Question

x-sinx*cosx*cos2x为什么是三阶无穷小？

Answer 1

x－sinxcosxcos2x
＝x－（1/2）sin2xcos2x
＝x－（1/4）sin4x
＝x－（1/4）（4x－（4x）³/3！＋…）〖对sin4x进行幂级数展开〗
＝32x³/3－…〖略去高阶部分〗
至此，就可以看这是一个三阶无穷小
〖解毕〗

Answer 2

x → 0 时
lim<x → 0>(x-sinx)cosxcos2x/x^k
= lim<x → 0> (x-x+x^3/6)cosxcos2x/x^k
= lim<x → 0> (x^3/6)cosxcos2x/x^k
= lim<x → 0> (x^3)/(6x^k) = C, k = 3

Answer 3

x->0
sinx.cosx.cos2x
=(1/2)sin2x.cos2x
=(1/4)sin4x
=(1/4)[4x -(1/6)(4x)^3 +o(x^3) ]
=x - (8/3)x^3 +o(x^3)
x-sinx.cosx.cos2x =(8/3)x^3 +o(x^3)