùi haïn .
Ví duï , Tìm caùc giôùi haïn sau 
a) limx→1x2+2x-32x2-x-1 b) limx→22-xx+7-3
c) lim→+∞2x3+3x-4-x3-x2+1; d) limx→-∞x2-x-4x2+12x+3
e) limx→01x1x+1-1 ; f) limx→-∞4x2-x+2x
Baøi giaûi :
a) limx→1x2+2x-32x2-x-1=limx→1x-1x+32x-1x+12=limx→1x+32x+1=43
b) limx→22-xx+7-3=limx→22-xx+7+3x-2=limx→2-x+7+3=-6
c) limx→+∞2x3+3x-4-x3-x2+1=limx→+∞2+3x2-4x3-1-1x+1x3=-2
d) limx→-∞x2-x-4x2+12x+3=limx→-∞x1-1x-4+1x22x+3=limx→-∞-1-1x-4+1x22+3x=12
e) limx→0-1x1x+1-1=limx→0-1-x+1xx+1=limx→0--1x+1=-1
f) limx→-∞4x2-x+2x=limx→-∞-xx4-1x-2x=limx→-∞14-1x+2=14
II. Moät soá daïngvoâ ñònh thöôøng gaëp vaø caùch bieán ñoåi .
Ñeå tính limx→x0u(x)v(x) khi limx→x0ux=limx→x0vx=0 . Ta laøm nhö sau:
Phaân tích töû vaø maãu thaønh nhaân töû . Sau ñoù giaûn öôùc nhaân töû chung :
limx→x0u(x)v(x)=limx→x0x-x0A(x)x-x0B(x)=limx→x0A(x)B(x)=A(x0)B(x0)=L (**)
Neáu u(x) vaø v(x) chöùa bieán soá döôùi daáu caên ,thì coù theå nhaân töû vaø maãu vôùi bieåu thöùc lieân hôïp ,tröôùc khi phaân tích chuùng thaønh tích ñeå giaûn öôùc .
Moät soá bieåu thöùc lieän hôïp thöôøng duøng :
1) a-1=a-1a+1 2) a+1=a-1a-1 3) a-b=a-ba+b
4) a+b=a-ba-b 5) 3a+1=a+13a2-3a+1 
6) 3a-1=a-13a2+3a+1 7) 3a-3b=a-b3a2+3ab+3b2
8 )3a+3b=a+b3a2-3ab+3b2 
9) na-nb=a-bnan-1+nan-2b+nan-3b2++na1bn-2+nbn-1
10 ) na+nb=a+bnan-1-nan-2b+nan-3b2-+na1bn-2+nbn-1
* Chuù yù : Trong (**) neáu A(x0)=B(x0)=0 ,ta laïi phaân tích tieáp chuùng thaønh :
AxB(x)=x-x1.C(x)x-x1D(x)=C(x)D(x)⇒limx→x0A(x)B(x)=limx→x0C(x)D(x)=C(x1)D(x1)=L1
* Khi u(x) hoaëc v(x) chöùa caên thöùc cuøng baäc : 
	Ta söû duïng phöông phaùp nhaân lieân hôïp ( nhö ñaõ cho ôû treân ) 
	Sau ñoù ruùt goïn laøm xuaát hieän thöøa soá chung .
	Giaûn öôùc thöøa soá chung ,seõ maát daïng voâ ñònh 
Ví duï1 . ( Baøi 4.57-tr-143-BTGT11-NC).
Tìm caùc giôùi haïn sau 
a) limx→-2x3+8x2+11x+18 b) limx→32x3-5x2-2x-34x3-13x2+4x-3
c) limx→0x+33-27x d) limx→03x2+x42x
e) limx→-2+xx+2x2+3x+2; f) limx→111-x-31-x3
Baøi giaûi :
a) limx→-2x3+8x2+11x+18=limx→-2x+2x2-2x+4x+2x+9=limx→-2x2-2x+4x+9=127
b) limx→32x3-5x2-2x-34x3-13x2+4x-3=limx→3x+32x2+x+1x-34x2-x+1=limx→32x2+x+14x2-x+1=1117
c) limx→0x+33-27x=limx→0xx+32+3x+3+9x=limx→0x+32+3x+3+9=27
d) limx→03x2+x42x=limx→0x3+x22x=limx→03+x22=32
e) limx→-2+xx+2x2+3x+2=limx→-2+xx+2x+1x+2=limx→-2+-xx+1=-2
Vì x→-2+, thì x+2<0 ,cho neân x+2=-x+2
f) limx→111-x-31-x3=limx→1x-12+x1-x1+x+x2=limx→12+x1+x+x2=1
Ví duï 2 ( Baøi 4.59-tr144-BTGT11-NC)
Tìm caùc giôùi haïn sau 
a) limx→1x+3-2x-1; b) limx→72-x-3x2-49
c) limx→3x2-2x+6-x2+2x-6x2-4x+3; d) limx→3-x-33-6x-x2
e) limx→2x+2-2x+7-3; f) limx→+∞3x2+x+1-x3
Baøi giaûi :
a) limx→1x+3-2x-1=limx→1x-1x-1x+3+2=limx→11x+3+2=14
b) limx→72-x-3x2-49=limx→77-xx-7x+72+x-3=limx→7-1x+72+x-3=-156
c) limx→3x2-2x+6-x2+2x-6x2-4x+3=limx→3-4x-3x-1x-3x2-2x+6+x2+2x-6
=limx→3-4(x-1)x2-2x+6+x2+2x-6=-13
d) limx→3-x-33-6x-x2=limx→3-x-33+6x-x2x-32=limx→3-3+6x-x2x-3=-∞
e) limx→2x+2-2x+7-3=limx→2x+7+3x+2+2.x-2x-2=limx→2x+7+3x+2+2=32
f)limx→+∞3x2+x+1-x3= limx→+∞x+13x2+x+1+x3
=limx→+∞1+1x3+1x+1x2+3=123=36
Ñeå tìm giôùi haïn :(Daïng :∞∞ )
limx→±∞u(x)v(x) khi limx→±∞ux=±∞ ;limx→±∞ux=±∞ ; 
Ta coù theå laøm nhö sau :
Chia töû vaø maãu cho xn, vôùi n laø soá muõ cao nhaát cuûa bieán soá x ( hay phaân tích töû vaø maãu thaønh tích chöùa nhaân töû xn ,roài giaûn öôùc ).
Neáu u(x) vaø v(x) coù chöùa bieán x trong daáu caên thöùc ,thì ñöa xk ra ngoaøi daáu caên ( vôùi k laø soá muõ cao nhaát cuûa x trong daáu caên ), tröôùc khi chia töû vaø maãu cho luyõ thöøa cuûa x . 
- Chuù yù ñeán caän : Khi x→+∞, nghóa laø x>0 ; coøn x→-∞ , nghóa laø x<0 
- Gioáng nhö ñoái vôùi daïng 00 , hoaëc ta phaân tích thaønh nhaân töû ,hoaëc ta nhaân lieân hôïp ,hoaëc ta ñöa x ra ngoaøi daáu caên thöùc ( phaûi chuù yù ñeán caän maø boû daáu trò tuyeät ñoái )
Ví duï 1. (Baøi 32-tr159-GT11-NC)
Tìm caùc giôùi haïn sau 
a) limx→+∞32x5+x3-12x2-1x3+x; b) limx→-∞2x+3x2+x+5 
c) limx→-∞x2+x+2x2x+3 ; d) limx→+∞x+1x2x4+x2+1
Baøi giaûi :
a) limx→+∞32x5+x3-12x2-1x3+x=limx→+∞32x5+x3-12x5+x3-x=limx→+∞32+1x2-1x52+1x2-1x4=1
b) limx→-∞2x+3x2+x+5=limx→-∞-2x+3-x1+1x3+5x4=limx→-∞-2+3x-1+1x+5x2=2
c)limx→-∞x2+x+2x2x+3= limx→-∞-x1+1x+2xx2+3x=limx→-∞2-1+1x2+3x=12
d)limx→+∞x+1x2x4+x2+1=limx→+∞xx+122x4+x2+1=limx→+∞1x+2x2+1x32+1x2+1x4=0
Ví duï 2. (Baøi 44-tr167-GT11NC)
Tìm caùc giôùi haïn sau 
a) limx→-∞x2x3+xx5-x2+3 ; b) limx→-∞x+x2+xx+10
c) limx→+∞2x4+x2-11-2x; d) limx→-∞2x2+1+x
Baøi giaûi :
a) limx→-∞x2x3+xx5-x2+3 =limx→-∞-x22x3+xx5-x2+3=limx→-∞-2+1x21-1x3+3x5=-2
b) limx→-∞x+x2+xx+10=limx→-∞-1+1+1x1+10x=-2
c)limx→+∞2x4+x2-11-2x=limx→+∞x.2+1x2-1x41x-2=-∞ 
d) limx→-∞2x2+1+x=limx→-∞x1-2+1x2=+∞
Ví duï 3. Tìm caùc giôùi haïn sau : 
ø giaûi:
. 
Ví duï 4. Tìm caùc giôùi haïn sau 
Baøi giaûi :
Baøi taäp töï luyeän
Tìm caùc giôùi haïn sau: 
 a) 	 b) 	 c)
	d) 
 e) f) g)	
	 h)
 i) j) 	 k) 	
	l) 	
 p) q)	 
Ñeå tính giôùi haïn :( Daïng ∞-∞ ) .
limx→x0ux-vx khi limx→x0ux=+∞ ;limx→x0vx=+∞ ; 
Hoaëc
limx→x0ux.vx khi limx→x0ux=0 ;limx→x0vx=±∞ ; 
Ta nhaân vaø chia vôùi bieåu thöùc lieân hôïp ( neáu coù bieåu thöùc chöùa bieán soá döôùi daáu caên thöùc ) hoaëc quy ñoàng ñeå ñöa veà cuøng moät phaân thöùc ( neáu chöùa nhieàu phaân thöùc )
Daïng voâ ñònh vaø daïng 0.∞
Ví duï 1. Tìm giôùi haïn cuûa caùc haùm soá sau 
Baøi giaûi 
Ví duï 2. Tìm giôùi haïn cuûa caùc haøm soá sau
a) limx→31x-131x-33 b) limx→111-x-31-x3
c) limx→-4-2x2+3x-4-3x+4; d) limx→-1x2+x+2-1-xx4+x
Baøi giaûi :
a)limx→31x-131x-33=limx→33-x3xx-33=limx→3-13.1x-32=+∞ 
b) limx→111-x-31-x3=limx→1x2+x-21-x1+x+x2=limx→1x-1(x+2)1-x1+x+x2
=lim-x→1(x+2)1+x+x2=-1
c)limx→-4- 2x2+3x-4-3x+4=limx→-4-2x-1(x+4)-3x+4
=limx→-4-2-3(x-1)x-1(x+4)=limx→-4-5-3x(x-1).1(x+4)=-175-∞=+∞
d) limx→-1x2+x+2-1-xx4+x=limx→-1x+12xx+1(x2-x+1)=limx→-1x+1x(x2-x+1)
=0.-13=0
Ví duï 3. ( Baøi 40-tr166-GT11-NC)1
.Tìm caùc giôùi haïn sau 
a) limx→-1+x3+1xx2-1; b) limx→+∞x+2x-1x3+x
c) limx→+∞x2+1-x; d) limx→-∞2x2+1+x
Baøi giaûi :
a)limx→-1+x3+1xx2-1=limx→-1+xx+12x2-x+12x-1x+1 
=limx→-1+xx+1x2-x+12x-1=0.1-2=0
b) limx→+∞x+2x-1x3+x=limx→+∞x+22x-1x3+x=limx→+∞x3+3x2-4x3+x
=limx→+∞1+3x-4x31+1x2=1
c) limx→+∞x2+1-x=limx→+∞1x2+1+x=0
d) limx→-∞2x2+1+x=limx→-∞x2+12x2+1+x=limx→-∞-x1+1x22+1x2-1=+∞
Baøi taäp töï luyeän
Tính caùc giôùi haïn sau: 
 	 e) 	 	 g) 
 h) 	 k) 
 l)	m) 
 	n)	 	 t)	
 o)	 	p) 	
q) 
 r)	 s) 	 	 
 v) w) 
Ñeå tìm giôùi haïn
limx→x0u(x)v(x)
Khi u(x) hoaëc v(x) chöùa caùc caên thöùc khoâng cuøng chæ soá .
Kheùo leùo theâm vaø bôùt vaøo töû soá hay maãu soá ( coù chöùa caên khoâng cuøng chæ soá ) moät soá hôïp lyù ( thöôøng laø theâm vaøo soá x0) 
Taùch giôùi haïn ñaõ cho thaønh hai giôùi haïn maø sao cho moãi giôùi haïn chæ chöùa caên thöùc coù cuøng chæ soá vaø aùp duïng caùc ñònh lyù ,hoaëc quy taéc tìm giôùi haïn ñaõ bieát .
Chaúng haïn ,ta tìm : limx→x0f(x)g(x)=m(x)-nv(x)g(x) ; mux0=nvx0=cgx0=0⟹limx→x0f(x)g(x)=m(x)-nv(x)g(x)=limx→x0mu(x)-cg(x)-limx→x0nv(x)-cg(x)
Chuù yù : Ñoâi khi ta phaûi theâm ,bôùt moät ñaïi löôïng h(x) sao cho h(x0)=c. Sau ñoù aùp duïng caùch phaân tích treân ñeå giaûi . ( Thoâng qua ví duï : limx→01+2x-31+3xx2 )
Ví duï minh hoaï
Ví duï 1. Tìm giôùi haïn cuûa caùc haøm soá sau
a) limx→021-x-38-xx; b) limx→13x-2-34x2-x-2x2-3x+2
Baøi giaûi :
a) limx→021-x-38-xx=limx→021-x-1x+limx→02-38-xx
=limx→02-11-x+1+limx→0-14+238-x+38-x2=-1312
b)limx→13x-2-34x2-x-2x2-3x+2= limx→13x-2-1x2-3x+2+limx→11-34x2-x-2x2-3x+2
=limx→13x-23x-2+1+limx→1-4x+3(x-2)1+34x2-x-2+34x2-x-22
=-32+73=56
Ví duï 2. Tìm caùc giôùi haïn sau.
a) limx→15-x3-3x2+7x2-1; b) limx→133x-2-4x2-x-2x2-3x+2
Baøi giaûi :
a) limx→15-x3-3x2+7x2-1=limx→15-x3-2x2-1x→0+limx→12-3x2+7x2-1
=limx→1-x3-1x2-115-x3+2+limx→1-14+23x2+7+3x2+72
=limx→1-x2+x+1x+15-x3+2-limx→114+23x2+7+3x2+72=-1124
b) limx→133x-2-4x2-x-2x2-3x+2=limx→133x-2-1x2-3x+2+limx→11-4x2-x-2x2-3x+2
=limx→13x-2[1+33x-2+33x-22]+limx→14x+32-x1+4x2-x-2
=-1+72=52
Ví duï 3. Tìm caùc giôùi haïn sau 
a) limx→13x+7-5-x2x-1 b) limx→03x+4-38+5xx
c) limx→031+2x-1+7xx
Baøi giaûi :
a)limx→13x+7-5-x2x-1= limx→13x+7-2x-1+limx→12-5-x2x-1
=limx→114+23x+7+3x+72+limx+12+5-x2x→1=112+12=712
b) limx→03x+4-38+5xx=limx→03x+4-2x+limx→02-38+5xx
=limx→033x+4+2+limx→0-54+238+5x+38+5x2=34-512=13
c) limx→031+2x-1+7xx=limx→031+2x-1x+limx→01-1+7xx
=limx→021+31+2x+31+2x2+limx→0-71+1+7x=23-72=-176
Ví duï 4. Tìm caùc giôùi haïn sau 
a) limx→1x+3-33x2+5x-1; b) limx→137+x3-3+x2x-1
Baøi giaûi :
a) limx→1x+3-33x2+5x-1=limx→1x+3-2x-1+limx→12-33x2+5x-1
=limx→11x+3+2+limx→1-1+x4+233x2+5+33x2+52=14-212=112
b) limx→137+x3-3+x2x-1=limx→137+x3-2x-1+limx→12-3+x2x-1
=limx→1x2+x+14+237+x3+37+x32+limx→1-11+x2+3+x2=312-24=-14
Ví duï 5. Tìm giôùi haïn sau :
limx→01+2x-31+3xx2
Giaûi :
Ta theâm ,bôùt moät haøm soá h(x)=1+x ,vôùi h(0)=1. Khi ñoù 
limx→01+2x-31+3xx2=limx→01+2x-1+xx2-limx→031+3x-1+xx2
=limx→0-11+2x+1+x-limx→0x+31+x2+1+x31+3x+31+3x2=-12+1
5 Ñeå tìm giôùi haïn : limx→x0u(x)v(x)
Khi u(x) hoaëc v(x) chöùa caùc caên thöùc khoâng cuøng chæ soá .( vôùi caên coù chæ soá cao hôn 3- töø 4 trôû ñi ).
Ta ñoåi bieán soá baèng caùch ñaët u=nf(x)⟹fx=un;khi x→x0 ;u→u0
Chuyeån giôùi haïn ñaõ cho töø bieán x trôû thaønh bieán u vôùi giôùi haïn môùi coù theå aùp duïng caùc ñònh lyù vaø quy taéc tìm giôùi haïn laø coù theå tìm ñöôïc ngay .
Ví duï1: minh hoaï ( ÑH-SP II-99).
Tìm giôùi haïn sau :
limx→142x-1+5x-2x-1
Baøi giaûi :
Ta coù : 
limx→142x-1+5x-2x-1=limx→142x-1-1x-1+limx→11+5x-2x-1
Ñaët :
u=42x-1⇒u4=2x-1⟺x-1=u4-12 ;Khi x→1;u→1 
⟹limx→142x-1-1x-1⟺limu→12u-1u4-1=limu→12u+1u2+1=12
Ñaët :
v=5x-2⟹v5=x-2 ;x-1=v5+1 ;Khi x→1 ;v→-1
⟹limx→11+5x-2x-1⟺limv→-11+vv5+1=limv→-11v4-v3+v2-v+1=15
Vaäy : 
limx→142x-1+5x-2x-1=12+15=710
Ví duï 2: Tìm caùc giôùi haïn sau 
A= limx→7x+2-3x+204x+9-54x+4; B=limx→138x6+19-4x4+5x-1
Baøi giaûi :
A=limx→7x-7+32-3x-7-3x-7+33-3x-74x-7+24-2x-7-54x-7+25-2x-7=16-127132-120=-56081
B=limx→138x6-1+27-3x6-1.x6-1x-1-4x4-1+9-3x4-1.x4-1x-1x-1+1-1x-1
Khi : limx→138x6-1+33-3x6-1=827; limx→1x6-1x-1=6 ; limx→14x4-1+32-3x4-1=23 ; limx→1x4-1x-1=4 ; limx→1x-1+1-1x-1=12; ⟹B=-169
6 Phaàn naâng cao . AÙp duïng giôùi haïn :
limx→0sinxx=1⟺limax→0sinaxax=1
Neáu giôùi haïn ñaõ cho chöùa caùc haøm soá löôïng giaùc , baèng caùch bieán ñoåi löôïng giaùc ,ta bieán ñoåi haøm soá caàn tìm giôùi haïn sao cho söû duïng ñöôïc giôùi haïn treân.
Neáu haøm soá tìm giôùi haïn chöùa hoãn hôïp caû caèn thöùc +löôïng giaùc ,hay ña thöùc vôùi löôïng giaùc thì ta phaûi theâm hay bôùt hoaëc taùch giôùi haïn ñoù thaønh hai giôùi haïn sao cho hai giôùi haïn naøy coù theå tìm ñöôïc ngay baèng caùc ñònh lyù vaø quy taéc tìm giôùi haïn ñaõ bieát .
Ví duï minh hoaï :
Ví duï 1. Tìm caùc giôùi haïn sau 
a) limx→01-2x+1+sinx3x+4-2-x; b) limx→01-2x2+11-cosx
Baøi giaûi :
a) limx→01-2x+13x+4-2-x+limx→0sinx3x+4-2-x
=limx→023x+4+2+x1+2x+1(x+1)-limx→0sinxx.3x+4+2+xx+1=-8
b) limx→01-2x2+11-cosx=limx→0-41+2x2+1sinx2x22=-2
Ví duï 2. Tìm giôùi haïn cuûa caùc haøm soá sau.
a) limx→5sinx-53-x2-16; b)limx→0cos4x-sin4x-1x2+1-1
Baøi giaûi :
a) Dat t=x-5 ⟹Khi x→5;t→0 ;x=t+5 ; x2=t2+10t+25
Vaäy : 
limx→5sinx-53-x2-16⟺limt→0sint3-t2+10t+9=limt→0-sintt.3+t2+10t+9t+10=-35
b)limx→0cos4x-sin4x-1x2+1-1=limx→0-2sin2xx2.x2+1+1=-4
III.Phaàn baøi taäp töï luyeän
Baøi 1. Tìm caùc giôùi haïn sau 
a) limx→2x-x+24x+1-3 98; b) limx→01-31-x3x 19
c) limx→13x-2+31-x-x2x2-1 13 d) limx→-13x+1x2+3-2 -23
Baøi 2. Tìm caùc giôùi haïn sau 
a) limx→+∞x-2x-31-2x -∞; b) limx→-1x+16x2+3+3x 1
c) limx→∞x33x2-4-x23x+2 29; d) limx→74x+9-2x-7 132
Baøi 3. Tìm caùc giôùi haïn sau 
a) limx→+∞x-x2+5x 52; b) limx→-∞x2-x-x2+1 12
c) limx→+∞3x3+1-x 0; d) limx→+∞x23x3+1-x 13
Baøi 4. Tìm caùc giôùi haïn sau
a) limx→12x2-1-1x-1 -12; b) limx→111-x-31-x3 -1
c) limx→0x+9+x+16-7x 7136; d) limx→13x2-23x+1x-12 19
Baøi 5. Tìm giôùi haïn cuûa caùc haøm soá sau 
a) limx→131+x-1-xx 56; b) limx→0x+1+x+4-3x 14
c) limx→22x+5-7+xx2-2x 12; d) limx→-234x+2x+2 13
III. Söû duïng ñònh nghóa ñaïo haøm ñeå tìm giôùi haïn cuûa haøm soá 
Theo ñònh nghóa ñaïo haøm : "Cho haøm soá y= f(x) coù D=(a;b)x0 laø moät giaù trò thuoäc D . Giôùi haïn cuûa tyû soá
limx→x0fx-f(x0)x-x0=f'x0 .
Goïi laø giaù trò ñaïo haøm cuûa haøm soá taïi ñieåm x0.
Neáu haøm soá f=f(x) toàn taïi ñaïo haøm taïi ñieåm x0 : f'(x0)≠ 0 , thì :
limx→x0f(x)g(x)=limx→x0f'(x)g'(x)
Moät soá coâng thöùc tính ñaïo haøm caàn bieát :
nf(x)'=f'(x)n..nf(x)n-1; f(x)g(x)'=f'x.gx-fx.g'(x)g2(x) ; fn(x)'=nfn-1xf'(x)
Ví duï aùp duïng
Ví duï 1. (ÑH-Thuyû lôïi -KA-2001).Tìm giôùi haïn sau 
I=limx→01+2x-31+3xx2
Baøi giaûi :
limx→01+2x-31+3xx2=limx→01+2x-1x2-limx→031+3x-1x2
=limx→02x1+2x+1-limx→03x31+3x2+31+3x+1
=limx→021+2x+1-1x-0-limx→0331+3x2+31+3x+1-1x-0
=limx→0fx-f(0)x-0-limx→0gx-g(0)x-0=f'0-g'0(*)
Vôùi :
fx21+2x+1⟹f0=1 ;gx=331+3x2+31+3x+1 ⇒g0=1
f'x=-21+2x1+2x+12;f'0=-12 ;g'x=-3231+3x2+31+3x31+3x231+3x2+31+3x+1⟹g'0=-1⟹I=12
Ví duï 2. Tìm caùc giôùi haïn sau 
a)A= limx→031-x-1x; b)B= limx→132x-1-3x-2x2-1
c) C= limx→0n1+3x-1x; d) D= limx→031+x2-41-2xx+x2
Baøi giaûi 
a)limx→0 31-x-1x=limx→0fx-f(0)x-0;fx=31-x⟹f'x=-1331-x2
⟺f'0=-13=limx→0fx-f(0)x-0=limx→0 31-x-1x
b) fx=32x-1-3x-2⟹f'x=2332x-12-323x-2;f1=0 
⟹B=limx→11x+1fx-f(1)x-1=12f'1=1223-32=-512
c) fx=n1+3x⟹f'x=3n(1+3x)n1+3x⟺f'0=3n;f0=1
⟹C=limx→0 fx-f(0)x-0=f'0=3n
d) fx=31+x2-41-2x⟹f'x=2x331+x22--2441-2x3
⟹f'0=12⟺D=limx→01x+1limx→0fx-f(0)x-0=f'0=1.12=12.
Ví duï 3. Tìm caùc giôùi haïn sau 
a) A=limx→132x-1-11-2-x2; b) B= limx→02x+1-3x2+1sinx
c) C=limx→1326x3+1-480x4+1x-1
Baøi giaûi 
a) fx=32x-1⟹f'x=2332x-12⟺f'1=23;f1=1
gx=2-x2⟹g'x=-x2-x2⟺g'1=1;g1=1
⟹A=lim-x→1fx-1gx-1=limx→1-fx-f(1)x-1gx-g(1)x-1=f'(1)-g'(1)=-23
b) fx=2x+1-3x2+1⇒f'x=12x+1-2x33x2+12 
⟺f'0=1;gx=sinx ⟹g'x=cosx ⟺g'0=1
⇒B=limx→0fx-0gx-0=limx→0fx-f(0)x-0gx-g(0)x-0=f'(0)g'(0)=11=1
c) fx=326x3+1-480x4+1⇒f'x=26x2326x3+12-80x3480x4+13
⟹f'1=-227 ;gx=x-1⟹g'x=12x⟺g'1=12 ;g1=0
⇒C=limx→1fx-f(1)gx-g(1)=fx-f(1)x-1gx-g(1)x-1=f'(1)g'(1)=-227.21=-427
* Chuù yù : Coù theå söû duïng moät soá keát quaû sau ñeå tìm giôùi haïn 
Keát quaû 1. Tìm giôùi haïn sau 
limx0n1+ax-1x ;Tổng quát : limx→0a2x+1.3a3x+1.4a4x+1.nanx+1-1x
Tìm giới hạn B=limx→0a2x+1.3a3x+1.4a4x+1x
a) Tìm A=limx→0n1+ax-1x. Đặt t=n1+ax ⟹tn-1=ax⟺x=t-1a. 
Khi x→0 thì t→1;fx=n1+ax-1x⟹A=limx→1atn-1+tn-2++t+1=an
Töø phaân tích : abc-1= (abc-ab)+(ab-a)+(a-1)=ab(c-1)+a(b-1)+(a-1). (1) Cho neân :
abcx=ab.c-1x+ab-1x+a-1x.Nếu coi a2x+1=a;b=3a3x+1;c=
c=4a4x+1, thì :B=a22+a33+a44
Tổng quát :C=a22+a33+a44+.+ann
Ví dụ . Tìm giới hạn sau 
A=limx→0ax+13bx+14cx+1-1x
Baøi giaûi :
Do (1)
Ta có fx=ax+13bx+14cx+1-1x+ax+13bx+1-1x+ax+1-1x
Nên A=limx→0fx=c4+b3+a2
Kết quả 2 .Tìm giới hạn sau
B=limx→0naux+bn-bu(x).với limx→0ux=0;Khi đó B= anbn-1
Chứng minh tương tự kết quả 1.
Ví dụ 1:
Tìm giới hạn sau B= limx→7x+2-3x+204x+9-54x+4
Bài giải :
Ta có B=limx→7x-7+32x-7-3x-7+33x-74x-7+24x-7-54x-7+25x-4= 16-127132-120=-56081
Ví dụ 2 :
Tìm giới hạn sau :C=limx→138x6+19-4x4+5x-1
Baøi giaûi :
Ta có ;C= limx→138x6-1+33-3x6-1.x6-1x-1-4x4-1+32-3x4-1.x4-1x-1x-1+1x-1
Với limx→138x6-1+33-3x6-1=827; limx→1x6-1x-1=6; limx→14x4-1+32-3x4-1=23
limx→1x4-1x-1=4 và limx→1x-1+1x-1=12.Cho nên C=-169
Moät soá baøi taäp töï luyeän
Baøi 1. Tìm giôùi haïn cuûa caùc haøm soá sau 
a) A=limx→2x3-2x2-4x+8x4-8x2+16; b) B= limx→01+x5-1+5xx2+x5
Baøi 2. Tìm giôùi haïn cuûa caùc haøm soá sau 
a) A= limx→axn-an-nan-1x-ax-a2; B= limx→01+mxn-1+nxmx2
Baøi 3. Tìm giôùi haïn cuûa caùc haøm soá sau
a) limx→0n1+ax-1x; b) limx→ax-a-a-bx2-a2
c) limx→0n1+ax-m1+bxx
Baøi 4. Tìm giôùi haïn cuûa caùc haøm soá sau
a) limx→01+4x31+6x41+8x51+10x-1x; b) limx→02x-1-3x2+1sinx
c) limx→0327x3+1-481x4+1x-1
BAØI TAÄP THAM KHAÛO - ÑEÅ LUYEÄN TAÄP
.Baøi 1. Duøng ñònh nghóa, CMR:
 a) 	b) 	c) 
Baøi 2. Tìm caùc giôùi haïn sau
 a) 	 b) 	c) 	 
 d) 
 e) f) 	 g) 
	 j) h) 	 k)	 
Daïng voâ ñònh 
Tìm caùc giôùi haïn sau:
 a)	 b) 	
 c) 	 d)
 e)	 f) 
	 g) h) 
 i) 	 j) 
	 k) 
 l) 	m) 
	 n) 	o)
	 p) 	q) 
	 r) 	s) 
 t) 	u) 
	 k) 
 2. Tìm caùc giôùi haïn sau:
 A = 	 B = 
 D = 	 C = 
 E = 	 G =	 
 H = 	 L = 
 I = 	 J = 	 
 N = 	 
 O = 	F = 
 	 P = 	 Q = 	 
 R = 	M = 
3. Tìm caùc giôùi haïn sau:
a) b) 	 c) 	 
 e) f) g) 
 d) 	 h) 	0) 
 i) j) 	 k) 
 o) 	 p) x)
 n) 	 
 q) 	 r) 	 s) 
 t) v) 	 w) 
4. Tính caùc giôùi haïn sau:
 a. 	b. 	 c. 
 d.	e. 	 f.
Daïng voâ ñònh 
1. Tìm caùc giôùi haïn sau: 
 a) 	 b) 	 c)
	d) 
 e) f) g)	
	 h)
 i) j) 	
	 l) k) 	
 m) n)	 
 o) 
 p) q)	 
	 r)
 s) t)
Giôùi haïn moät beân
1. Tìm caùc giôùi haïn sau
 a) 	 b) c) 	 
 d) e)
 f) g) h) 
 i) 	 j) 
 k) l) g) 
 h) i) 
2. Tìm giôùi haïn beân phaûi, giôùi haïn beân traùi cuûa hs f(x) taïi xo vaø xeùt xem haøm soá coù giôùi haïn taïi xo khoâng ?
3. Tìm A ñeå haøm soá sau coù giôùi haïn taïi xo:
 a)	 vôùi x0 = 1 	 
 b) vôùi x0 = 3
Giôùi haïn haøm löôïng giaùc
1. Tính caùc giôùi haïn sau:
 a) 	 b) 	 c)
	d) e)	 f)
 g) 	 h) 
D¹ng 1: x ® a
Bµi 1: Thay vµo lu«n. 
1) 	2) 	3) 	 
 4) 	 5) 6) 	 
Bµi 2: Ph©n tÝch thµnh nh©n tö.
1) 	 2) 	
3) 	4) 
5) 	 6) 	 	 
 7) 
 8) 	 9)
10) 	11) 	
12) 	13) 
14) 	15) 	
16)	17) 
Bµi 3: Nh©n l­îng liªn hîp (cã mét c¨n bËc hai) 
 1) 	2)	 3) 
 4) 	 5) 
 6) 	 7) 
 8) 	 9) 
10) 	 11) 	 
 12) (n ÎN, n ³ 2) 	 13) 
14) 	15) 	
16) 	17)
Bµi 4: Nh©n l­îng liªn hîp (cã hai c¨n bËc hai)
1) 	 2) 	
3) 	 	 4) (a > 0)
5) 	 6) 	
7) 	 	 10) 
8) 	 9) 	
Bµi 5: Nh©n l­îng liªn hîp (cã mét c¨n bËc ba)
a) 	b) 	
c) 	d) 
Bµi 6: Nh©n l­îng liªn hîp (c¶ tö vµ mÉu)
1) 	2) 	3) 	
4) 	5) 	9) 
6) 	7) 	8) 	
Bµi 7: Nh©n l­îng liªn hîp (cã c¶ c¨n bËc hai vµ c¨n bËc ba)
1) (§HQG – KA 97) 2) 	
 3) 
 	4) 	 5) 	
6) 	 7) 
D¹ng 2: Giíi h¹n mét bªn
1) 	 2) 
 3) 	 	 4) . 
5) .	 T×m ; 	 
 6).