課件mth019微積分無窮小以及階

SupplementaryInfinitesimals yOrderof（無窮小以及無窮小的階(參閱中 Ch1.5,pp.32-TheconceptofInfinitesimal無窮小Def.Ifafunctionf(x)fi0asxfic(xfi￥),thenf(x)iscalledaninfinitesimal(infini ysmall)asxfic(xfi￥).

x2,sinx,tanx,1-cosxareallinfinitesimalsasxfix-1isaninfinitesimalasxfi =1isaninfinitesimalasnfi￥.

Anynonzeroconstant,howeversmallitis,say,10-100,cannotberegardedasaninfinitesimal.TheoremTheoremxfi(xfi￥limf(x)=Lifandonlyxfi(xfi￥lim(f(x)-L)=Thatisto f(x)-Lisaninfinitesimalasxfic(xfi￥Forexample,sincelimsinx= sinx-1isanxfi asxfiTheorem2(1)Thesumoffinitenumberofinfinitesimals

Theproductofaninfinitesimalandaboundedfunctionisaninfinitesimal.

Findlim(x 1xxfi 1xSolutionsinx

isaboundedfunction

￡1(x?andxisaninfinitesimalasxfi0,sobytheaboveweconcludethatxsinx

isaninfinitesimalasxfiThat lim(xsin1)=xfi lima(x)=0xfi(xfi￥xfi(xfi￥ limb(x)=0,thenwesaythatb(x)isanxfi(xfi￥ofhigherorderthana(x),denotedbyb(x)=o(a(x))asxfi(xfi￥a(x limb(x)=1,thenwesaythatb(x)anda(x)arexfi(xfi￥infinitesimals,denotedbyb(x)~a(x),or,)b(x)asxfi(xfi￥a(xx2xItiseasytocomputethat

= so

isanxfi0ofhigherorderthan2x,denotedbyx2=o(2x).Sincelim1-cosx= 1-cosxisaninfinitesimalofhigherxfi thanx,denotedby1-cosx=o(Severalimportantequivalent

1Sol:(1)xfi

xfiHence,sx0,

= tfi0Sol:Letu=ex-1,x=ln(1+u),xfi0ufi

xfi

e-1= x x

AsAsxfi0ex-1~Wehavefoundthatlimsinx=1,limtanxxfi xfi limarcsinx=1,limarctanx=1,limln(1+x)=1xfi

xfi

xfi andlim1-cosx=1,sowegetthexfi

x2/equivalentinfinitesimalsasxfi

ex-x~sinx~tanx~arcsinx~arctanx~ln(1+xex-1-cosx 2WeaddtwousefulequivalentinfinitesimalsintheWeaddtwousefulequivalentinfinitesimalsinthe(x/(x/ 1+x

-1~ (xfi1+11+1+(x/ 1+x1+1+1+1+

-1=lim

+1)=

xfi x/2 1+ 1+x+12xfi1+-1~1+2

xfi(xfi

ufiIngeneral,wecanshowthefollowingAsxfi0,(1+x)a-1~ax.For (1+x)3-1~3x(xfi

-1~x(xfi3131+Equivalentinfinitesimalsasxfiex-x~sinx~tanx~arcsinx~arctanx~ln(1+x)ex-1-cosx 21+11+12ax+xaaxAclevermethodforfinding Theorem

Supposethata1(x),a2(x),b1(x),b2(x)areinfinitesimalsanda1(x)~a2(x),b1(x)~b2(x)asxfic(xfi￥ limb2(x)exists.xfi(xfi￥

a2(limb1(x)=limb2(x)

xfi(xfi￥

a1(

xfi(xfi￥

a2( limb1(x)=lim a2(x)=limb1(x)limb2(x)lima xfi

xfi

a2

xfi

xfi

a2

xfi

a1(x)=1limb2(x)1=limb2(x)xfica2 xfica2ExamplesFindthefollowinglimsin

limln(1+xfi0sin tan

xfi

ex-xfi0

+

xfi

3sinx2

31-x-

xfi 2

xfi lnlim(ex-xfilimsin5x Sincesin5x~5x,sin3x~3xas 0sinwe limsin5x=lim5x=5. 0sin

0 limln(1+x)=limx=1 ex- 0limtanx Sincetanx~x,as 0x3+3xwe

=

=1

0x3+ 0x3ln(1-x2 = =-

0x2+3 3sin 0 (4)xfi33xx

.31-31-x- x

-1~

3

asxfix=-lim1-x-1=lim 1x=-xfi

xfi limarcsin(1-x).Sincearcsin(1-x)~1- lnx=ln[1+(x-1)]~x-1,as 1,we2

limarcsin(1-x)=lim1-x=- 1x-lim(ex- Letu=2.Thenx=2, 0as ￥. 2

=

u=lim(ex-1)x=

0Examplelimtanx-sinxfi Warning:Ifthenumeratorordenominatoristhesumseveralterms,theningeneral,anyoneofthetermscannotbere cedbyitsequivalentinfinitesimal.SuchsubstitutionExample

tanx-sinx=limsinx(1-cos

xfi

xfi

x3cos

=limsin 1-cos =111=1

xfi0

cosx Butifyou cetanxandsinxbytheirinfinitesimalyouwillgetthewrongres