1.极限lim』—(提示：令y=k2x2)(B)不存在(C)等于1(D)存在且不等于0或2、设函数f(x,y).1.1xsm—+ysin—0,即有x2+y2<1.故所求函数的定义域为D二{(x,y)1x2+y2<1}，图形为图⑵要使函数z二ln(x+y)有意义，必须有x+y>0.故所有函数的定义域为D={(x,y)lx+y>0},图形为图(3要使函数z=有意义，必须有ln(x+y)丰0,即x+y>0且x+y丰1.ln(x+y)故该函数的定义域为D={(x,y)1x+y>0,x+y丰1},图形为图⑷要使函数z=ln(xy-1)有意义，必须有xy-1>0.故该函数的定义域为D={(x,y)Ixy>1}，图形为图图图2、求极限limysin2xxT0\,xy+1—1yT0解:lim42L=limy沁xQE+D=4xT0xT0yT0yT0x3、求极卩限lim_1sin(xy).xT0x3y2yT0解:原式=limxTox3y2(1+Jx2y+1)yT0sin(xy)=limxT01+px2y+1yT0sin(xy)x-1~xye4、求极限解:lim=lim秤(4)=-8xT0+xyxT0yT0yT05、设u=xsiny+yco...
