,000lim)0,0(0xfxx,000lim)0,0(0xfyy])0,0()0,0([)]0,0(),([lim00yfxffyxfyxyx2200)()(limyxyxyx,00)(1sin)(lim)0,0(220xxxfxx,00)(1sin)(lim)0,0(220yyyfyy])0,0()0,0([)]0,0(),([lim00yfxffyxfyxyx22222200)()()()(1sin))()((limyxyxyxyx)0,0(),(yx2222221cos21sin2),(yxyxxyxxyxfx,01sin2lim22)0,0(),(yxxyx2222)0,0(),(1cos2limyxyxxyx)0,0()1,0()1,0()1,1()0,0()1,1()1,1(fffffff.211)1(),0()1,(yxff.1.11.1),(yxyxf,0)1,1(,2.2)1,1(,0)1,1(fff)0,0()0,1()0,1()1,1()0,0()1,1(ffffff.101)1()0,(),1(xyff1)0,0()1,1(ff22)0,0(),(22)0,0(),(2222)0,0(),(lim),(lim)(),(limyxyxyxfyxyxyxfyxyxyx1),(lim22)0,0(),(yxyxfyx【解1】直接法0),(lim)0,0(),(yxfyx,0)0,0(f),(yxf)0,0(则,若在点连续,否则不连续。故(A)不正确。【解法2】排除法,,0,10,),(222222yxyxyxyxf02),(lim2222)0,0(),(yxyxyxyxfyx,0)0,0(f02),(lim22)0,0(),(yxyxyxfyx【解】知,且则,0]2[)0,0(),(lim22)0,0(),(yxyxfyxfyx)(2)0,0(),(yxfyxf,1)0,0(,2)0,0(yxff【解】验证法:(A)(C)(D)都不满足xxfy)0,(,故应选(B).222yz),(22xydyyz,2xyyz),()2(2xxyydyxyz.12xyyz0),(lim2200yxyxfyx,0)0,0(f,0),(22yxyxf,0)0,0()0,(lim0xfxfx,0)0,0()0,(lim0xfxfxyyxxzddd,)(2122Cyxz【解1】【解2】)(ln)(yfxfxz)()()(yfyfxfyz)(ln)(22yfxfxz,,,yxz2)()()(yfyfxf)()]([)()()(2222yfyfyfyfxfyz)()(ygxfxz)()(ygxfyz)()(22ygxfxz)()(2ygxfyxz)()(22ygxfyz,,,,.2222222)]0()0([)0()0()0()0(gfgfgfyxzyzxz).0()0()0()0(gfgfyxFFxy)(0)(0xy2)(...