高等数学:1无穷小的等价代换:证明下列关系式

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第1个回答  2019-09-20
(4)
x->0
tanx = x+(1/3)x^3+o(x^3)
sinx = x-(1/6)x^3+o(x^3)
tanx -sinx =(1/2)x^3+o(x^3)
√(1+tanx) -√(1+sinx)
=(1+tanx) -(1+sinx) /[√(1+tanx) +√(1+sinx) ]
=(tanx -sinx)/[√(1+tanx) +√(1+sinx) ]
=[(1/2)x^3+o(x^3)]/2
=(1/4)x^3 +o(x^3)
x->1
1+cos(πx)
=1+ cos[π+π(x-1)]
=1- cos[π(x-1)]
=1 - [ 1 - (1/2)π^2.(x-1)^2 +o((x-1)^2) ]
= (1/2)π^2.(x-1)^2 + o((x-1)^2)本回答被提问者和网友采纳
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