证明在点(0,0)处f(x,y)连续且偏导数存在,但不可微f(x,y)=x^2y^2/(x^2+y^2)^(3/2)
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证明在点(0,0)处f(x,y)连续且偏导数存在,但不可微f(x,y)=x^2y^2/(x^2+y^2)^(3/2)
证明在点(0,0)处f(x,y)连续且偏导数存在,但不可微
f(x,y)=x^2y^2/(x^2+y^2)^(3/2)
证明在点(0,0)处f(x,y)连续且偏导数存在,但不可微f(x,y)=x^2y^2/(x^2+y^2)^(3/2)
教材上应该有类似的例题,依样画葫芦即可:
1)由于
|[(x^2)(y^2)]/(x^2+y^2)^(3/2)|
<= {{[(x^2)+(y^2)]/2}^2}/(x^2+y^2)^(3/2)
= [(x^2+y^2)^(1/2)]/4 → 0,(x,y)→(0,0),
可知
lim[(x,y)→(0,0)]f(x,y) = 0 = f(0,0).
2)由
lim(x→0)[f(x,0)-f(0,0)]/x = lim(x→0)(0-0)/x = 0,
知 fx(0,0) = 0,同理,fy(0,0) = 0.
3)若 f(x,y) 在 (0,0) 可微,应有
[△f(0,0)-df(0,0)]
= f(0+△x,0+△y)-f(0,0)-[fx(0,0)*dx+fy(0,0)*dy]
= [(△x^2)(△y^2)]/(△x^2+△y^2)^(3/2)
= o(ρ) (ρ→0),
其中,ρ=(△x^2+△y^2)^(1/2),但
lim(ρ→0)[△f(0,0)-df(0,0)]/ρ
= lim(ρ→0)[(△x²)(△y²)]/(△x²+△y²)²
= lim(ρ→0)[(△x)(△y)/(△x²+△y²)]²
不存在,矛盾.因此 f(x,y) 在 (0,0) 不可微.