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1、 1 Answer to section 12.3()()()222222222,0,000 Let ,then sin sinsin limlimlim1.x yrtxyrxyrtxyrt+=+=+7.()()()()222242222,0,0,0,0 Take ,then limlim 1 Since this limit depends on ,thus,the limit does not exist.x yx yykxxykxkxyxk xkk=+16.()()()()2,0,022,0,00 Along the-axis(0):then lim0;01 Along :lim 22
2、Hence,the limit does not exist because for some points near the origin(,)is gettx yx yxyxxyxxf x y=+=35.1ing closer to 0,but for others it is getting closer to.220232200 Along the-axis(0):then lim0;01 Along :lim 2 Hence,the limit does not exist.xxxyxxxyxxx=+=+36.()()()()()()()()()2324222240002242440
3、042242,0,0 a lim limlim01 b lim lim2 c lim does not exist.xxxxxx yxmxmxmxxm xxmxmxxxxxxxxx yxy=+=+37.2()()()()()()()()()022222222000222200,0,a 0,lim0 limlim.,0,0 b ,0lim limxhhhyhhfh yfyfyhhyhyy hyhyyhhyf xhf xfxhxhxhxh+=+=+=+=42.()()()()()()()()2222000000lim.0,00,0 c 0,0lim0 lim1.0,00,0 d 0,0lim0 l
4、im1.Thyyyxhhxxxyhhxhxxhxhfhffhhhfhffhhh=+=+=()()herefore,0,0 0,0.xyyxff ()()()()333,0,0,0 If we approach the point 0,0,0 along a straight path from the point ,we have1 lim0,3 Since the limit does not equal to (x x xx x xx x xxxxf=+47.0,0,0),then the function is not continuous at the point(0,0,0).3()()()()22,0,00,0,0 If we approach the point 0,0,0 along the-,we get0 lim011 0 Since the limit does not equal to (0,0,0),then the function is notxx axisxxf+=+48.continuous at the point(0,0,0).