请教一道关于无穷小量与无穷大量的比较的证明题
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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)
o(g(x))+o(g(x))=o(g(x)) (x->x0)
![请教一道关于无穷小量与无穷大量的比较的证明题](/uploads/image/z/15002453-29-3.jpg?t=%E8%AF%B7%E6%95%99%E4%B8%80%E9%81%93%E5%85%B3%E4%BA%8E%E6%97%A0%E7%A9%B7%E5%B0%8F%E9%87%8F%E4%B8%8E%E6%97%A0%E7%A9%B7%E5%A4%A7%E9%87%8F%E7%9A%84%E6%AF%94%E8%BE%83%E7%9A%84%E8%AF%81%E6%98%8E%E9%A2%98)
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))
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))