CAP SO CONG VA CAP SO NHAN ^'
A. KIEN THOC CAN NHd
5. Quy tac ve gidi han vo circ
a) Quy tdc tim gidi hqn eua tichf{x).g{x) lim /(x)
X-^XQ
L>0 L<0
lim g{x)
X-^XQ
+ 0 0
—00 + 0 0
—00
lim f{x)g{x)
. X-^XQ + 0 0 - 0 0
—00 + 0 0
b) Quy tdc tim gidi hqn cua thuang f{x) 8{x) lim /(x)
X^XQ
L L>0 L<0
lim g{x)
X^XQ
± 0 0
0 0
Dd'u cua g{x) Tuyy
+
-+
-X^XQ g{x)
0
+ 0 0
—00
—00 + 0 0
(Ddu eua ^(x) xet tren mdt khoang K nao dd dang tfnh gidi han, vdi x ^ XQ).
B. VI DU
• Vidul
Cho ham sd f{x) = 2x^ + X - 3 x ^ n • Dung dinh nghia ehiing minh ring lim /(x) = 5.
x->l
153
Gidi Ham sd da cho xdc dinh trdn R \ {1}.
Gia sfl (x„) la day sd bdt ki, x„ 9^ 1 vd x„ -^ 1.
2x^+x - 3 2(x„ - l)(x„ + - ) 3
lim /(x„) = lim ± 5 2 J 3 L _ ^ = lim 2_
n->+oo n->+oo X„ — 1 n—>+co X^ — 1
= Um 2(x„ + 1 ) = 5. Do dd, Um/(x) = 5.
• Vidu 2 Cho ham sd
Diing dinh nghia chiing
fix)
minh
fx , nlu X > 0 [l - X, ndu X < 0.
ring ham sd fix) khdng edgidi han khi x-> 0.
Gidi Ham sd da cho xdc dinh tren R.
Ldy day sd (x„) vdi x„ = —.
Ta ed x„ -> 0 va lim /(x„) = lim x„ = lim — = 0. (1)
n—»+oo - n-»+oo n—>+oo /2
Ldy day sd {y„) vdi y„ = — .
Ta cd >'„ ^ 0 vd lim f{y„) = lun (1 - y„) = Um (1 + - ) = 1. (2)
n->+oo n->+oo n->+oo /2
Tfl (1) va (2) suy ra ham sd/(x) khdng cd gidi han khi x -> 0.
^ Nhan xet
De dung djnh nghTa chflng minh hdm sd y =fix) khdng cd gidi han khi x -» XQ, ta thudng lam nhu sau :
• Chgn hai day sd khae nhau (a„) v^ ( i „ ) thoa man : a„ v^ b„ thugc tdp xae djnh cOa ham sd y =fix) va khae XQ ; a„ -> XQ ; i „ -> XQ ;
154
• Chflng minh rang lim / ( a „ ) ^ lim f{b„) hoSc chflng minh mdt trong cac gidi
n-»+oo n-»+oo
han nay khdng tdn tai.
^ Luu y : Trudng hgp x ^ Xg, x -> XQ hay x -> ±00 chflng minh tUOng tU.
• Vi du 3 Tfnh
a) lim ( V x ^ + 5 - 1 ) ; b) lim ^ ^ ; c) lim (-x^ + x^ - x + 1);
x^-2 \ J x^3~ X - 2 A:->-CO
,. ,. 1 - X . ,. 2 x - l d) h m ; e) lim - .
•^-^'^ ( x - 4 ) x^3- x-3
Gidi
a) lim (yfx^ + 5 - lj = yl{-2f + 5 - 1 = 2;
b) lim ^^— = — — = 4 ; x^r X-1 i - 1
c) lim (-x^ + x^ - X + 1) = lim x^(-l + r- + -r-) = -H» .
d)Taed lim (l - x) = - 3 < 0. (1) l i m ( x - 4 f = 0 v a ( x - 4 f > 0 v d i m g i x ^ 4 . (2)
x^A
f{x)
Ap dung qui tic vl gidi han vd cue ddi vdi thuong ^^-rr, tfl (1) va (2) suy ,. 1 - x
ra lim = - 0 0 .
^^^x-Af
e) Ta cd lim (2x - l) = 5 > 0, lim (x - 3) = 0 va (x - 3) < 0 vdi
x-^3 x^3 2 x - l
moi X < 3 . D o dd, lim — = - 0 0 . x^3~ X-3
^ Nhan xet
Trong cac vf du tren ta da dung true tidp cac dinh If v^ gidi han cOa tdng, hieu, tfch, thuong va can cOa cac ham sd hoSc cac quy tac vi gidi han vd cue.
155
• Vi du 4
Tfnh cdc gidi han sau : a) lim x'^ + 2x - 3
^^1 2x^ - X - 1 2x^ + 3x - 4
x^ + 1 c) lim
J:->+«) —X
e) Um - | - ^ - l x-^o'-^V-^ + l J
b) lim •
x^2 Vx + 7 - 3
^, ,. Vx^ - X - ylAx^ + 1 d) lim r——5
AT—>-00 Z X + J
f) Um {\1AX^ - X + 2x).
Gidi
, ,. x ' ^ + 2 x - 3 ,. ( x - l ) ( x + 3) ,. x + 3 4 a) U m — = lun — = Um = - .
- 1 2 x 2 - x - 1 - i 2 ( ; , _ i ) ( ^ + | ) -->i2x + l 3 2 - X (2 - x)(Vx + 7 + 3) ^ b) lim -^^ J-^ = lim ^ - ^ ; ^ = l i m - (Vx + 7 + 3] =
J : ^ 2 VX + 7 - 3 A:^2 X - 2 ;c->2
J 4_
, ,. 2 x ^ + 3 x - 4 ,. "^r2 v3 c) lim — = lim ^ ^ = - 2 .
x-^+00 _;c - X +1 x^-^ i 1 1
^ x^
d) lim
. yfx^-yl^
Ax^ + 12 x + 3 = lim
-j r - > - o o
\x\Jl IxL 4 + —r 2 x + 3
= lim
-- x J l -- -- + x j 4 + —r
2 x + 3 = lim
--Jl-i..4.J.
2 + 1
X
2
e) lim —
;c->0" X r
x + 1 - 1 = lim l - ( x + l)
= lim - 1
;t^o- x{x + 1) ;,^o- {x + 1) = - 1 156
.2 „^ A Jl
f) lim ( V 4 x 2 - x + 2 x ) = lim ^^^ ""^ "^"^
^ ^ ^ ^^-^ yJAx^ - X - 2x
= lim , = lim , = lim , = —.
X-^^x> j 1 ;c->-oo / 1 At->-oo / 1 4
^ Nhan xet
Khi tfnh gidi han md khdng the ap dung true tidp djnh If ve gidi han trong sach giao khoa, ta phai bien ddi bieu thflc xae djnh h^m sd ve dang ap dung dugc cac djnh If nay.
Sau day la mgt sd each bien ddi thudng dugc dtjng.
• Tinh lim — - khi lim u{x) = lim v(x) = 0
x-*Xf) v ( x ) X-^XQ X-*XQ
- Phdn tfch tfl vd mdu thdnh tfch cdc nhdn tfl vd gian udc. Cu thi, ta biln ddi nhu sau :
,. M(X) ,. ( X - X A ) A ( X ) ,. A(x) . , , ,. A(x)
lim -7-^ = lim "; ; [ = lim — ^ va tfnh Irni —7^.
X-^XQ V ( X ) X-^XQ ( X — X o ) B ( x ) X->XQ B ( x ) X-^XQ B ( x )
- Ne'u M(X) hay v(x) ed chfla bidn sd dudi dd'u cdn thi cd thi nhdn tfl va mdu vdi bilu thflc lien hgp, trudc khi phdn tfch chflng thdnh tfch dl gian udc.
• Tinh lim khi lim u{x) = ±00 va lim v(x) = ±00
J:->±CO V ( X ) X^XQ X^X^
- Chia tfl vd mdu cho x" vdi n la sd mu bdc cao nhd't eua bidn sd x (hay phdn tfch tfl vd mdu thanh tfch chfla nhdn tfl x" rdi gian udc).
- Ndu M(X) hay v(x) cd chfla bidn x trong dd'u can thflc, thi dua x ra ngoai ddu cdn (vdi fe la sd mu bdc cao nhdt cua x trong dd'u cdn), trudc Ichi chia tfl va mdu cho luy thfla eua x.
• Tinh lim [M(X) - v(x)] khi lim M(X) = +00 va lim v(x) = +00
X->XQ X->XQ X^XQ
hoac lim M(X).V(X) khi lim M(X) = 0 va lim v(x) = +00.
X^>XQ X^XQ X->.XQ
Nhan vd chia vdi bieu thflc lien hgp (neu cd bieu thflc chfla bien sd dudi ddu cdn thflc) hoSc quy dong mau de dua ve cung mot phan thflc (neu chfla nhieu phan thflc).
157
C. BAI TAP
x2 x ^ .
b) lim —
x^+co x'^ +1
, ndu X > 0 - 1 , ndu X < 0 . 2.1. Dung dinh nghia tim cdc gidi han
. ,. x + 3 a) lim:r ;
x^5 3- X 2.2. Chohamsd'/(x)=<
a) Ve dd thi eua ham sd/(x). Tfl dd du dodn vl gidi han cua/(x) khi x -> 0.
b) Dung dinh nghia chiing minh du dodn trdn.
2.3. a) Chiing minh ring ham sd y = sinx khdng ed gidi han khi x —> +oo.
b) Giai thich bing dd thi kit ludn d cdu a).
2.4. Cho hai hdm s6 y = fix) va y = g{x) cung xdc dinh trdn khoang (-oo ; a).
Dung dinh nghia chiing minh ring, ndu lim /(x) = L vd lim g{x) = M
x->-ao .)C->-oo
thi lim /(x).g(x) = L.M.
2.5. Um gidi han eua edc ham sd sau :
^)fix) = ; khi X ^ 3 ; b) h{x) = khi x -> - 2 ;
^ - 1 {x + 2f
c) fe(x) = sAx^ - x + 1 khi X ^ - 00 ; d)fix) =x + x^ + 1 khi x -> -oo ; X —15
-e) h{x) = khi x ^^ - 2 va khi x —> - 2 . 2.6. Tfnh eae gidi han sau :
, ,• ^ + 3 , , ,. (1 + x)^ - 1
a) lun -^ ; b) lim-!^ ;
— - 3 x 2 + 2 x - 3 .^^0 X e) lim -^ ; d) lim
^->+°o X^ - 1 ' .>:^5 Vx - >/5 ' . ,. JC-5 ^ ,. \lx^+5-3
e) Iim -j= 7= ; f) lim
;c->+oo V x + V5 x^-2 X + 2 158
g) lim V x - 1 x-^i Vx + 3 - 2
i)
lim-T-h) lim 1 - 2x + 3x'*
x-^-w X -9
;c->Ojc^ U ^ + l 1 ; j) lim (x^ - 1)(1 - 2x)^
' x'^ + X + 3 2.7. Tfnh gidi han eua cdc ham sd sau khi x -> +oo va khi x -> -oo
Vx^ - 3x a)Ax)= ^ ^ , , c)/(x) = V x ^ - x - Vx^ + 1.
b)/(x) = X + Vx^ - X + 1 ;
2.8. Cho hdm sd
fix) = 2x^ - 15x + 12 X - 5x + 4 cdddthinhuhinh4.
a) Dua vao dd thi, du doan gidi han eua ham s6fix) khi x -> 1"^;
x - ^ l ; x - > ' 4 ' ^ ; x - > 4 ; x->+Qovakhi x->-oo.
b) Chflng minh du dodn trdn.
2.9. Cho ham sd _ 1 fix) =
y
3 2 0
M
1 4
/ ^ " " ^
/ '''
•, ne'ux>l
^ - 1 x ^ - l
mx + 2, ne'ux<l.
Hinh 4
Vdi gia tri nao eua tham sd m thi ham ^6 fix) cd gidi han khi x —> 1 ? l i m gidi han ndy.
2.10. Cho khoang K,XQeK va ham s6y =fix) xdc dinh tren K\ {XQ}.
Chflng minh ring ndu lim /(x) = +QO thi ludn tdn tai ft nhd't mdt sd c thude
X^XQ . •
Ar\{xo} sao cho/(c) > 0.
159
2.11. Cho ham sd y =/(x) xdc dinh trdn khoang {a ; +00).
Chiing minh ring ndu lim J
J : ^ + C C
thude {a ; +00) sao cho/(c) < 0
Chiing minh ring ndu lim /(x) = -00 thi ludn tdn tai ft nhdt mdt sd c
J : ^ + C C
§3. Ham so iien tuc
A. KIEN
THQC CAN N H 6 1. Ham so lien tuc• Cho ham sd y =/(x) xdc dinh tren khoang ^ vd XQ e K.
y =fix) lien tuc tai XQ khi vd chi khi lim / ( x ) = /(JCQ) .
X^XQ
• y =fix) Uen tuc tren mdt khoang ndu nd Udn tuc tai mgi dilm cua khoang dd.
• y = fix) lien tuc tren doan [a ; b] nlu nd lien tuc tren khoang (a ; b) va Um /(x) = f{a), Um /(x) = f{b).
x^>-a* x^^b
^ Nhan xet : Dd thj ciia ham sd lien tuc tren mdt khoang la mdt "dfldng lien" tren khoang dd.