设f(x,y)在(0,0)处连续,,则( ) A.(A) f(x,y)在(0,0)处不可偏导B.(B) f(x,y)在(0,0)处可偏导但不可微C.(C) f'x(0,0)=f'y(0,0)=4且f(x,y)在(0,0)处可微分D.(D) f'x(0,0)=f'y(0,0)=0且f(x,y)在(0,0)处可微分 相关知识点: 试题来源:...
解析:(特殊函数法) 由于即当(x,y)→(0,0)时f(x,y)与一(x2+y2)是等价无穷小.取f(x,y)=-x2-y2,则f(x,y)满足题目条件. f’x=-2x, f’y=-2y, f’’xx=-2,f’’xy=0,f’’yy=-2、 显然f’x(0,0)=0,排除A、 B. 在驻点(0,0)处, A=-2, B=0, C=-2, B2...
设函数f(x,y)在点(0,0)连续,下列命题正确的是(A.若lim_(x→0)(f(x,y)-xy)/(|x|+|y|) 存在,则函数f(x,y)在点(0,0)可微B.
【题目】5.设函数f(x,y)在(0,0)处连续,那么下列命题正确的是()(A)若极限lim_(x→0)(f(x,y))/(|x|+|y|) 存在,则f(x,y)在(0,0)处可微(B)若极限lim_(x→0)(f(x,y))/(x^2+y^2) 存在,则f(x,y)在(0,0)处可微(C)若f(x,y)在(0,0)处可微,则极限lim_(x→0)(f(x...
因此△ z=(∂ f(0,0))(∂ x)△ x+(∂ f(0,0))(∂ y)△ y+o(√ (△ x^2+△ y^2)).从而f(x,y)在点(0,0)可微. 并且df(x,y)|_((0,0))=(∂ f(0,0))(∂ x)△ x+(∂ f(0,0))(∂ y)△ y=0. ...
123【答案】C【分析】(lim)/((x+y)→(0,0))(∫(x+3))/(|x|+1) (f(x,y))/((x-1,+∞)=由lim1,及(x,y)→(0,0)lim_(x→m)1/x (x,0|x|+|y|=0 知limf(x,y)=0,又f(x,y)在点(0,0)处连续,则f(0,0)=0.(x,y)→(0,0)由(xlim_((x))lin(0,0)+(f(x...
【解析】(1)由于当(x,y)→(0,0)时,-|||-In cos./r2+y2~cos√r2+y2-1~,-|||-x2+y2-|||-2-|||-代入原极限式可得-|||-lim-|||-f(,y)+11-|||-zy→0x2+y2-|||-2-|||-可知limf(x,y)=-1.-|||-7-|||-由于f(x,y)在点(0,0)处连续,因此f(0,0)=-1....
【题目】设f(x,y)在点(0,0)处连续,且lim_((x,y)→(0,0))(f(x,y)-1)/(e^(x^2+y^2-1))=2 ,求和 (∂f(0,0))/(∂
设函数f(x,y)在点(0,0)处连续,且,则A.fx(0,0)存在且不为零.B.fx(0,0)不存在.C.f(x,y)在点(0,0)处取得极小值.D.f(x,y)在点(
【题目】设f(x,y)在点(0,0)处连续,且lim_(x→)=f(t_1y_0)g_2=2+y^2-1 求f(0,和0,0y,并讨论f(,y)在(0,0)处是否可微,若