Question

大一的高数，求解的具体过程？

第（6）

Answer 1

分子等价无穷小，分母泰勒展开。

Answer 2

等价替换，望采纳

Answer 3

x->0
tanx = x+(1/3)x^3 +o(x^3)
ln( 1+ tanx)
=ln[ 1+x+(1/3)x^3 +o(x^3)]
= [x+(1/3)x^3] - (1/2)[x+(1/3)x^3]^2 +(1/3)[x+(1/3)x^3]^3 +o(x^3)
= [x+(1/3)x^3] - (1/2)[x^2+o(x^3)] +(1/3)[x^3+o(x^3)] +o(x^3)
=x -(1/2)x^2 +(2/3)x^3 +o(x^3)
sinx = x-(1/6)x^3 +o(x^3)
ln( 1+ sinx)
=ln[ 1+x-(1/6)x^3 +o(x^3)]
= [x-(1/6)x^3] - (1/2)[x-(1/6)x^3]^2 +(1/3)[x-(1/6)x^3]^3 +o(x^3)
= [x-(1/6)x^3] - (1/2)[x^2+o(x^3)] +(1/3)[x^3+o(x^3)] +o(x^3)
=x -(1/2)x^2 +(1/6)x^3 +o(x^3)
ln[(1+tanx)/(1+sinx)]
=ln(1+tanx) -ln(1+sinx)
=[x -(1/2)x^2 +(2/3)x^3 +o(x^3) ] -[x -(1/2)x^2 +(1/6)x^3 +o(x^3)]
=(1/2)x^3 +o(x^3)
lim(x->0) [(1+tanx)/(1+sinx)]^(1/x^3)
=lim(x->0) e^{ ln[(1+tanx)/(1+sinx)] /x^3 }
=lim(x->0) e^[ (1/2)x^3 /x^3 ]
=e^(1/2)

Answer 4

望采纳

追问

这两步，x∧3原本在ln外面，为什么可以变到里面还是分母

不应该这样吗？

追答

分子是用等价无穷小变的，分母没改，望采纳

追问

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