§2二元函數(shù)極限1、試求下列極限(1)lim(2(0,0)X2+y2分析：對(duì)趨近于原點(diǎn)且含有X2+產(chǎn)類的極限問題，采用極坐標(biāo)變換較為簡(jiǎn)單。解：(1)對(duì)函數(shù)自變量作極坐標(biāo)變換：x=rcos0,y=rsin0這時(shí)(x,y)f(0,0)orf0由于|f(x,y)-由于|f(x,y)-0X2y2X2+y2=r2sin20cos20<r2因此，對(duì)Ve>0，取3=個(gè)￡.當(dāng)0<r=Jx2+y2因此，就有|f(x,y)-0|<r2<slim上，二0(x,y)f(0,0)x2+y21+x2+y2(x,y)f(0,0)x2+y2解：令x=rcos0,y=rsin0i 1+x2+y2 1+r2lim - =lim =+8(x,y)f(0,0)x2+y2 rf0r2⑶limJ+y2(x,y)f(0,0)\：1+x2+y2—1分析：可以先分母有理化，再使用極坐標(biāo)變化。解：令x=rcos0,y=rsin0limx2+y!——=lim "1+x2+y2+1)(x,y)f(0,0)1+x2+y2—1(x,y)f(0,0)=lim(%1+r2+1)=2rf0

xy+1lim (一(0,0)x4+y4解：令x=rcos9,y=rsin0,(x,y)-0or—0不妨限制0<r<1fi勺，則VM>0,當(dāng)0<r<min{1,4;-'-}時(shí)42Msin40+cos40=4(3+cos40)>—>M

4r4xy+1 r2sin0cos0+14+2r2>—>M

4r4x4+y4r4(cos40+sin40)"(3+cos40)xy+1lim =+8(…(a。)x4+y4(5)lim(x,y)-(1,2)12x-yx2+y2x2+y2解：對(duì)Ve>0,當(dāng)鄧f,|y|<f時(shí)有(x+y)sin一1一x2+y2<|x|+|y|<f解：對(duì)VM>0,當(dāng)|x-1|<擊,卜-2|<2M且(x,y)w(1,2)時(shí)有-1+12-y|2x-y |2(x-1+12-y|1\lim =8(x,y)—(0,0)2x-y「. lim(x+y)sin——i——=0(x,y)-(0,0) x2+y2sin(x2+y2)lim (x,y)—(0,0) x2+y2解：令令x=rcos0,y=rsin0,(x,y)—0or—0sin(x2+y2) sinr2lim =lim =1(x,y)—(0,0) x2+y2 r—0r22、討論下列函數(shù)在點(diǎn)(0,0)處的重極限與累次極限(1)f(x,y)=(1)f(x,y)=y2x2+y2解：lim—y2—=lim上網(wǎng)W=limsin20,重極限不存在(x,y)f(0,0)X2+y2r-0 r2 r-0而limlim—y2—=lim0=0,limlim—y2—=lim^2=1xf0y—0X2+y2 xf0 yf0x-0X2+y2y—0y2(2)f(x,y)=(x+y)sin—sin—xy解：對(duì)V￡>0,38='，當(dāng)|x|二B,|y|<B,(x,y)中0時(shí)TOC\o"1-5"\h\z2 1 1(x+y)sin—sin—<|x|+|y|<28-8xy「. lim(x+y)sin—sin-=0(x,y)f(0,0) xy而limlim(x+y)sin—sin—與limlim(x+y)sin—sin」均不存在xf0yf0 xyyf0xf0 xy(3)f(x,y)-a、x2y2+(x—y)2解：lim—二--lim—比一-11xf0x2y2+(x—y)2 xf0k2x2+(1-k)2 [0y-kx所以重極限不存在而limlimxf0yf0x2y2x2y2+(x-y)2-lim0-0,limlim xf0 yf0xf0x2y2+(x-y)2-lim0-0yf0(4)f(x,y)-x3+y3x2+yx3+y3 x3+x6TOC\o"1-5"\h\z解：lim -=lim 0..xf0x2+y xf0x2+x2?y-x2「 x3+y3 x3+x9-3x8+3x7-x64lim -二lim 1xf0 x2+y xf0 x3y-x3-x2所以重極限不存在x3+y3 x3+y3而limlim ——limx—0,limlim(x+y) ——limy2-0xf0yf0x2+yxf0 yf0xf0 x2+yyf0(5)f(x,y)-ysin1x解：limlimysin—=lim0=0,limlimysin，不存在xf0yf0 xxf0 yf0xf0 x

二1ysrn-x對(duì)V￡>0取b=￡,當(dāng)|x|<B,|y|<B(x,y)二1ysrn-x<|y|<B="nlimysin—=0(x,y)-(0,0) x(6)f(x,y)=X2y2X3+y3解：limX2y2=lim^^二0..x-0X3+y3 x-02x3y=x「 x2y2 x2(x4—2x3+x2) 1lim 二lim 二-x-0x3+y3 x-0x3+(x6—3x5+3x4—x3)3y=x2—x所以重極限不存在而limlim'、殍=lim0=0,limlim'、殍=lim0=0x—0y—0x3+y3x—0 y—0x—0x3+y3y—0,_、°, 、ex—ey(7)f(x,y)=--sinxyex—ey0解：lim =lim=0x—0sinxy x—0sinx2y=xex—ey ex[1—e-x2] x2 .lim =lim =lim =1x—0sinxy x—0sin(x2—x3) x—0x2—x3y=x—x2所以重極限不存在而limlime^~ey與limlim(x+y)e^~ey均不存在x—0y—0sinxy y—0x—0 sinxy3、證明：若1 limf(x,y)存在且等于A；2y在b的某領(lǐng)域內(nèi)有o o(x,y)—(a,b)limf(x,y)=q)(y)則limlimf(x,y)=Ax—ay—bxx—a證：由題設(shè)，玄］>0當(dāng)|y-b|<q證：由題設(shè)，玄］>0當(dāng)|y-b|<q時(shí),有l(wèi)imf(x,y)押(y)n對(duì)VQ0T">0x——a當(dāng)。^^^1,^-^以]時(shí)，有|f(x,y)-q)(y)|<又因?yàn)閘imf(x,y)=A,所以我>0,當(dāng)0<(x,y)—(a,b)有|f(x,y)-A|<8x-a|<52,0<|y-b]<b2及(x,y)手(a,b)時(shí)取b=min仿,5',5取b=min仿,5',5},則對(duì)該8>0，

1 1 2limf(x,y)—A=取y)—A|<^(y)—A|+|f(x,y)-A|<28x—anlimlimf(x,y)=Ay—bx—a4、試用8-5定義證明lim x2y =0(x,y)—(0,0)x2+y2

證法1：對(duì)Ve>0,取b=￡,々x=rcos9,y=rsin9,當(dāng)0<r=q'x2+y2<3時(shí)TOC\o"1-5"\h\z即p(P,P)=\，奴2+y2即時(shí)，(P(x,y),P(0,0))有0 0f(x,y)=X2y=rsin0cos20<r<3=8X2+y2\o"CurrentDocument"x2y 八nlim =0(…(0,0)x2+y2證法2:所以對(duì)V8>0,取3證法2:所以對(duì)V8>0,取3=8x2y

x2+y2y)||x<3,|yx2y

x2+y2y)||x<3,|y<3,(x,y)豐(0,0)}xy

x2+y21<-x<8,2x2y -以lim———=0。(x,y)f(0,0)x2+y25、敘述并證明：二元函數(shù)的唯一性定理，局部有界性定理與局部保號(hào)定理.(1)二元函數(shù)極限的唯一性定理：若limf(x,y)存在，(x,y)f(a,b)則此極限是唯一的.證：設(shè)有l(wèi)imf(x,y)=A,limf(x,y)=B(x,y)f(a,b) (x,y)f(a,b)n對(duì)V8>0,33,3>0,記P(a,b),1 2 0當(dāng)P￡5(,3?時(shí),|f(x,y)-A|<8,當(dāng)P￡U0(P0,3J時(shí),|f(x,y)-B|<8,取3=min{3,3},則當(dāng)P￡U0(P,3)時(shí)，上兩式同時(shí)成立n|A-B|<|A一f(x,y)|+|f(x,y)-B|由的任意性，推得這就證明了唯一性.⑵二元函數(shù)的局部有界性定理：若limf(x,y)存在，(x,y)f(a,b)則f在P(a,b)的某個(gè)空心領(lǐng)域U0(P)內(nèi)有界.0 0證：設(shè)limf(x,y)=A，取8=1，則33>0，對(duì)VPeUo(P，3)，(x,y)—(a,b) 0有|f(x,y)-A|<1n|f(x,y)<f(x,y)-A|+|A|<1+|A|nf(x,y)在U0(P，3)內(nèi)有界.0⑶二元函數(shù)的局部保號(hào)定理：若limf(x,y)=A>0(或<0),則對(duì)(x,y)f(a,b)任何正數(shù)r<A(或r<-A)，存在Uo(P),使得對(duì)VPeUo(P,3),有f(x,y)>r>0(或f(x,y)<-r<0)證：設(shè)A>0,對(duì)Vr￡(0,人)，取￡二人-上則孤>0.使得對(duì)VPeUo(P，3)有f(x,y)>A-e=r對(duì)A<0的情形可類似地證明，所以結(jié)論成立.6、試寫出下列類型極限的精確定義lim f(x,y)=A(x,y)—(+s,+s)解：若對(duì)Ve>0，mM>0，當(dāng)x>M,y>M時(shí),有|f(x,y)-A|<s則稱(x,y)f(+8,+s),f(x,y)=Alimf(x,y)=A(x,y)f(0,+s)解：若對(duì)Ve>0,孤,M>0,當(dāng)0<|x|<3且y>M時(shí)，有|f(x,y)-A|<s則稱當(dāng)(x,y)f(0,+6,f(x,y)以A為極限，記為limf(x,y)=A(x,y)f(0,+s)7、試求下列極限(1)lim三±匕(x,y)f(+8,+8)x4+y4提示：采用極坐標(biāo)變換再放不等式。解：對(duì)Ve>0，33=4，當(dāng)x>M,y>M時(shí)，記x=rcos9,y=rsinex2+y2x4+yx2+y2x4+y41r2(cos40+sin40)4

r2(3+cos40)4<—2r22x2+y22<—2M21 =8M2limx2+y2=0(x,y)(x,y)f(+8,+s)x4+y4(2)lim (x2+y2)e-(x+y)(x,y)—(+<?,(2)lim (x2+y2)e-(x+y)(x,y)—(+<?,+<?)解：令x=rcos9,y=rsin9,則(x,y)T(+s,+s)等價(jià)于0<0<—,rt+s.2因?yàn)?x2+y2)e-(x+y)=r2e-r(sine+cose)=r2e-2rsin(°+/而—<e+—<3—^亙<sin(e+—)<14 44 2 4r2 —c、r2故——<r2e-2sin(4+e)<—.e'2r err2 r2但是1加——=lim =0r—8e2r r—8erlim(x2+y2)e-(x+y)=limr2e-2rsin(e+4)=0(x,y)t(+8,+8)rt+80<e<―2⑶lim (1+—)xsiny(x,y)t(+s,+s) xy分析：變形利用重要極限的結(jié)論。解：5(1+)xsiny=--——ln(1+——)xyxyyxy當(dāng)(x,y)T(+8,+8)時(shí),乂丫T+8,所以lim ln(1+—J—)xsiny=lim ———ln(1+——)xy=0.1=0(x,y)t(+8,+8) xy(x,y)t(+8,+8)y Xy1x lim ln(1+工)xsinylim(1+—)xsiny—e(x,y)t(+8,+8) xy=e0=1(x,y)t(+8,+8)xy1x2(4)lim(1+-)x+y(x,y)t(+8,0) x分析：變形利用重要極限的結(jié)論。解：因?yàn)閘im(x,y)T(+8,0)x+y—=1推得8、試作一函數(shù)f(x,y)使當(dāng)xT+s,yT+s時(shí)(1)兩個(gè)累次極限存在而重極限不存在解：取f(x,y)二x2±I2X2-y2則limlimx+1y=lim(T)=Tlimlimlim1=1xT+8yT+8X2-y2xT+8 yT+8xT+8xT+8由定理16.6推論2知重極限不存在⑵兩個(gè)累次極限不存在，而重極限存在.解：取f(x,y)=(L+-)sinxsinyxy因?yàn)閘im工sinxsiny不存在，但lim』sinxsiny=0xT+8y xT+8Xnlimlimf(x,y)不存在，同理limlimf(x,y)也不存在.xT+8yT+8 yT+8xT+8_, 2,但對(duì)Ve>0,3M=—,當(dāng)x>M,y>M時(shí),有8f(x,y)=(—+—)sinxsinyxynlim(—+—)sinxsiny=0(x,y)t(+?,+?)xy⑶重極限與累次極限均不存在.解：取f(x,y)=(x2+y2)sin(x2+y2)則設(shè)函數(shù)當(dāng)xT+8,yT+8時(shí)，重極限與累次極限均不存在.(4)重極限與一個(gè)累次極限存在，另一個(gè)累次極限不存在.解：mf(x,y)解：mf(x,y)=Isinx,則lim(x,y)t(+8,+8)1—sinx=0ylimlim1.lim0=0,,口limlim1.丁十十sinx= 但 $19乂不存在.xT+8yT+8y xT+8 yT+8xT+8y9、證明定理16.5及其推論3.D的任一子集(1)定理16.5：limf(P)=AD的任一子集只要P0是E的聚點(diǎn)，就有l(wèi)imf(P)=A.PTP0P6E分析：必要性顯然，充分性采用反證法。證：“必要性”，若lim戈戶二人且￡uD,P是E的聚點(diǎn)P2P0n對(duì)V8>0,35>0當(dāng)PcU0(,5)nD,就有f(p)-A|<8WU0(P,5)pEw°nUo(P,5)nEuUo(P,5)nD故也有|f(x,y)-A|<8nlimf(P)=A.P—P“充分性”，若對(duì)D的任一子集E,只要P是E的聚點(diǎn)，就有olimf(P)=A.P—P0采用反證法，假如limf(P)wA,即反>0,對(duì)V5>0,P—P0都不cUo(P0,5)nD,使|f(P)-A|>80,取b=1,n=1,2,...,則相應(yīng)得PcUo(P,5)nDnn n 0n使6并互不相同，(iif(P)-A|由此得到￡=但｝uD,因limp(P,P)=0.n—g所以P是E的聚點(diǎn)，但這與應(yīng)有l(wèi)imf(P)=limf(P)矛盾,P—P n—g nPcE所以必