洛必达法则LHopitals rule.docx
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洛必达法则LHopitalsrule
洛必达法则(L'Hopital'srule)
ThesecondsectionL'HospitalRule
XM.Am
1.(6)LIM(a=0);
X.>axn.An
.
Solution:
LimXMN
.amn=LimMX
NM
.
.
Eleven
=mamn(a=0)
X.>ax.Ax.>anxn
Lntan7x
(7)Lim
X.>0+lntan2x
1.Sec27x.72
Solution:
Lim,lntan7,x=Lim,Tan7,x=Lim,Tan,2,x,SEC,7x,
X.>0+,LN,tan2,xx.>0+,1,.Sec,2,2x,.2,x.>0+,Tan,7,x,SEC,2,2x,
Tan2,X
2sec27x7
=Lim,X.2,.=1
X.>0+7sec2x2
X
One
Ln(1+)
(9)Limx
X.>+,arccot,2,X
Eleven
11(.
X2)1
Solution:
Lim
Ln(1+
X)
=Lim
1+
.
X
Two
=Lim,1+,x2
=Limx2
+
+
Eleven
=1
X.>+,arctan,2,xx.>+,1,x.>+,,x,x.>+,X2,1
1+XX
Twenty-one
(14)LIM(2.);
X.>1,x,.1,x,.1
11
Solution:
LIM(22.1)=Lim2
X1=Lim
.1.
X.>1,x,.1,x,.1,x.>1,x,.1,x.>12x,
One
(16)LIM()Tanx
X.>0+x
.
1lim+tanx..Ln1xlim+1x
(1)
1tanxx.>+0
Solution:
LIM()=e
LimTan
Ee00200lnxxxxxxxx.>.>
+.>
.
===elimx
E==1
X.>0+x
Twenty-one
Xsin
Thereare3.verificationlimitLimx,butnotwithl'Hopital'srulethat.
X.>0SiNx
X21'11
(SIN)2xsin.Cos
Card:
Lim=LimbecauseXXdoesnotexist,itcannotusel'Hopital'sruletosolvethelimit,
X.>0(sin,x)(x)',x.>0,cos,X
Butyoucanusethefollowingmethodstofindthelimit:
One
X2sin1
Lim,x=,LIM(x.Xsin1)=Lim,x,.Lim,xsin,1,=10.=0
X.>0,sin,xx.>0,sin,xxx.>0,sin,xx.>0,X
ThirdsectionTaylorformula
3.seekfunctionf(x)=
X3orderTaylorformulawithLagrangeremainderbypowerexpansionof(X.,4)
357
.
Solution:
F'(x)=1
That'f'(x)=1X2,f'''(x)=3X2,f(4)(x)=.15x2
Two
X4816
F(4)=2,f'(4)=1
F','(4)1.
F''(4)=3
432256
112135.74
ByTaylorformula:
X=2+(X.4).(X.+4)(X.4).2(X.4).
464512384
(Eisbetweenavaluebetween4andx)
4.thenorderTaylorformulawithPenanoremainderforthepowerexpansionofthefunctionf(x)=lnXby(X.,2)
Solution:
becauseFN()=(-1)n
Xn,f()
(2)(-1)n
2n
)!
()x(-1)(-1)=NN=(-1)n(-1)
Therefore:
N
F''
(2)2F()
(2)(X-2)n
N
LN,xf=,F',X(X-2+...+...
(2)+
(2)(-2)+)+o[(-2)]x
2n!
One
)
13)2(-1)n-11
Nx)NXn
=ln2+(X-2-(X-2+...+)(-2,+o[(-2)]
22N2
5.seekfunctionf(x)=1
3orderTaylorformulawithLagrangeremainderbypowerexpansionof(x+1)
X
Solution:
becauseFN()==(-1)
Nn+1
!
FN.
()xn()(-1)=-n,so...
X
F''(-1)(x+1)2,f'''(-1)(x+1)3
()=(-1)+f'x)
FXf(-1)(+1+++++)
23!
Nnn+
(
(1))
(f)(-1)(x+1)fn.
N!
+
(n+1)!
(x+1)+1,hasavaluebetween.1andX
N+1
(=)
One
X
=-[1+(+1)+(+1)x2+..Xn(-1)n+1(x)
.
+1
N+2
)avaluebetween
FX+X(+1)]+is.1andX.
10.usetheTaylorformulatofindthelimit
Two
(1)LIM(3,x3+3,X2,.4,x4,.2x3)
X++
Thirty-four
Jie:
maket=
One
Theoriginal=lim+
1+3t.1.2t
XT-0t
Because31+=+X113
Xox+(),41+=+X114
Xox+(),
So313+t==+(),4,.T=.,1,t,ot,
1tot121+()
Two
Thirteen
T(+++)]
Primary=lim
[1T0()][1..
Two
Tot
=lim2
Tot+()
=3
T,0+,t,t,0+,t,
Two
.x
(2)LimcosXe2
.
X=0,x2[x+ln(1,.X)]
24x2242
解:
cosx=x
2.
+X
4.
+,and(4).
2=x
2
+X
8
Ox+4),LN(1x)=XX
2
Ox+2)
1(OX1..(
24244
[Ox(X+X+4.X+X+(4xOX4
)][1])+(OX).
原极限=lim2!
4.
Xxx
2
)]
8=lim
.
12
Ox
=1
X→02x22x→0x4
46[..Ox((++)
22
122
1+1+XX
(3)lim2
2x→0(COSx.ex)x2
解:
1+x2+X+1=.3xOx(4.X)=cosx+EX2OX2)=++x22
1(OX(242)
(1),
24.2
WithoutX2=x2
1234412
所以原极限=lim
1+
2
X+1.
2
X.
4.
X+o(X)
=1
X→0[1.1+(x2)2or1or2+12]X2(x2)
X..X
2
第四节函数的单调性与曲线的凹凸性
3.(4)=ln(1+X+x2)
1+X
2
解:
y=
1+
1
X2
+
+
XX
=
1+
1
X2
∴函数在(>0,,Life.
LIFE+)内单调递增.
Nx
(7)(N>0,x≥0Yxe)
=
3
'..1.XnXn.
解:
=NX.XEandE=1x(.),(N>0,x≥0)
XENX
当x∈(0,n)时,and'>0,当x∈(n+时LIFE),and'<0,
∴函数在[0,n][n+内单调递增,在,内单调递减LIFE).
(3)当0x4.
No
2
Withoutsuchx>X+2
<<时,X
令f解:
x=sin(X)+(0,sox.2x,x∈
No,∴fx)=cosx+2sec
X.2('),
2
F''(X)=x+2.2secxtanx=sinX(32secX.1x∈(0)>0,,
D).
2
因此f'(x在(0)
上单调递增故当x∈(0,2),
2)时,FX>F(0)()
ππ
(')'=0,从而fx
在(0,
No
(2)上单调递增,即)>F(0)=0,sox.2x>亦即sinx+(0,0,x∈
No
2
FX),所以
No
Withoutsuchx>X+2;xx∈(0,).
2
(4)当0X
No
2
So>+13
X3
<<时,XX
解:
记(XX)=x3so(0,1,x∈
D)
FX.
32
'
(1)=SEC2xX2=2=2(suchasXXXX(XX))
FX..+
由'=(XX)=X.1=SEC2as2x>0,X)=知gxso内单调递增,XX
.在GX(0,
().()
So即gx(XX)=g(0)
>=0.
故fx'(>0),(0,x∈
No,从而fx在(0)
(d)内单调增加,因此fx)>F(0),(0,x∈
No
()).
222
即当0X
No
2
Soxx13
3>0,从而tanxx13
(0,3,x∈
No
2
).
时>+<(5)当x时>4,2>x2.
2ln42Ln2and
解:
取f(TLN)T=2.2ln,∈(4)'(X)=ln2=.>>0,.
TT,FT
T2x24
故当x时>4,F()x单调递增,从而fx>F(4)
()=0,即
.2lnXLN2x>0,亦即2xX2(x>>4).
5.讨论方程lnx=ax(其中a>0)有几个实根?
解:
取函数()=ln+Xa,∈LIFE),FX()=1.,得驻点x=1,
FX.XX',
X
4
当0<<1
时,FX>0,因此函数x在(0,1
'(X)F(内单调增加));
AA
1<<'Life
XF()1
当x+时,F()<0,因此函数x在(LIFE+)内单调减少.
从而f()为最大值又limFX,FX=lim=.∞,.∞,故
1+()()(AA)
Ax→0x→+life
111
..
当f..=ln(1==0,即a时,曲线y=lnxAX与x轴仅有一个交点,这时原方程
..Aae
有惟一实根.
当f1=ln1.即0<<1>0,
时,曲线yLNx=Ax与x轴有两个交点,这时原方
To1.
Toe
..
程有两个实根.
当f1=ln1<0,即a>1.
时,曲线yLNx=Ax与x轴没有交点,这时原方程没有
1.
..Aae
实根.
8.(3)and(4xx1)
=++E
解:
y'=4(X+1)3former,and''(x1)=1220,formerLife.
++>,
Therefore,thecurveisconcavein(++),andthecurveisnot
Thereisaninflectionpoint
(4)y=ln(X2,+1);
'2x'2(x21)2x.2x2(x1)
The'x'.+1).
Solution:
y=
X2+1;y=
(X2,+1)2=
(X2,+1)2
Makey=0,getx1=.1,X2=1.
WhenH-infinityx.1,y'<0',thecurveisin(,
Theinteriorofaninfinity.1]isconvex,
When1X1y'>0',thecurveisconcavein[-1,1];
When,
When1x+,theY'<0,thecurvein[1,++)isconvex,andthecurvehastwoinflectionpoints,respectively
Theinfinity
(1,LN2),(1,LN2)
.
11.askawhyBisworth,thepoint(1,3)istheinflectionpointofthecurvey,=ax3,+bx2
Solution:
y'=3ax2,+2bx,y',=6ax,+2b,
Thea+b=3,6a:
+2b=0,:
a=.
Twenty-three
B=92
12.trytodeterminethecurvesy,=ax3,+bx2,+cx,+dina,B,C,D,sothatthecurveatx=.2hasahorizontaltangent,
Five
(1,.10)istheinflectionpoint,andthepoint(.2,44)isonthecurve
Solution:
y',=3ax2,+2bx,+c,y',=6ax,+2b
Thus12a,.4b,+c,=0,6a,+2b,=0,a,+b,+c,+d,=.10,.8a,+4b,,.2c,+d,=44,
Geta,=1,B,=.3,C,=.24,D,=16
14.y=f(x)hasacontinuousderivativeoforderthreeinaneighborhoodofX=x0,iff'(x0)=0,f'''(x0)=0,test
Askwhether(x0,f(x0))isaninflectionpoint?
Why?
Solution:
'f''(x0)=0maybe'f''(x0)>0,and'f'(x),thedelta>0,whenx(x0.X0,Delta,delta)when
'f'(x)>0,thef''(x)increasedsothatwhenx(x0.Delta,x0)atf'<0'(x),whichisin(x0.6,x0).
Y=f(x)graphisconvex,whenx(x0,x0+8)when'f'(x)>0in(x0,x0+delta)onY=f(x).
Thegraphisconcave,so(x0,f(x0))istheinflectionpoint
Theextremevalueandthemaximumvalueoffifthfunctions
3x2+4x+4
1.(6)y=;
X2x1
++
(++
Solution:
y==(6x,+4)(X2,x)
(x
1)
Two
.
X
(2x
1)
+
21)(3x2,+4x,+4)=
(x
.
Two
Xx
X
+2)
1)2
Makey'=0getstuck
++++
X1,=.2,X2,=0.
Whenx2y'<0,sothefunctionismonotonicallyreducedin(=.2]);when2,x,0,y'>
2.<<.,,0,
Itfunctionsin[2,0.increaseswhen0Itcanbeseen
(2)8
Forminimumvalues,y(0)=4
Y.
Three
Formaximumvalue
One
(9)y=.32(x,+1)3
Solution:
whenx=.1,y'=.2.1
Thefunctionisundefinedwhen2/3,<0,andX=.1,sothefunctionisknown
3(x,+1)
Thereisnoextremumin((=+++))
Theinteriorismonotonicallyreduced,sothefunctionis(
(10)YXTan
X.
Solution:
thefunctiongivenbyY=+1sec2>0ismonotonicallyincreasedin(+++),sothatthefunctionisinfinitein(++)
Six
Value.
2.testsprovethatifthefunctiony,=ax3,+bx2,+cx,+dsatisfytheconditionB2,.3ac,<0,thenthisfunctionhasnoextremum,
License:
y'=3ax2+c.+2bxB2.3ac<0bya=0,C=0.ytwo'isthreetype,
==
(2)B2.,a,.C,=4(B2.AC)
4(3)3<0.
Whena>0,theimageof'y'isopenedupandabovetheXaxis,soy'>0',sothegivenfunctionisin(+++)
Monotonicallyincreasing,whena<0,theimageof`y'isopeneddownwardandbelowtheXaxis,soy'<0',sothatthegivenfunctionisin
(B23<0issetupandthegivenfunctionismonotonein(++++)
TheinterioroftheboundisreducedmonotonicallysothattheconditionisAC
Functionin,
Thereisnoextremuminthe+++)
3.,askwhyavalue,functionf(),=a,sin,x,+1,sin,3
PI
X
Three
X=x=
Three
Whereistheextremevalue?
Isitamaximumor?
Minimumvalue,andfindtheextremevalue
.
Solution:
F=acosx+cos3'(x),functionextremumatx=f'..=.
Xpi
Three
PI30
.
Acos
PI
Three
+cosPI=0,soa=2.
PiPi
"(=.2sin)x.3sin3,f2Sin.3sin3<0..=.PI,
FXX..
Thirty-three
.
PiPi1
.
Therefore,f..=2sin+sinpi=3ismaximum
..333
5.y=2x3.6x2.18xaskedthefunction.7(x=1~4)getsthemaximumvaluewhere?
Andcalculateitsmaximumvalue.
Solution:
thefunctionisderivableon[1,4],andy'=6x2,.12,x,.18,=6(x,+1)(x,.3)
TheY'=0,x1=.1(left)tothestagnationpoint,X2=3,havefunctiontoobtainthemaximumvalueatx=1,andthemaximumvalueis
Y|x1=.29
=
9.tobuildacylindricaltank,volumeV,theirradiusisequaltothenumberofRandhighsurfaceareatomaketheminimumatbottom