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GII TCH 1
HONG HI HBCH KHOA TPHCM
17th June 2013
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1 GII HN DYPhp ton v gii hn dy
BI TP THC HNH2 HM S
Hm s c bn
Gii hn hmVCBV cng lnV. Hm lin tc
3 O HMCc php ton o hmQuy tc LHospitale
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4 KHO ST HM SCc tr hm sTim cn
5 TCH PHN BT NH
Phng php tnh v cc dng tch phnTch phn hu tTch phn lng gicTch phn v t
6 TCH PHN SUY RNG
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CHNG I: DY S
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I. Cc php ton v gii hn dy
Hai dy an, bnc lim an =a, lim bn =bth:
lim(an bn) =a blim(anbn) =ablim an
bn= a
bnu bn
=0, b
=0
lim anbn =ab nu an >0
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I. Cc php ton v gii hn dy
Hai dy an, bnc lim an =a, lim bn =bth:
lim(an bn) =a blim(anbn) =ablim an
bn= a
bnu bn
=0, b
=0
lim anbn =ab nu an >0CH :
1 a = , a
=0, a
0 = (a =0),a+= +(a>1), a=0(a>1).
2 Cc dng v nh:0
0
,
,
, 0.
,
0, 00, 1.
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II. Phng php tnh gii hn
VCB-VCL
Dy s{an}l VCL nu lim |an| = +, l VCB nulim an =0.
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II. Phng php tnh gii hn
VCB-VCL
Dy s{an}l VCL nu lim |an| = +, l VCB nulim an =0.Tng ng VCL
Hai VCL{an},{bn}gi l tng ng nu lim anbn
=1.K hiu : an bn
So snh bc VCLVCL{an}c bc nh hn{bn}nu liman
bn=0. K hiu:
an
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So snh bc cc VCL
ln
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Ta c ng thc sau: Nu an, bn, cn, dnln lt l ccVCL, v an
cn, bn
dn, th:lim
an
bn=limcn
dn.
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Gii hn kpNu un
xn
vn, lim
nun = lim
nvn =A. Khi :
limn
xn =A
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Gii hn kpNu un
xn
vn, lim
nun = lim
nvn =A. Khi :
limn
xn =A
V d 1.1a. lim
nsin x
x =0, ( >0)do1
xsin x
x 1
x
b. limn
2008
n
n=0 do 00limxx0
g=1 TH lnf lng
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0
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Cng thc tnh gii hn dnh cho dng 00
Nu f , g khi x x0 TH limxx0 fg = limxx0
Cc tng ng thc c bn khi x
01/ ln(1+x) x 2/ ex 1 x
3/ (1+x) 1 x 4/1 cosx x2
2
5/ arctanx x 6/ shx x7/ chx 1 x
2
2
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Tm hm g(x) =Axn sao cho f gkhi x 01. f =sin
2x+
x+x2 x5
2. f =2x2
1
3. f = 4 x4 +x2 24. f =ln(1+sinx) x2
5. f =2sinx tan2x
6. f = x2artanx
x5 +x2 +17. f =ex2+x e2x
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p dng tng ng VCB tnh cc gii hn sau
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1. limx x
3 1x 1
2.
limx03
1+3x
5
1+2x1+5x 41+2x
3.
limx0ex
2
1
1+sin2x 1
4.lim
x1+xx x
ln(1+
x2 1)
5. limx1ex
2+x
e2x
cosx
26.
limx0 2x+2arctan3x+3x2
ln(1+3x+sin2x) +xex
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Gii hn dng m 1
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1/ limx0
cosx+arctan2x
1arctan2x 2/ lim
x1lnx x+1
x xx
3/ limx
e1/x +1xx
4/ limx+
2 arctanxx
5/ limx0(chx)1
1 cosx
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IV. VCL
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NH NGHA(x)c gi l VCL khi x x0nu lim
xx0|(x)| =
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IV. VCL
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NH NGHA(x)c gi l VCL khi x x0nu lim
xx0|(x)| =
So snh cc VCLXt 2 VCL (x), (x)khi x x0v t s : limxx0
=k.
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IV. VCL
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NH NGHA(x)c gi l VCL khi x x0nu lim
xx0|(x)| =
So snh cc VCLXt 2 VCL (x), (x)khi x x0v t s : lim
xx0
=k.
a. k=0
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IV. VCL
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NH NGHA(x)c gi l VCL khi x x0nu lim
xx0|(x)| =
So snh cc VCLXt 2 VCL (x), (x)khi x x0v t s : lim
xx0
=k.
a. k=0=o()
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IV. VCL
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NH NGHA(x)c gi l VCL khi x x0nu lim
xx0|(x)| =
So snh cc VCLXt 2 VCL (x), (x)khi x x0v t s : lim
xx0
=k.
a. k=0=o()
b. k=
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IV. VCL
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NH NGHA(x)c gi l VCL khi x x0nu lim
xx0|(x)| =
So snh cc VCLXt 2 VCL (x), (x)khi x x0v t s : lim
xx0
=k.
a. k=0=o()
b. k= l VCL cp CAO hn .
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IV. VCL
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NH NGHA(x)c gi l VCL khi x x0nu lim
xx0|(x)| =
So snh cc VCLXt 2 VCL (x), (x)khi x x0v t s : lim
xx0
=k.
a. k=0=o()
b. k= l VCL cp CAO hn .c. kHH. v l 2 VCL cng cp. Nu k=1, ta c
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CC TNH CHT CA V CNG LN TNG NG
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CC TNH CHT CA V CNG LN TNG NG
a. Nu f TH f .
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CC TNH CHT CA V CNG LN TNG NG
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CC TNH CHT CA V CNG LN TNG NG
a. Nu f TH f .
b. Tng cc VCL khc cp s T v cng ln cp CAO
nht.
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CC TNH CHT CA V CNG LN TNG NG
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CC TNH CHT CA V CNG LN TNG NG
a. Nu f TH f .
b. Tng cc VCL khc cp s T v cng ln cp CAO
nht.c.
limxx0
f =A TH f A
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CC TNH CHT CA V CNG LN TNG NG
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CC TNH CHT CA V CNG LN TNG NG
a. Nu f TH f .
b. Tng cc VCL khc cp s T v cng ln cp CAO
nht.c.
limxx0
f =A TH f A
d. f g, f, g>0limxx0
g
=1 TH lnf lng
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TH T SO SNH CC VCL HAY GPlnx
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TH T SO SNH CC VCL HAY GPlnx
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f sau:
a. f = x2arctanx
x5 +x2 +1, x0= +
b. f =
x
x+2 2x+1+ x, x0= +
c. f =
x2 +1
x, x0=
d. f =
x2 +1 31 x2, x0= HONG HI H (HBK TPHCM) GII TCH 1 17th June 2013 31 / 85
TNH GII HN BNG CCH THAY TNG NGV CNG LN:
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V CNG LN:
1. limx(x2 +8x+3 x2 +4x+3)
2. limx
+
(2x 4x2 7x+4)
3. limx
arcsin(
x2 +x+x)
4. limx
x+
x3
+2x2
x+1
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5. limx+
( 6
x6 +x5 6x6 x5)
6.(*) limx+ ex
(sh
x
2
+x sh
x
2
x)
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HM LIN TC
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Hm lin tca. Hm f(x)l lin tc ti x0nu tha mn:f(x0+) =f(x0) =f(x0)
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HM LIN TC
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Hm lin tca. Hm f(x)l lin tc ti x0nu tha mn:f(x0+) =f(x0) =f(x0)b. Hm f(x)lin tc trn tp Dnu n lin tc ti miim ca D.
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HM LIN TC
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Hm lin tca. Hm f(x)l lin tc ti x0nu tha mn:f(x0+) =f(x0) =f(x0)b. Hm f(x)lin tc trn tp Dnu n lin tc ti miim ca D.c. Cc hm s cp lin tc trn tp xc nh.
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HM LIN TC
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Hm lin tca. Hm f(x)l lin tc ti x0nu tha mn:f(x0+) =f(x0) =f(x0)b. Hm f(x)lin tc trn tp Dnu n lin tc ti miim ca D.c. Cc hm s cp lin tc trn tp xc nh.d. Kho st tnh lin tc mt hm s( thng l hm
ghp) trn tp xc nh, ta ch cn kho st ti cc imghp.
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Tm gi tr a, b cc hm sau l hm lin tc trn tp
xc nha. f(x) =
e1x , x=0
a, x=0
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Tm gi tr a, b cc hm sau l hm lin tc trn tp
xc nha. f(x) =
e1x , x=0
a, x=0
b. f(x) =
(x
1)3, x
0
ax+b, 0
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Tm gi tr a, b cc hm sau l hm lin tc trn tp
xc nha. f(x) =
e1x , x=0
a, x=0
b. f(x) =
(x
1)3, x
0
ax+b, 0
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CHNG III: O HM
HM S
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I. Cc php ton o hm v vi phn
o hm hm hp
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o hm hm hpf =f(u), u=u(x)
f
(x) =f
(u)u(x)
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I. Cc php ton o hm v vi phn
o hm hm hp
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o hm hm hpf =f(u), u=u(x)
f
(x) =f
(u)u(x)
o hm tham sHm tham s c cho bi: x=x(t),y=y(t). y(x) = y(t)
x(t) .
o hm cp 2: y(x) =[y(x)]
t
x(
t)
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I. Cc php ton o hm v vi phn
o hm hm hp
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o hm hm hpf =f(u), u=u(x)
f(x) =f(u)u(x)
o hm tham sHm tham s c cho bi: x=x(t),y=y(t). y(x) = y(t)
x(t) .
o hm cp 2: y(x) =[y(x)]
t
x(
t)
o hm cp cao
Cng thc Lebnitz:(fg)(n) =n=
Ckn f(k)gnk
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Cc cng thc o hm cp cao c bn
1. f =ax, f(n) =axlnna2. f = 1
x+a, f(n) = (1)n n!
(x+a)n+1
3. f =ln(x+a),
f(n) =
1x+a
(n1)
= (1)n1(n 1)!(x+a)n
.
4. f =sin(ax),
f(n) =ansin(ax+n
2
)
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Tnh o hm cc hm s sau:
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Tnh o hm cc hm s sau:
1. y=aarcsin(1+x2)
x , a>0
2.y
= (ln
(1+x
))
sinx
3. y=
xnsin1
x
, x
=0, n
N
0 x=0
4. y=arctan(2x)
5. y=x|x2
5x+6
|6. x=ln(1+t2),y=t arctant
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Gi s f(x), g(x)c o hm vi mi x. Tnh ohm ca:a. y=f(x)g(x), f(x)>0
b. y=f(ex)ef(x)
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Vi phn
Vi phn cp n ca hm y = f (x) ti x0
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Vi phn cp n ca hm y=f(x)ti x0
dn
f(x0) =fn
(x0)dxn
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Vi phn
Vi phn cp n ca hm y = f (x) ti x0
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Vi phn cp n ca hm y f(x)ti x0
dn
f(x0) =fn
(x0)dxn
Quy tc vi phn
d(fg) =fdg+gdf
d(f
g) =
gdf fdgg2
d(f
) =f
1
dfd(af) =aflnadf, d(lnf) =df/f
Trong ,f,g l hm ph thuc x. CH : d2x=0.HONG HI H (HBK TPHCM) GII TCH 1 17th June 2013 41 / 85
Tnh df(x0):
a f =1 + ln x 1 x0 = 1
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a. f =x
+ln
x , x0= 1
b. f = x22xxx
, x0=2
Vit biu thc dytheo du v dv:
a. y=euvb. y= uv
u2 +v2
Tnh vi phn cp 2 ti x0ca hm:f =arctan
2+x22 x2
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Tnh d2ytheo dx2 cc hm tham s:t2 t3
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1. x= t
1+t3
,y= t
1+t3
2. x= (t2 +1)et,y=t2e2t ti t=0
Tnh cc o hm cp cao sau: yn(
x0)
1. y= x2
1 x, n=8, x0=0.
2. y= (x2
2x)cos3x, n=101, x0=0.3. y=x2lnx, n=100, x0=1
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Quy tc LHospitale
0
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Xt gii hn dng 0
0
hoc
Nu limxx0
f(x)g(x)
tn ti th:
limxx0 f(x)g(x)= limxx0 f(x)
g(x)
CH :
i vi cc dng v nh cn li, mun dng LHopitale,phi bin i v hai dng trn trc.
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Tnh cc gii hn sau:1 lim xarcsinx
2
2 lim (x+1)ln(x+1) x
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1. limx0 xcosx
sinx
2. limx0 ex
x
1
3. limx0+
3+lnx2 3lnsinx 4. limx+
3xln(lnx)3
2x+3
lnx
5. limx
0+(xx
1)lnx 6. lim
x
+
(
2arctan
x)
x
7. limx0
1sinx
1x
8. lim
x0
1x 1
ex 1
9. limx1 x 1x1
10. limx0+(arcsinx)
tanx
11. limx
+
(3x2 +3x)1x 12. lim
x
0+
xxx1
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II. KHAI TRIN TAYLOR
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nh nghaHm f(x)lin tctrn[a, b]v kh vi n cp n-1. Tnti f(n)(x0)vi x0 [a, b]. Khai trin Taylor ca f(x)ncp n ti x0l :
f(x) =n
k=0
f(n)(x0)
k! (x x0)k +o((x x0)n).
Trong : o((x
x0)n)l phn d Peano.
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II. KHAI TRIN TAYLOR
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nh nghaHm f(x)lin tctrn[a, b]v kh vi n cp n-1. Tnti f(n)(x0)vi x0 [a, b]. Khai trin Taylor ca f(x)ncp n ti x0l :
f(x) =n
k=0
f(n)(x0)
k! (x x0)k +o((x x0)n).
Trong : o((x
x0)n)l phn d Peano.
Vix0=0 : Khai trin MACLAURINT.
HONG HI H (HBK TPHCM) GII TCH 1 17th June 2013 46 / 85
Khai trin Maclaurint cc hm s cp n cp n:1 ex = 1 + x + x
2
+x3
+ +xn
+ o(xn)
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1. e =1+x+ 2! + 3! +...+ n! +o(x )
2. ln(1+x) =x x2
2 +x3
3! +...+ (1)n1x
n
n +o(xn)
3. (1+x) =1+x+( 1)2! x2 +...+( 1)...( n+1)
n! xn +o(xn)
4. sinx=x x3
3! +x5
5! +...+ (1)n x
2n+1
(2n+1)!+o(x2n+1)
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Khai trin TAYLOR trong ln cn x0n cp ncc
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hm s sau:
1. y=ln(2x+1), x0=12
2. y=
x, x0=1
3. y= x2 +4x+4
x2 +10x+25 , x0= 2
4. y=e2x2
+2x1
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THC HIN KHAI TRIN MACLAURINT N CP n:
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1. y=tanx, n=3
2. y=
1 2x+x3 31 3x+x2, n=3
3. y= 1
x2 +3x+2 , n=44. y=e
1+2x, n=2
5. y=sin(arctanx), n=4
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Tnh o hm cp n:f(n)(0)
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1. y=ex2, n=6
2. y= 11+x+x2 , n=32
3. y= 11 x4 , n=60
4. y= (x2 +2x)cos(x2
x)
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Tm bc cc v cng b sau khi x 0
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Tm bc cc v cng b sau khi x 0
1. y= (1+x2)sinx arctanx
2. y=ln(
1+2x ln(1+x))
3. y=ln(1+sinx) tanx
4. y=tanx xex2
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P DNG KHAI TRIN MACLAURINT TNH GIIHN HM S:f (x)
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Dng thng gp tnh l I
=limx0f(x)
g(x)vi dng vnh00
Khai trin Maclaurint c t v mun bc b nht:f(x) =axn +o(xn), g(x) =bxm +o(xm).Xy ra cc trng hp sau:
1
Nu n=mTH I = a
b.2 Nu nmTH I =0
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TNH CC GII HN SAU:
1 i 2
1 i 2
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limx0
1+sin2x
1
sin2x
ln(1+sinx) tanx
limx0
arctanx arcsinxtanx
sinx
limx0
ex2+x 1+2x x2ex
2 1limx x x
2ln(x+1
x
)limx0
3
1 2x2 xcotxxsinx
HONG HI H (HBK TPHCM) GII TCH 1 17th June 2013 54 / 85
li ( )cot2x
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limx
0(cosx)cot x
limx0
(x+1)x+1 xln(cosx)
limx0 1+xcosx 1+2xln(1+x) sinx
limx
0
ecosx e 31 4x2
(1/x)arcsin2x 2coshx2
HONG HI H (HBK TPHCM) GII TCH 1 17th June 2013 55 / 85
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CHNG IV: KHO ST
HM S
HONG HI H (HBK TPHCM) GII TCH 1 17th J 2013 56 / 85
CC TR HM S
1. TX.2 Gii phng trnh y=0 + tm nhng im y
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2. Gii phng trnh y =0 + tm nhng imy
KHNG TN TI. Tp hp nhng im ny gi l IMNG.3. Lp BBT kho st cc tr: Sp cc IM NG trn
BBT v xt du y(x0). Nu qua x0, y i du t + sang- th ti hm t cc i. Ngc li, hm t cc tiu.Qua x0y khng i du th hm khng c cc tr ti .Hoc:Xt du y(x0). Nu y(x0)>0 hm t CT. Nu
y(x0)
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TM TIM CN HM Sy
=f
(x
)
TIM CN NG : x=al TC nu limxa
=
HONG HI H (HBK TPHCM) GII TCH 1 17th J 2013 58 / 85
TIM CN
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TM TIM CN HM Sy
=f
(x
)
TIM CN NG : x=al TC nu limxa
=
TIM CN NGANG: y=bl TCN nu limxy=b
HONG HI H (HBK TPHCM) GII TCH 1 17th J 2013 58 / 85
TIM CN
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TM TIM CN HM S y=f(x)
TIM CN NG : x=al TC nu limxa
=
TIM CN NGANG: y=bl TCN nu limxy=bTIM CN XIN: y=ax+bl TCX nu
a= limxy
xb= limx(y ax)
HONG HI H (HBK TPHCM) GII TCH 1 17 h J 2013 58 / 85
Tnh li lm, im un
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Tm im unGii nghim y hoc im y khng tn ti.Xt du y qua cc im trn, nu qua x0, y i duth l im un.
Tnh li lmHm y=f(x) li trn khong (a,b) nu trn , f(x)
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1. y=x+ 4x2
+1
2. y= (x+3)e1x
3. y= x2
|x| +4
4. y=
1+1xx
5. y=1 xe1
|x| 1x
6. y=xx1
x
x
7. y=x
x
x+4
8. y=1+xe3x
HONG HI H (HBK TPHCM) GII TCH 1 17 h J 2013 60 / 85
TM CC TR CC HM S SAU:
1 3
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1. y= 5 x2 4x 2. y= 1+3x4+x2
3. y= 3
x2 x 4. y= |x 5|(x 3)3
TM GTLN-GTNN ca hm y=f(x)trn khong [a,b]:y=x 2x, x [0, 5]
y= |x2
+2x 3|, x [1/2, 2]
HONG HI H (HBK TPHCM) GII TCH 1 1 h J 2013 61 / 8
KHO ST V V CC HM SAU
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y=x+
3 x
y=e
1
x
x(x+2)y=
lnxx
HONG HI H (HBK TPHCM) GII TCH h J 6 /
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CHNG I: TCH PHN
BT NH
O G ( C ) G C /
I. Phng php tnh v cc dng tch phn
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i binXt tp:
f(x)dx. t x=(u). Khi , tp tr thnh:
f(x)dx= f((u))(u)du
( ) /
I. Phng php tnh v cc dng tch phn
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i binXt tp:
f(x)dx. t x=(u). Khi , tp tr thnh:
f(x)dx= f((u))(u)duTng phn
udv=uv vdu
Tnh cc tp bt nh sau
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1. 1
1+ 3
x+1dx
2.
dxex +e
x
3. arcsinx
x2 dx
4.
dx(x2 +4)2
II. Tch phn hu t
Xt tp hu t: P(x)Q(x)
dx, vi degP(x)
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1. Phn tch mu thnh tch cc nhn t cha cc nhthc v tam thc bc 2 v nghim.
2. Gi s ta c: Q(x) = (x
a)(x
b)s1(ax2 +bx+c).
3. Khi :
HONG HI H (HBK TPHCM) GII TCH 1 17th June 2013 66 / 85
II. Tch phn hu t
Xt tp hu t: P(x)Q(x)
dx, vi degP(x)
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1. Phn tch mu thnh tch cc nhn t cha cc nhthc v tam thc bc 2 v nghim.
2. Gi s ta c: Q(x) = (x
a)(x
b)s1(ax2 +bx+c).
3. Khi :P(x)
Q(x)=
A
x
a+
B1
x
b+...+
Bs1(x
b)s1
+ Cx+D
ax2 +bx+c
4. Tnh Cx+D
ax2 +bx+cdx
HONG HI H (HBK TPHCM) GII TCH 1 17th June 2013 66 / 85
Tnh Cx+D
ax2 +bx+cdx.
Phn tch: Cx+Dax2 + bx + c
= C
2a 2ax+b
ax2 + bx + c+
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ax +bx+c 2a ax +bx+c
+(D bC2a) 1
a
(x b2a )
2
+4ac b2
4a2
.
Khi , tch phn cn tnh l:C
2aln(ax2
+bx+c) +
2Da
bC
2a21Karctan
x
b/2a
K
. Vi K2 =4ac b2
4a2HONG HI H (HBK TPHCM) GII TCH 1 17th June 2013 67 / 85
Tnh cc tch phn hu t sau
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1.
xdx2x2 x+1
2. dx(x+2)(4x2 +8x+7)
3.
4x2 +4x 11(2x
1)(2x+3)2
HONG HI H (HBK TPHCM) GII TCH 1 17th June 2013 68 / 85
III. TCH PHN HM LNG GIC
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R(sinx, cosx)dx
t t=tan x
2
cho TH TQ.
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III. TCH PHN HM LNG GIC
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R(sinx, cosx)dx
t t=tan x
2
cho TH TQ.
Trng hp c bit1 R(sinx, cosx) = R(sinx, cosx), t=cosx2 R(sinx,cosx) = R(sinx, cosx), t=sinx3 R(
sinx,
cosx) =R(sinx, cosx),
t=tanx
HONG HI H (HBK TPHCM) GII TCH 1 17th June 2013 69 / 85
Dng c bit
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A1sinx+B1cosx+C1Asinx+Bcosx+C
dx.
Tch:A1sinx+B1cosx+C1=(Asinx+Bcosx) +(Acosx Bsinx) +
HONG HI H (HBK TPHCM) GII TCH 1 17th June 2013 70 / 85
Tnh cc TPLG sau
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1.
cosxdxsin3x+cos3x
2. dxsin3xcos5x
3.
dx
3 2sinx+cosx
4.
cosx 2sinxcosx+sinx
dx
5.
tan3xdx
HONG HI H (HBK TPHCM) GII TCH 1 17th June 2013 71 / 85
IV. Tch phn v t c bn
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1.
dxa2 x2 =arcsin xa +C
2. dxx2 a2 =ln|x+
x2
a2
|+C
3. dx
x2 +a2=ln|x+ x2 +a2| +C
HONG HI H (HBK TPHCM) GII TCH 1 17th June 2013 72 / 85
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R(x,
ax+b
cx+d
p1q1 ,
ax+b
cx+d
p2q2 )dx.
t
ax+b
cx+d =ts
vi sl BCNN ca q1, q2.
HONG HI H (HBK TPHCM) GII TCH 1 17th June 2013 73 / 85
V d:
4 3 x 6 x
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1
x
x
x( 6x+1)dx
x 1 1x+1+1 x 3x+2
x+ 3
x+2 x
4
x3(4 x)
HONG HI H (HBK TPHCM) GII TCH 1 17th June 2013 74 / 85
TP EULER
R(x,
ax2 +bx+c)dx
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Phng php 1: Lng gic.
aR(x,
ax2 +bx+c)v mt trong 3 dng sau:
R(u,
u2 A2). t u= Asint
hocu=Acht(Acosht)
R(u, u2
+A2
). t u=AtantR(u,
A2 u2). t u=Asint
HONG HI H (HBK TPHCM) GII TCH 1 17th June 2013 75 / 85
Phng php 2: Euler tng qut.
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1 t
ax2 +bx+c=x
a+tnu a>02
ax2 +bx+c=xt+
cnu c>0
3 ax2 +bx+c= (x x1)tnu x1l nghim ca tamthc bc 2.
HONG HI H (HBK TPHCM) GII TCH 1 17th June 2013 76 / 85
Tnh cc tch phn v t sau:5
dx
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3/4
2+3x 2x2
1
0
dx
2x2 +2x+51
0
x
2
+2x+5dx
HONG HI H (HBK TPHCM) GII TCH 1 17th June 2013 77 / 85
(x+2)dx
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3x2 +2x+2 (x+2)
2x2 x+1dx
21
e2xe2x +ex
HONG HI H (HBK TPHCM) GII TCH 1 17th June 2013 78 / 85
Tch phn dng: dx2
.
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(x x0)
ax +bx+c
t t= 1x x0 . V du:
(3x+2)dx(x+1)
x2 +3x+3
HONG HI H (HBK TPHCM) GII TCH 1 17th June 2013 79 / 85
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dx
(x+2)
3x2 +5
dxx1 ln2x
HONG HI H (HBK TPHCM) GII TCH 1 17th June 2013 80 / 85
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CHNG VI. TCH PHN
SUY RNG
HONG HI H (HBK TPHCM) GII TCH 1 17th June 2013 81 / 85
I. Tch phn suy rng loi 1
im k di
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im x0c gi l im k d ca hm f(x) nu:limxx0
f(x) =
HONG HI H (HBK TPHCM) GII TCH 1 17th June 2013 82 / 85
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Cng thc tnh
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+a
f(x)dx= F(x)|+a =F(+) F(a).
Tp c gi l hi t nu gi tr ca n hu hn, ngcli tp phn k.
HONG HI H (HBK TPHCM) GII TCH 1 17th June 2013 83 / 85
Tiu chun so snh
Xt hai tch phn sr loi 1: I=+
a
f(x)dx, J=+
a
g(x)dx,
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trong f(x), g(x) 0, x [x0, +), x0 a. Xt:k= lim
x+f(x)
g(x).
Nu k hu hn th I,J cng bn cht(THBk=1,I J)k=0, ch kt lun c I nu J hi t, khi I hi t.
k=+, ch kt lun c I nu J phn k, khi Iphn k.
HONG HI H (HBK TPHCM) GII TCH 1 17th June 2013 84 / 85
Ch
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1. +a>0
dx
xhi t khi >1.
2. +
a
dxxlnxhi t khi >1 hoc =1, >1
HONG HI H (HBK TPHCM) GII TCH 1 17th June 2013 85 / 85
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