高数多元分段函数在分界点出的连续性问题

2024-12-23 02:03:35
推荐回答(2个)
回答1:

如果函数在原点处有极限,那么它在任意趋近于原点的方向上都存在极限,且值都等于函数在原点处的函数值。
反之,如果能找到一个方向,函数沿此方向趋近于原点的极限不存在或虽然存在但不等于函数在原点处的函数值,则函数在原点处就不连续。

令 y=x^3 ,只是让点沿着曲线 y=x^3 趋近于原点,此时可求得极限为 1/2 ,不等于 f(0,0) ,所以函数在原点处不连续 。
这实际上是特殊与一般地辩证关系 。

回答2:

若f(x,y)在原点有极限,则(x,y)沿任何方式趋于原点(0,0)时,f(x,y)都有同样的极限值。
注意上面是以任何方式。因此经常用这个结论的逆否命题来证明f(x,y)在(0,0)没有极限。
就是:找两个(x,y)趋于原点的方式,使得f(x,y)在此两种方式下收敛到的极限值不同,
这就能说明f(x,y)在原点没有极限。
与之类似,只要能找到一种方式,使得f(x,y)在此种方式下的极限值与函数值不同,
就能说明f(x,y)在原点不连续。观察函数表达式可以知道,
取y=x^3时,函数极限是1/2,不等于函数值f(0,0)=0,因此函数不连续。

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