多元函数的极限
设 `alpha, beta ge 0`, `alpha + beta gt 1`, `m, n` 为正整数, 则
`x^(n alpha) y^(m beta) = o(sqrt(x^(2n)+y^(2m)))`,
`quad (x,y) to (0,0)`.
特别取 `m = n = 1` 和 `m = n = 2` 得
`x^alpha y^beta = o(sqrt(x^2+y^2))`,
`quad x^alpha y^beta = o(root 4(x^4+y^4))`.
由不等式 `|x| le root(2n)(x^(2n)+y^(2m))`,
`|y| le root(2m)(x^(2n)+y^(2m))` 有
`(|x|^(n alpha) |y|^(m beta))/(sqrt(x^(2n)+y^(2m)))`
`le (x^(2n)+y^(2m))^(alpha/2 + beta/2 - 1/2)`.
记 `r = sqrt(x^2+y^2)`, 则 `max{|x|, |y|} le r le 1` 时有
`x^(2n)+y^(2m) le x^2+y^2 = r^2`, 从而上式小于等于 `r^(alpha+beta-1)`.
令 `r to 0` 即得结论.
- `x y`, `x^2`, `x^(1+p)` (`p gt 0`) 等函数皆是 `sqrt(x^2+y^2)`
的高阶无穷小.
- `x y = (x^(3/4) y^(3/4))^(4/3) = o(root 4(x^4+y^4))^(4/3)`
`= o(root 3(x^4+y^4))`.