整理多元函數(shù)的極限與連續(xù)習題

1、多元函數(shù)的極限與連續(xù)習題1.用極限定義證明:lim(3x2y)14。x2yi2.討論下列函數(shù)在(0,0)處的兩個累次極限，并討論在該點處的二重極限的存在性。(i)f(x,y)f(x,y)(x3.f(x,y)f(x,y)求極限(i)3x-2xJysin-xlim(x2x0y022y2)*xy;、.1.1y)sin-sin；xy(2)limx0y022xy(3)lim(xx0y0y)sin-2(4)limx0y0sin(x2y2)4.試證明函數(shù)f(x,y)ln(1xy)xyx0、x0在其定義域上是連續(xù)的x01.用極限定義證明：lim(3x22y)14。x2yi因為x2,y1,不妨設|x2|0,|y

2、1|0,有|x2|x24|x2|45,|3x22y14|3x2122y2|3|x2|x2|2|y1|15|x2|2|y1|15|x2|y1|0,要使不等式|3x22y14|15|x2|y1|成立取min一,1,于是300,min,10,30(x,y)：|x2|,|y1|且(x,y)(2,1),有|3x22y14|,即證2.討論下列函數(shù)在（0,0）的存在性。處的兩個累次極限，并討論在該點處的二重極限(Df(x,y)xlimlim丫1,xylimlimx0y0xy二重極限不存在。xy八或lim0,x0xyyxy0x0x.xylim-x0xyy2x1.1(2) f(x,y)(xy)sinsin一；x

3、y110|(xy)sin-sin|x|y|xy可以證明lim(|x|y|)0所以因此同理limf(x,y)0。x0y01.1,.0時，f(x,y)(xy)sinsin一極限不存在,xylimlim(xx0y0、.1.1-y)sinsin一不存在,xylimlim(xy0x0、.1.1y)sin-sin不存在。xy一x3f(x,y)xlimf(x,y)x0yxlim2x30,x當P(x,y)沿著yx2x3趨于（0,0）時有l(wèi)imx0yx23/323x(xx),f(x,y)lim232-1,x0xxxx3所以limf(x,y)不存在;x0y0limlimx0y0f(x,y)0,limlimf(x,

4、y)0。y0x0一、,1f(X,y)ysin一X10lysin一|y|Xlimf(x,y)x0y00,1limlimysin一1-limlimysin一不存在。3.求極限(Dlim(x2x0y0y0x022)Xyx2y2ln(x2y2)|(X2y2)22廣11n(x22)l，,22、又limx04y02-ln(X22、y)limt03ntlim(x(2)limy2)x2X2X2y22y2ylime(x,y)e22xyln(x(0,0)y2)1。limX0V12X2Xy21limX0y0(X22y1)(,1x2-y2d2。1y)siny2，(3) lim(xx0y01.,l(xy)sin2l|x

5、y|,xy而lim(xy)0x0y0lim(xx0y0y)sin工2xy(4)limx0y0sin(x2rcosyrsin(x,y)(0,0)時,r0,limx0y0.z2sin(xy2)2sinr/lim1。r0r21n(|1n(1xy)0|y1n(1xy)xy|x1|y|1n(1xy)xy|2|y|,于是，無論x0,x0,當|x|,|y|時，都有1imf(x,y)0f(0,0)x0y0(2)(3)在(0,y)處。(y0)1當x0時,|f(x,y)f(0,y)|y1n(1xy)xyy|刈)x04.試證明函數(shù)f(x,y)Xx0在其定義域上是連續(xù)的yx0證明：顯然f(x,y)的定義域是xy-1.當x0時，f(x,y)是連續(xù)的，只需證明其作為二元函數(shù)在y軸的每一點上連續(xù)以下分兩種情況討論。(1)在原點(0,0)處f(0,0)=0,當x0時f(x,y)1n(1xy)0y01y1n(1xy)xyy01由于lim1n(1xy)xy1x0y01不妨設11n(1xy)xy1|1,1|1n(1xy)xy|2,從而0,取3,當0|x|,0|y|時,1|y(ln(1xy)xy1)(yy)|i|y|ln(1xy)xy1|yy|當x=0時，|f(x,y)f(0,y)|yy|,1注意到，當y。時limln