由高阶无穷小的定义有
lim( o(g(x)))+o(g(x)) )/(g(x))
= lim o(g(x))/(g(x)) + lim o(g(x))/(g(x))
=0+0=0
所以o(g(x))+o(g(x))=o(g(x))
请教一道关于无穷小量与无穷大量的比较的证明题
由高阶无穷小的定义有
lim( o(g(x)))+o(g(x)) )/(g(x))
= lim o(g(x))/(g(x)) + lim o(g(x))/(g(x))
=0+0=0
所以o(g(x))+o(g(x))=o(g(x))