请教一道关于无穷小量与无穷大量的比较的证明题o(g(x))+o(g(x))=o(g(x))(x->x0)

Question

请教一道关于无穷小量与无穷大量的比较的证明题
o(g(x))+o(g(x))=o(g(x)) (x->x0)

Answer 1

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))