Thiết kế bài giảng đại số và giải tích 11 nâng cao (tập 2) phần 2

B. GlCfl HAN CUA HAM SO. HAM SO LIEN TUC
§4. D i n h n g h i a v a m o t s o d i n h l i v e
g i d i h a n c u a h a m so'
( t i e t 7, 8, 9 )
I. MUC TIEU
1. Kie'n thurc
HS ndm dugc :
• Dinh nghTa gidi ban ciia ham so
Djnh If ve gidi han hiiu ban.
2. KT nang
Sau khi hgc xong bai nay HS cdn giai thanh thao cac dang toan ve gidi
ban ciia ham so.
Van dung td't cac quy tac tim gidi ban cua ham sdi
3. Thai do
•
• Tu giac, tfch cue trong hgc tap.
Bie't phan biet rd cac khai niem ca ban va van dung trong tirng trudng
hgp cu the.
- Tu duy cac van dd ciia toan hgc mdt each Idgic va he thd'ng.
II. CHUAN BI CUA GV VA HS
1. Chuan bi ciia GV
• Chuan bi cac cdu hdi ggi md.
• Chudn bj phdn mau va mdt sd do dung khac.
2. Chuan bj cua HS
• Cdn dn lai mdt sd kie'n thiic da hgc ve gidi ban day sd.
III. PHAN PHOI T H 6 I L U O N G
Bai nay chia lam 3 tiet :
Tii't 1 : Tic ddu din hit phdn 1.

138

Tii't 2 : Tii'p theo di'n hi't dinh li 1.
Tii't 3 : Phdn con lgi vd bdi tap.
IV. TIEN TRINH DAY

HOC

A. OAT VAN DE
Cau hdi 1
Tim gidi han ciia cac day so sau ddy
n-3"
1-3"
a) Iim
b) Iim
2" +'3"
n + 3"
Cau hoi 2
Tfnh cac tdng sau
a) Sn = l + - + - + .
"
2 4

b ) S n = l - - + - + .... .
"
3 9

B. BAI Mdl
HOATDONCl
1. Gidi han cua ham so tai mgt didm
• GV neu bai toan:

GV treo bang
X

jr, = 10

fix) Axi)

=6

H =9

A:^

M)

M)

•«4=3

...

..^=1,9

/(JC4)

...

M)

^2

> 7

Sau dd GV dua ra cac cau hdi sau
Hoat dgng ciia GV
Cau hoi 1

Hoat ddng ciia HS
Ggi y tra ldi cau hdi 1

Xac dinh f(Xn)
/(^«) = ^ T ^

= 2(^n + 2) vdi

mgi n.

13<


Cau hoi 2

Ggi y tra ldi cau hdi 2

Tirn limf(Xn).

Iim/(Xn) = lim2(j:n + 2)
= 2(IimjCn + 2) = 2(2 + 2) = 8.

GV neu djnh nghTa :
Gid sic (a ; b) la mpt khodng chiia diem XQ vdf Id mpt hdm so xdc dinh

tren tap hgp (a ; b)\{xQ}. Ta ndi rdng hdm so fed gidi hgn Id sdtlncc L
khi X ddn din XQ (hodc tgi diem XQ) ni'u vdi mgi ddy sd(x^) trong tap ligp
(a ; b) \{XQ} (ticc la Xj^ e(a ; b) vd x^ ^XQ vdi mgi n) md limx^ = XQ, ta
diu cd limf(x^) = L.
Khi dd ta vie't
lim fix)

= L hoac/(AT) -> L khi JC -> XQ.

Thuc hien vf du 1 trong 3 phiit.
Hoat ddng cua HS

Hoat ddng cua GV

Ggi y tra Idi cau hdi 1

Cau hdi 1
Vdi mgi day so

(JC„)

ma

/(x„) = x„cos—,•

JC,, ^ 0 vdi mgi n hay xac
dinh f(Xn)
Cau hdi 2
Vdi lim Xji = 0 hay xac dinh

Ggi y tra Idi cau hoi 2
1

|/(^H)| = k l c o s —
X..

^w

va lim|x„| = 0

Iimf(Xn)'
ndn

Iim/(j:„) = 0. Do dd

Iim fix) = lim jccos— = 0.
;c-*0

GV dua ra mdt so cau hdi ciing cd :
140

x-^0\

XJ


HI. Neu mdt VI du khac ve vf du ham sd.
H2. Ham sd khdng xac djnh tai a nhung cd gidi ban tai a. Diing hay sai?
• GV dua ra nhan xet:
lim X = XQ ; lim c = c, vdi c Id hang so.

H3. Tim gidi ban ham sd sau bdng djnh nghTa:
2x + l
f(x) = —
khi X ddn de'n 1.
X +X + 1
• Thuc hien [HIJ trong 5'
Hoat ddng cua HS

Hoat ddng ciia GV

Ggi y tra Idi cau hdi 1

Cau hdi 1
Hay rut ggn f(x).' -'^

X^+3A: + 2

/W=-

ix + l)ix + 2)

;
x+l
x+l
Ggi y tra ldi cau hdi 2

Cau hdi 2
Hay xac djnh Iimf(Xn).

;

=

= x+2

lim/(;cn) = lim (Xn + 2) = - 1 + 2 = 1. i
,. J;^ + 3X + 2
Vay hm
:
= 1.
Jf^-l

x +l

GV ndu nhdn xet:
a) Ni'u fix) = c vdi mgi x e R, trong dd c Id mdt hdng sd, thi vdi mgi
XQ

e R, lim fix) - lim c = c.

b) Ni'u g(x) = X vdi mgi x e R thi vdi mgi XQ e R,
lim gix) = Iim A: = XQ.
h) Gi&i hgn vo cixc

IA:


H4. Gidi ban vo cue ciia ham sd tai mdt diem dugc djnh nghTa tuong tu nhu gidi
ban hiiu ban cua ham sd tai mgt di^in. hay phat bieu djnh nghTa dd.
• GV ndu va hudng ddn HS thuc hien vf du 2.

HOATDONC 2
2. Gidi han cua ham sd tai vd cue
• GV neu djnh nghTa 2:
• Gid sic ham sdfxdc dinh tren khodng (a ; +ao). Ta ndi rdng hdm sdf
cd gidi hgn Id sdthuc L khi x ddn din +oo ne'u vdi mgi ddy sd(x^)
trong khodng (a ; +cc) (ticc ldXn> a vdi mgi n) md limx^y = +oo, ta diu
cd
limfix^) = L.
Khi dd ta vii't
lim fix) = L liodc fix) -> L khi x -^ +oo,
• Cdc giai hgn
lim fix) = + 00, lim fix) = - co, lim fix) = L,
lim fix) = + 00 vd lim fix) = - oo dicgc dinh ngliia tuang tie.
• GV neu vd hudng ddn HS thuc hien vf du 3.
• GV neu nhan xet:
Ap dung dinh ngliia gidi hgn ciia hdm sd,cd the chieng minh dicgc rdng :
Vdi mgi sd nguyin dicaiig k, ta cd
;.

.. '
a)

hm

Jf

= + 00 ,•

.v^+oo

,

.

b)

hm

.v->-<»

c) lim -V = 0 ;

A-»-90jf*

HOATDONC 3
3. Mdt so djnh li ve gidi han huru han

142

f+°o ne'u A: Chan
=<

[-00 ne'u ^ l e

d) lim - ^ = 0.

,V—>+QO_f

• GV neu dinh If 1:

t
X


Gid sic Iim fix) = L vd lim ^(Jf) = M (L, M e R). Khi dd
a) lim [fix) + gix)] =L+ M:,
b) lim [fix)-gix)]

=L-M;

.v->jr„

c) lim [fix)gix)]

^LM;

Ddc biet, ni'u c Id mdt hdng sdthi
,'

.

lim

[C/(JC)]

= cL ;

x^.x^^,

d) Ne'u M^O thi Iim 4 4 = ^ •

A-*.r„ gix)

M

H5. Hay phat bieu bang ldi djnh If trdn.
• GV neu nhan xet:
Ni'u k la mdt sd nguyen duang vd a Id mpt hang sdthi vai mgi XQ e R,
ta cd
k

k

k

lim ax = Um a. lim x. Um x... lim x =ai lim x) = OA^ •
Ar->..v,)

•f->-*'i)

'f^-^'o

-<;—>A'()
k thita .so

•>•—>-<() •

•'^:"^-*'o

• GV neu va hudng ddn HS thuc hien vf du 4.
• Thuc hien |H2| trong 5'

Hoat ddng cua GV

Hoat ddng ciia HS
Ggi y tra Idi cau hdi 1

Cau hdi 1

Hay tim mien xac dinh eua Ham so da cho xac djnh vdi mgi x 7^ 0
ham so.
va x ^ _2.
Cau hdi 2
Tim lim
JC->-l

Ggi y tra Idi cau hdi 2
fix).

4
,. 2A:2-jr + l
=
hm
jr-^-i x^+2x
-I

4.

I A3


• Thuc hien vi du 5 trong 5 phiit.

Hoat ddng ciia GV

Hoat ddng ciia HS
Ggi y tra ldi cau hoi 1

Cau hdi 1
Chia ca tir so va mdu so cho
3

2x^ -x + 10
jc^ + 3JC - 3

X ta dugc sd nao?

2
1
X x^^
, , 3
X

2

10
x^
3 "
X

3

Ggi y tra ldi cau hoi 2

Ap dung dinh ll 1. HS tu tinh.

Cau hdi 2
^, ^ ,.
2x^ -x + m
' Tmh ;r^+oo
hm —
X + 3JC - 3 .

HOATDONC 4
• Thuc hien |H3| trong 5'
Hoat ddng cua G V
Cau hoi 1

Hoat ddng cua HS
Ggi y tra Idi cau hdi 1

Chia ca tir so va mSu so cho HS tu tim.
3
X ta dugc sd nao?
Cau hoi 2
2x^-x^+x
Tinh hm —j
.
jr-^-«>x^+2jc^-7
• GV ndu dinh If 2
Gid sic Iim fix) - L. Khi dd

a) Um \fix)\ = \L\ ;
b) Iim 4fOi) = 41 ;

144

Ggi y tra ldi cau hoi 2
,.
lim

2x^-x^+x
4

2

x-^-^ x^ +2x^ -1

^
=2.


c) Ne'u fix) > 0 vdi mgi x e A (XQ) . trong dd J Id mpt khodng ndo dd
chica XQ, thiL>Ovd

lim V / U ) = 4L.
x->x^•|

GV neu va hudng ddn HS thuc hien vf du 6.
• Thuc hien |H4| trong 5'
H o a t ddng cua H S

H o a t ddng cua GV

Ggi y t r a Idi cau hdi 1

Cau hdi 1

lim(A'^+7x) = - 8

Tim lim (x + Ix)

Ggi y t r a ldi cau hdi 2
Cau hdi 2
lim x^+lx
Tfnh lim x" +lx va
lim 4x^ + Ix .
x->-\

=8

x-^-\

va lim ylx^+7x = 4^

= -2

x^-l

HOATDONC 3

TOM TfiT Bfil HOC
1. Gia sii (a ; b) la mdt khoang chiia diem XQ v a / l a mdt ham sd xac djnh tren tap
hgp (a ; ^)\(xo}. Ta ndi rang ham s d / c d gidi ban la so thUe L khi x dan den XQ

(hoac tai diem XQ) neu vdi mgi day sd (Xn) trong tap hgp ia;b)\

{XQ} (tiic la Xn e(a ; 6)

ya Xn ^ XQ vdi mgi n) ma limXn = Xg, ta deu cd lirri/(Xn) = L.
Khi dd ta viet
lim fix)

= L hoac/(x) -^ L khi x -> XQ.

2. a) Ne'u/(x) = c vdi mgi x G M, trong do c la mdt hdng sd, thi vdi mgi XQ G R ,

10-TKBGDSVGTIINCT2

145


lim fix) = lim r = c.
•V >.v„

.V ->.v„

b) Neu g{.\) F A N'di moi .v 6 M ihi vdi mgi xg e R,
lim ^v(-v) = lim x = XQ.
,V >.V„

.\--->.V||

3. • Gia su' ham so/'xac djnh tren khoang (c/ ; '+00). Ta ndi vdng ham ,sd/c6 gidi
han la sd ihuc /. khi x dan den +co neu vdi mgi day sd (x^) Irong khoang (c/ ; +00)

(tiic la Xn > a vdi moi 11) ma lim.Vn = +co. la deu cd
!im/(Xn) = L.
Khi dd la vicl
iim /(x) =Lhoac / ( x ) -^ L khi x —> +QO.
.\

• >

* ' '

•'

•

• Cac gioi ban
lim /'(x) = + a;, lim /'(x) = - c o ,
•

> <• J'

.\

->tX

lim / ( x ) = / , .

lim /(x) = + cc va

(

V ->--^

> - cc

lim /'(.r) = - cc dugc dinh nghla luong lu.
4. Gia sir lim f(.\) = /. va lim ^!,'(-v),= M (/.. M e R), Khi dd
a) lim [fi.\-) + gix)} = L + Al;
.V ->.v„

b) lim [/(x)-^(,'(.v)l
A- ->A„

=L-M:

•

c) lim [/(.vX!>(x)] =LiV/;
.V >.V|,

Dac bicl, neu r la mdl hdng sd ihi

lim [ r / ( x ) ] = cL ;
X >.1„

d) Neu M ^ 0 thi

146

lim ^ ^ = — •
V

. > r„ gi.x)
M'


5. .Neu k la mol sd nguyen duong \'a a la mol hdng so thi vdi mgi-xg J R, la cd
lim ax^ = lim c/. lim .v. lim A;... lim x =«( lim x)^' = C/.VQ .
•i->.f||

v ->.r|,

.v-->.V||

,->-->.vji

' v->-Vn

•>'->-»(i

/: thfra s o

6. Gia sir Um /(x) = L.Khidd
.v->.v„

a) lim |/(x)| = |/.| ;

b) lim 4M. = 4L ;
c) Nc'u/(x) > 0 vdi mgi X e A {xol, trong dd / la mol khoang nao do chUa XQ.
Ihi L > 0 va lim J/Cv) = 41.

.

HOATDONC 6

MQT SO Cfia HOI TR^C NQHIEM ON TfiP Bfil 4

X

x+ 1 ,bang:
> 1 x --2
(b>- 2

(a)l;


:

Trd ldi. (b).
Cdu 2. lim
X^->1

2x + l

bang:

x2-2

(a) 2

(b)2;

(c) I ;

(d) 1.

Trd ldi. (c).

147


,., ,
Cau 3.

r
X + V2
Iim__-^;
bang:
- / 2 X^

x-»

(a)l

(b)2;
1

(c)

(d) 42.

2V2 '

Trd ldi. (c).
X-1
Cdu 4. lim-^;
x^ix^-1

bang:

(a) 3

(b)2;

(c)

Trd loi. (d).
Cdu 5. 3 la gidi ban cua day so nao sau day:
2x + l
(b) lim
x->l3x,,-2

, . ,. .3x + l
(a) h m —
x^lx-2
. . ..
3x + l
(c) h m
x->l 3 x - 2

3x + l

.
(d) h m
x->l x - 3

'Trd ldi. (b).
Cdu 6, _3 la gidi ban cua day sd nao sau day:
, , ,. - 3 x + l
—
(a) h m
x-»i x - 2
, , ,.
3x
(c) h m
;

(b) h m
x-^1 ^ x - 2
3x + l
(d) h m - — — .
x->l x - 2

X-^-i x - 2

Trd ldi. (c).
Cdu 7. lim

bang:

x->r x-1
(a) + 00

148

(b) - cx).



(c)0;

Trd ldi. (a).
„ , „ ,.
3x +1 , ^
Cau 8. h m
bang:
x-1
x-^ r
( b ) _ CO.

( a ) + 00

(0 i ;
Trd Idi. (b).
Cdu 9.

lim -:
bang:
r 2x-l

x-^-

(a) + 00

( b ) _ 00.

(0 i ;
Trd ldi. ia.).
Cdu 10. lim
bang:
1 2x-l
(a) + 00

(b)-».

(d) I
Trd ldi. (b).


HOATDONC 7

nao^NG D^r4 Bfii Tfi? sficn GI^O KHOfi
Bai 21. Hifdng ddn. Sir dung djnh nghTa gidi ban ciia ham sd.
a)

Hgat ddng ciia HS

Hoat ddng ciia GV
Cau hdi 1

Ggi y tra ldi cau hdi I

Hay lim tap xac dinh ciia ham Vdix5^-i,
sd. Riil ggn cong thiic ciia
,, , x2-.3x-.4,. (x + l)(x- 4 )
ham so
x+1
x+l

.V.

= x-4.
Cau hdi 2
Tim gidi ban ciia ham sd.

Ggi y tra idi cau hdi 2
Vdi limXn = - 1 , lacd
limy(Xn) = lim(Xn-4) = - l - 4 = - 5 .
Do dd

x^ - 3 x - 4 -

lim-^-.v-T>-l

'•— = Tt5.

x + 1

b)
Hoat ddng ciia GV
Cau hdi 1

Hay lim tap xac djnh ciia ham
sd. Riil ggn cdng thdc cua
bain sd
Cau hdi 2

Hgat ddng ciia HS
Ggi y tra Idi cau hdi 1
Ham sd fix) = - ,
xac djiih iren
.
V5 - V
khoang (-00 ; 5).
Ggi y tra Idi cau hdi 2

T m gidi han ciia ham sd.
Vdi lim Xn = I, tacd

im.


lim/(Xn) = lim
1
Do dd lim ,
.f->iV5-x

1
2

Bai 22. Hifdng ddn. Sir dung djnh nghla gidi ban ciia ham sd.
Hoat ddng cua H S

H o a t ddng ciia GV

Ggi y t r a ldi cau hdi 1

Cau hdi 1
T m gidi han ciia cac day so

K),c.v;;),(/(x;,)) va (/(x;;))..

lim xJ, = 0,

lim x\] = 0, lim /(x'„) =

lim cos2/7;r= 1,
n = 0.
lim y( x;;) = Um cos (2/7 + I) —

Cau hdi 2

Ggi y t r a ldi cau hdi 2

Ton tai hay khdng lim cos— ?
.v->0

X

lim/(x;,) ^ Um/(x;;). Do dd khdng
.
. ,.

1
ton tai lim cos—.
.V-+0

X

Bai 23.11 uang ddn. Sir dung cac tfnh chat cua gidi ban ham sd.
Dap sd:
a) 3 7 ;

b) 0 ;

c)
H o a t ddng ciia G V
C a u hdi 1
Hay tim tap xac djnh cua ham
sd. Riit ggn cohg thdc cua
ham sd
C a u hdi 2

Hoat ddrig ciia H S
Ggi y t r a Idi cau het i l
Vdi mgi Jc^i 0, la cdx

I

x)

= x-l.

Ggi y t r a ldi cau henl

15


T m gidi ban cua ham so.
lim X 1--

lim(x-l) = - l .

d)

Hoat ddng cua GV
Cau hdi 1

Hoat ddng cua HS
Ggi y tra Idi cau hdi 1

Hay tim tap xac djnh cua ham Vdi mgi x > 0 va x T^ 9, ta ed
so. 'Rut ggn cdng thiic ciia
VX - 3
V-^ - 3
ham sd.
9x-x^
x(9-x)
4x-3
" "
-I
xi4x+3)
xi3-4x)i3 + 4x)

Cau hdl 2
Tim gidi ban cua ham so.

Ggi y tra ldi cau hdi 2
,. 4^-3
hm
A-^9 9x - x^

e ) - - ' ^

,.
= - Um

I
1.
1=—— =

•v-»9 x(V Jr + 3)

•-••^ ^

Hoat ddng ciia GV
Cau hdi 1

Hoat ddng ciia HS
Ggi y tra Idi cau hdi 1

Hay tim tap xac djnh cua ham . Ham sd xac djnh vdi mgi x.
sd. Riit ggn cdng thUc cua
ham sd

Cau hdi 2
Ggi y tr^ ldi cau hdi 2
Tirn gidi ban cpa haiti so.
Um x'-A = 1.
x^S

0
Hoat ddng ciia GV
Caii hdi 1
Hay tim tap xac djnh ciia bam
152

Hoat ddng ciia HS
Ggi y tra ldi cau hdi 1

54


so. Rut ggn ; cong thdc cua Ham sd xac dinh vdi
bam so
x^ + 3 x - l _ . ,. x ^ + 3 x - l „
~
=3
> 0 va lim
2x - 1
^->2 2 x ^ - 1
Cau hdi 2

Ggi y tra Idi cau hdi 2

T m gidi ban bua ham so.
lim
'V^2V

x^+3x-l

= 4^.

2x^-1

Bai 24. Hicdng ddn. Sir dung cac tfnh chat cua gidi ban ham so.
Dap so :
a) 0 ;

-r-

b) 2 ;

d) Vdi mgi X < 0, ta cd

Vx^+2

I'fB

3x^-1

3x^-1

, lim
,. 4x^+2

Do dd

.V^-OO 3 ; ^ ^ _ 1

-'H -H
1

3x^-1

1+
lim

X-^-CO

I

J.
3

Bal 25. Hicdng ddn. Sii dung cac tfnh chdt cua gidi ban ham so.
HS tu giai.

153


§5. Gi6i h a n mot ben
(tiet 10)
1. M U C T I E U
1. Kien thuc

,HS ndm dugc :
• Djnh nghla gidi ban ben Irai va gidi ban ben phai ciia ham sd
Mdi quan lie giQa dao ham mdl ben va dao ham ciia ham so.

•

?

2. KT nang
--Sau khi hgc xong bai nay IIS can giai thanh thao cac dang loan vS gidi
han mol ben ciia ham so.
Van dung dinh If de chung minh ham sd cd gidi han hoac khdng cd gidi
ban lai mgt diem.
3. Thai do
- Tu giac, tich'cuc trong hgc lap.

•

• Bie't phan biei rd cac kliai niem co ban va van dung trong lung trudng,
hgp cuthe.
"• Tu duy cac van de ciia toan hgc mdt each Idgic va he thd'ng.
II. C H U A N BI CUA G V VA HS
1. Chuan bi ciia GV
Chudn bj cac cau hdi ggi md.
Chudn bj phan mau va mdl sd dd diing khac.
2. Chuan bi ciia HS
Can on lai mgt sd kie'n ihiic da hgc ve gidi ban day sd, ham sd.
III. P H A N P H O I THCII L U O N G
Bai nay chia Iam 1 tie'l

154


IV. TIEN TRINH DAY

HOC

A. OAT VAN OE
Cau hdi 1
Tim gidi han ciia cac.ham so sau day
1-x
1-x
b) lim
a) lim
X >l x ^ - 1
X >l x ' ' - 3 x + 2
Cau hdi 2
Tim gidi ban cua cac ham so sau day

1-Vx

l-Vx
b) lim
x->ix^-3x + 2

a) lim
x->l x ^ - 1

B.

BAI Mdl
HOATDONC 1

1. Dinh nghla
•

CJV

dai van de :

Tim lap xac djnh ciia ham sd sau :
2x-3
f(x) =

khi

x>l

x^-3x + 2
khi X < 1
x-l

HI. Tfnh f(I).
H2.Tim h m ( 2 x _ 3).

113. Tim lim

x^-3x + 2
x-1

H4. Lieu cd tirn dugc gidi ban ciia ham sd khi x —>l hay khdng?
• GV neii dinh nghTa 1 :
Gid sif hdm sdfxdc dinh tren khodng (.XQ ; h) (XQ e R). Ta ndi rang
hdm sd fed gi&i hgn ben phdi Id sdiliu'c L khi x ddn de'n XQ (hoac lgi
155


diem XQ) ne'u vdi mgi day sd(xj trong khodng (XQ ; h) md lirhx^ = XQ, ta
diu cd limfix^) = L.
Khi dd ta viet
lim fix)

= L hodc fix) ->Lkhix->

XQ.

X -^ x^.,

H5. Hay xac djnh gidi ban ben pbai tai x = 1 cua ham sd tren.
• GV neu djnh nghTa 2:
Gid sic hdm sdfxdc dinh tren khodng (a ; XQ) (XQ e R)'. Ta ndi rdng
hdm sdfcd gioi hgn bin trdi Id sdthicc L khi x ddn di'n XQ (hodc tgi
diem XQ) ne'u vdi mgi ddy sd'(Xji) trong khodng (a ; XQ) md limx^^ = XQ, :.
ta diu cd limfix^) = L.
Khi dd ta vie't
Iim fix)-L

hodc fix)—>L khi X—> XQ.

H6. Hay xac djnh gidi ban ben trai tai x = 1 ciia ham sd tren.

H7. Ne'u hani sd cd gidi han tai XQ thi cd gidi ban ben phai va gidi ban ben trai tai
dd hay khong?
• GV neu nhan xet:
1) Hien nhien ne'u lim fix) - L thi hdm sdfcd giai han ben phdi vd
^^^0

' ' ', '

gidi hgn bin trdi tgi diim XQ vd Iim fix) = Iim fix) = L.
X —^ Xn

X —> Xn

2) Ta thica nhdn diiu ngicgc lgi ciing dimg, nghia Id
Niu

lim / ( x ) = lim fix) = L thi hdm sdf cd gidi hgri tgi diim XQ

vd lim fix) = L .
3) Cdc dinh li 1 vd dinh li 2 trong §4 vdn dimg khi thay x —> XQ bai
X —> XQ"" hogc.x

156

—> XQ^


GV hudng ddn HS thuc hien vf du 1.
• Thuc hien \H^\ trong 4'
Hoat ddng ciia HS

Hoat ddng cua GV

Ggi y tra ldi cau hdi 1

Cau hdi 1

•3

T m gidi ban bdn trai cua ham
sd.

Um

fix)'=

x^(-ir

lim

x

Jr->(-l)-

=
( - ly) ^tra
= -Idi
l . cau hdi 2
Ggi

Cau hdi 2
T m gidi ban ben trai dua ham
so.

lim

/(x)=

x^{-l)*

lim (2x - 3 )
x^(-l)*

= 2.(-l)^-3 = - l .
Cau hdi 3

Ggi y tra ldi cau hdi 3
Hai gidi ban nay bang nhau. do dd

Ket Iuan.

Iim/(x) = - l .
x^-l

HOATDONC 2
2. Gidi han vd cue
• GV eho HS neu djnh nghTa gidi ban tai vo cue.
• Thuc hien vfdu 2 trong 5',
a)
Hoat ddng ciia HS

Hoat ddng ciia-GV
1

Cau hdi 1
T m gidi ban ben trai cua ham
sd.
Cau hdi 2
T m gidi ban ben trai cua ham
sd.

Ggi y tra ldi cau hdi 1
lim — = - CO.
x-^O' X

Ggi y tra ldi cau hdi 2
1
hm
x-^O'

— = + 00
X

15'


Cau hdi 3

CJgi y tra ldi eau hdi 3

Kcl luan.

Vi

lim ~ 9t lim -V->0

X

nCn, khong

>()• X

ton lai lim —
.v->0 X
b)
Hoat ddng cua HS

Hoat ddng cua GV

Ggi y tra ldi ciiu hdi 1

Cau hdi 1
Tim gidi han ben trai ciia ham
sd.

I
=
lim
,v->0 I -^ I

+00.

Ggi y tra ldi cau hdi 2
Cau hdi 2
T m . gidi ban ben trai ciia bam
sd.

1
U m -— =
.v->0^ I X

+00

Ggi y tra ldi cau hdi 3
C;;au hdi 3

Vi

lim pT= lim 7~-T = +co nen i

Ke'l luan.
l i m j—r = +00
,.v ->0 X

• Thuc hien H2 Irone 4
Hoat ddng ciia GV
Cau hdi 1

Ggi y tra Idi cau hdi 1

ChUng minh
I-V ->2 yJ2-

Hoat ddng cua HS

1

X

Dal fix) — —
V2-X

(x„ ) Irong khoang (-oo, 2) ma lim x„
= 2, ta cd

158

Vdi moi day sd


lim/(x„) = lim
C a u hdi 2
Tim

1
lim -^ -X >2 -J2 .V

+0C .

^2-x„

(Jgi y trsi ldi cau hdi 2
lim / ( x ) = lim
x->2

X'^2

,

1

= +co.

v 2 -..V

HOATDONC 3
TOM TfiT Bfil HOC
1. Gia sir ham so/xac dinh tren khoang (xg ; b) (XQ G R ) . Ta ndi iing ham s d / c d
gidi ban ben phai la sd ihuc L khi x dan de'n XQ (hoac lai diem XQ) neu \'di moi day
.so (.x'n) trong khoang (XQ ; h) ma limXn = XQ, la dSu cd lini/(.rn) = L.
Khi dd la vicl
lim fi-x) = Z. hoacy(x) -> L khi x -^ XQ .
2. Gia sir ham so /'xac djnh Ircn khoang ia ; XQ) (XQ e R). 'fa ndi rang ham s d / c d
gidi ban ben trai la sd ihuc L khi x dan de'n XQ (hoac lai diem .XQ) neu vdi moi day
sd (Xn) trong khoang (c7; XQ) ma limXn = XQ, la deu cd lirri/(Xn) = LKhi dd ta viel
lim / ( x ) = L hoac/(x) -^ L khi x -^ .XQ.
.V ->.v„

HOATDONC 4
MOT SO Cfia HOI TRfiC NQHIEM ON TfiP Bfil 5

Cc////. Cho ham so f(x) = 2x 1. Khi do, lim f(x) bdng:
x >l

159


(a)l;

(b)-2;

(0-1 ^

(d)|

Chgn cau tra ldi diing.
Trd Idi. (a).
Cdu 2. Cho ham sd f(x) = 2x _ 1. Khi dd lim f(x) bang:

x-»r
(b)_2;

(a)l;
(c)-Chgn cau tra ldi diing.
Trd ldi: (a).
Cdu 3. Cho ham so f(x) = 2x _ 1.

(a) Ham so da cho cd gidi ban trai va gidi han phai tai 1 bang nhau ;
(b) Ham sd da cho cd gidi ban trai va gidi han phai tai 1 bdng 1 ;
(c) Ham sd da chd ed gidi ban tai 1 ;
(d) Ca ba khang djnh tren deu sai.

Chon cau tra ldi sai
Trd Idi. (d).

Cdu 4. Cho ham sd f(x) =

x-1
x+1

khi

x>0

khi

x<0

Khi dd Iim f(x) bdng:

(a)-l;

(b) 2 ;

(c)l;

(d) Mgt ke't qua khac.

Chgn cau tra ldi diing.
Trd ldi. (e).

160

Cdu 5. Cho ham sd f(x) = x-\
x+l

khi

x>0

„ , . ,,

khi x < 0

(a) 1 ;
(c) I ;

,.

f/ X u-

. Km do hm f(x) bang:
^"^°"

(b) 2 ;
(d) Mdt ke't qua khac.

Chgn cau tra ldi diing.
Trd Idi. (a).
khi x > 0
Cdu 6. Cho ham sd f(x) =

x + 1 khi x < 0
(a) Ham sd da cho cd gidi han trai va gidi ban phai tai 0 bang nhau ;
(b) Ham sd da cho cd gidi ban trai va gidi ban phai tai 0 khac nhau ;
(c) Ham sd da cho ed gidi ban tai 0 ;

; /„ t

(d) Ca ba khang djnh trdn d|u sai.
Chgn cau tra ldi diing.
Trd Idi. (b).
1
khi X > 0
.Khidd lim f(x) bang:
Cdu 7. Cho ham so f(x) = 2 x - l
x - 1 khi x''~^^~
(a)-l;

(b) 2 ;

(c)l;

(d) Mdt ke't qua khac.

Chgn cau tra Idi diing.
Trd ldi. (a).
khi x>0 ^,, . ,, ,. . .
Cdu 8. Cho ham sd f(x) = 2 x - l
.Khi do hm r(x) bang:
x - 1 khi x<0

''"^°*
(a) - 1 ;

n.TKBGDSVGTllNCT2

(b) 2 ;

161


(d) Mgt ke't qua khac.

(c)l;
Chgn cau tra ldi diing.
Trd ldi. ia).
Cdu 9. Cho ham sd .f(x)

kh i X > 0
x-1
x+1

khi

x<0,

(a) Ham sd da cho cd gidi ban trai va gidi ban phai tai 0 bdng nhau ;
(b) Ham sd da cho cd gidi ban trai va gidi ban phai tai 0 khac nhau ;
(c) Ham so da eho khdng cd gidi ban tai 0 ;
(d) Ca ba khang djnh tren deu sai.
Chgn cau tra ldi diing.

Trd ldi. (a).
Cdu 10. Cho ham so f(x) = |x - ll. Khi dd lim f(x) bdng:
•

x->r

(a) - 1 ;

(b) 0 ;

(c) 1 ;

(d) Mdt ke't qua khac.

Chgn cau tra ldi diing.
Trd ldi. (b).
Cdu 11. Cho ham sd f(x) = | Jf - 1 | • Khi dd lim f(x) bang:
'
x->r
(a) -1 ;

(b) 0 ;

(c)l;

(d) Mdt ke't qua khac.

Chgn cau tra ldi diing.
Trd ldi. (b).
Cdu 12. Cho ham sd f(x) = |-« - 1 | •

(a) Ham sd da eho cd gidi ban trai va gidi ban phai tai 1 bdng nhau ;
(b) Ham sd da eho cd gidi ban trai va gidi ban phai tai 1 khac nhau ;
(e) Ham sd da cho khdng cd gidi ban tai 1 ;
162