微积分a2第1次习题课答案

微积分A(2)第一次习题课参考答案（第三周（1）AB(A

（2）(AB)A （3）AA\证明：（1）aABaAaBaA，那么存在0B(a,AB(a,AB，从而aAB)；若aB（2）aAB，那么0,均有U(a,AB)且U(a,AB)c因此[U(a,A[U(a,B]且U(a,AcBc)，从第二个表达式可以看出，U(a,Ac，U(a,Bc。由于[U(a,A[U(a,B那么U(a,A,U(a,B至少有一个不是空集.不妨设U(a,AaA，进一步可以得到aAB（3）AAAAAAAAA。事实上，aAA，可以得到aAaA。由aA非空，只能U(a,Ac，从而有U(a,A，即aAf:RnRm是续映射，求证：Rm中的开集G，它的原像f1(GxRn|f(xGRn证明:由于f:RnRm 续映射，由定义，有aRn，0,0，dn(xa时，有dmf(x),f(aaf1(G)f(aG。由于GRn00Uf(a),0G00利用连续的定义，00dn(xa0 df(x),f(axU(a,)f(xUf(a),)Gxf 即U(a,0)f1(G)f1(G)Rn中的开集，证毕(x,y

xy(xy)xy1 (x,y

(xy)ln(x2y2) (x,y

x2y

)x20 (x,y

x3；x2（5）

xx 2x

xyxy 1(xy (xy)xy1 (1(xy1))x e2

(x,y (x,y (xy)ln(x2y2)0(x,y

1x2x2

(x,y

x2y

x20令y x3y3lim (x,y)(0,0)x2 x0x2再令yx2 (x,y

x3y3x2

x3(x2x3lim 当(xy00)0| x ||

|

|x

|yx2xy x2xy

x2xy

3x2(x

3y2(y4(11

3|x |yx因此x

2xy 讨论下列函数的累次极限limlimfxylimlimfxy与二重极限limfxyxsin1

y01

x0y

x2y2（1）fx,y=

y xy

x2y2（3）f(x,y) x2 x2y2(xlimlimfxylimlimfxy0，而二重极限limfxyy0x0 x0y0 limfxy0，故limlimfxy0limlimfxy0 y0 x0yx趋于(0,0)limfxy1y0趋于(0,0)xylimfxy0，故limfxyyz

fx,y在R2上连续, x2y2

fxyfR2u2vlimfx,ydu2v 存在QB x,yx2y2d2:fQ x,y

fx,

d:fu,vM显然,fQx,y

fxy0f(x,y)f(x0,y0)f(x,y)f(x0,y)f(x0,y)f(x0,y0fx(,y)xfy(x0,)(x)2(M(x)2(

M(xy) x(x)2(取 时，f(x,y)f(x(x)2(M点处可微，并写出全微分dfx,y)。f(0,0)0，f(xy)(x,y)f(0,0)(xy)(x,因为(xy在点（0,0）处连续，(xy(0,0)(xy（(xy0(xy0fxy)(0,0)xy)(xy)(xy)(xy)(x)2 0，所以f(0,0)x(x)2,df(xy)(0,0)dx(0,0)dyfx(x0y0fy(x，y在点x0y0f(xy在点x0y0处可微.分析:证明函数fx,y)在某点可微的关键,是利用题目条件,将函数改变量ff(x0xy0yf(x0y0表示成AxByx2y2x2y时, 比较是高阶x2yAxByxyx2y20时,00ff(x0x,y0y)f(x0,y0f(x0x,y0y)f(x0x,y0)f(x0x,y0)f(x0,y0

因为fy(x,y)在点(x0,y0)连续，所以，根据一元函数的日微分中值定理，f(x0x,y0y)f(x0x,y0fy(x0x,y0y)y[fy(x0,y0)()] x2y其中， ，()为当0时的无穷小量.又由已知，fx(x0,y0)存在x2ylimf(x0x,y0)f(x0,y0)

f(x,y

f(x0x,y0)f(x0,y0)

f(x,y)其中，(x是当x0

即f(x0x,y0)f(x0,y0)fx(x0,y0)x(x) ffx(x0,y0)xfy(x0,y0)y(x)x()x2y20时,(x和都是无穷小量,所以根据函数在一点可微性定义推出fx,y在点x0,y0处可微.z

x，求dzy 设函数z(x2 f(uzx

f(xyyf(xy

2；x 2

x

，

2(x

2

2z2

0 ( v

1，求df(xy) u

42)dx

22)dy

f f A. B.（8）z ， 解：z 2

C.i D.ixxx 2x x x xx

f(x,y)f(0,x0x0

y x xf(xy在(x0y0点可微,且全微分在点(x0y0fx0fy

0的是（Df(xy在点(x0y0的全增量f

xx2yx2yf(x,y在点x2y

sin(x2y2f(xy在点(x0

)的全增量f(x2y2) x2y(1)fx(x0y0fy(x0y00(2)fx0,y0dfx0,y0 fx0,y0xx2yx2y条件(1)都成立，只有D中条件(2)成立,故选设f(x,y) ,则在(0,0)点( (C)可微 解题思路f(x,y) ,在(0,0)的两个偏导数都等于零.如果f(x,y) ,在x2y可微,则df(0,0)0,从而f(x,y)x2y若zf(x,y)在点P0(x0,y0)处的两个偏导数存在，则（ （A）f(xyP0（B）zf(xy0zf(x0yyy0xx0（C）f(xyP0点的微分为dz

dx

dyz（D）f(x,y)在P0点的梯度为gradf(P0)(x,y 0f(xy在点(x0y0（Cf(xy在点(x0y0f(xy在点(x0y0沿任何方向vf(xy在点(x0y0f(xy在点(x0y01(1exy分别函数f(x,y)

xx

(1 ) x0

f(0y)f(xyx

x

(1

xy)

yex 1(1exy)x0时 f(0,x)

lim

2

1(1exy)yexyy

f(0,x0 x0

x 设u(xyC2u0,证明u(x,y)f(x)(yu xyuxy“u 已知xyu