1.设函数x²+y²≠0时,f(x,y)=xy/(x²+y²) ；当x²+y²=0时,f(x,y)=0；问f(x,y)在（0,0）处是否连续?计算计算f’x (0,0)和f’y (0,0)；
当x²+y²=0时,必有x=0且y=0,此时f(x,y)=f(0,0)=0,即函数z=f(x,y)在原点有定义0；
当动点P沿x轴趋近原点时,x➔0limf(x,0)=0；当动点P沿y轴趋近原点时,y➔0limf(0,y)=0；
可见动点无论沿x轴还是沿y轴趋近原点该函数都有极限0,且此极限=f(0,0)=0；但若动点不沿坐标
轴,而沿任意方向趋近原点,情况就不一样了!
令y=kx,k∈R；则【x➔0,y=kx➔0】limf(x,y)=【x➔0,y=kx➔0】limf(x,kx)
=【x➔0,y=kx➔0】limkx²/(x²+k²x²)=k/(1+k²),可见极限值与k有关；由此即可判断该函数f(x,y)在原点没有极限,极在原点不连续.f’x (0,0)=0,f’y (0,0)=0.
【在一元函数里,有导数必连续；但在多元函数里,所有偏导数都存在,但不一定连续,此即
为一例】
2.将函数f(x)=sin(x/2)展开成x的幂级数.
f(0)=0；f'(x)=(1/2)cos(x/2)=(1/2)sin(π/2+x)；f''(x)=-(1/2²)sin(x/2)=(1/2²)sin(π+x/2)；
f'''(x)=-(1/2³)cos(x/2)=(1/2³)sin(3π/2+x/2)；f''''(x)=(1/2⁴)sin(x/2)=(1/2⁴)sin(2π+x/2)；
.；f⁽ⁿ⁾(x)=(1/2ⁿ)sin(nπ/2+x/2).
f'(0)=1/2；f''(0)=0；f'''(0)=-1/2³；f''''(0)=0；.；f⁽ⁿ⁾(0)=(1/2ⁿ)sin(nπ/2)；
故sin(x/2)=(1/2)x-[1/(2³▪3!)]x³+[1/(2⁵▪5!)]x⁵-[1/(2⁷▪7!)]x⁷+.
3.将1/(5-x)展开成为x-2的幂级数.并指出收敛域.【自己作吧!】
4.求函数f(x,y)=e^(x-y)(x²-2y²)的极值.
令∂f/∂x=e^(x-y)(x²-2y²)+2xe^(x-y)=[e^(x-y)](x²+2x-2y²)=0,得x²+2x-2y²=0.(1)
再令∂f/∂y=-e^(x-y)(x²-2y²)-4ye^(x-y)=-[e^(x-y)](x²+4y-2y²)=0,得x²+4y-2y²=0.(2)
(2)-(1)得4y-2x=0,故得x=2y,代入(1)式得4y²+4y-2y²=2y²+4y=2y(y+2)=0,故得y₁=0,y₂=-2；
相应地,x₁=0,x₂=-4；即有驻点P₁(0,0)；P₂(-4,-2)；
∂²f/∂x²=e^(x-y)(x²+2x-2y²)+e^(x-y)(2x+2)=e^(x-y)(x²+4x-2y²+2)
∂²f/∂x∂y=-e^(x-y)(x²+2x-2y²)-4ye^(x-y)=-e^(x-y)(x²+2x+4y-2y²)
∂²f/∂y²=e^(x-y)(x²+4y-2y²)-e^(x-y)(4-4y)=e^(x-y)(x²+8y-2y²-4)
对P₁(0,0)：A=∂²f/∂x²=2；B=∂²f/∂x∂y=0；C=∂²f/∂y²=-4；B²-AC=0＋16>0,
故P₁不是极值点；
对P₂(-4,-2)：A=(1/e²)(16-16-8+2)=-6/e²；B=-(1/e²)(16-8-8-8)=-8/e²；
C=(1/e²)(16-16-8-4)=-12/e²
B²-AC=(64-72)/e⁴=-8/e⁴