Question

这道题对了吗？极限

如果有错或者步骤错了请指出来谢谢

Answer 1

lim(x->0) { e^x -e^[ln(1+x)]} /[(1-x^2)^(1/3) -1]

不对

应该是

分母

x->0

(1-x^2)^(1/3) =1 -(1/3)x^2 +o(x^2)

(1-x^2)^(1/3) -1= -(1/3)x^2 +o(x^2)

故此，分子应该展开到2阶

分子

e^x =1+x+(1/2)x^2 +o(x^2)

ln(1+x) = x-(1/2)x^2+o(x^2)

e^[ln(1+x)] 

= e^[x-(1/2)x^2+o(x^2)]

=1+[x-(1/2)x^2+o(x^2)] +(1/2)[x-(1/2)x^2+o(x^2)]^2 +o(x^2)

=1+[x-(1/2)x^2+o(x^2)] +(1/2)[x^2+o(x^2)] +o(x^2)

=1+x +o(x^2)

e^x -e^[ln(1+x)] =(1/2)x^2 +o(x^2)

故此根据上述结果

lim(x->0) { e^x -e^[ln(1+x)]} /[(1-x^2)^(1/3) -1]

=lim(x->0)(1/2)x^2 /[-(1/3)x^2]

约掉 x^2

=-3/2

Answer 2

利用展开式eˣ=1+x+x²/2!+...+xⁿ/n!+...
利用等价无穷小 (1+x)ⁿ-1~nx
首先分母是³√(1-x²)-1~⅓(-x²)
那分子部分利用展开式只需保留到二阶部分，即eˣ=1+x+½x²+o(x²)
∵e^ln(1+x)=1+x,
∴原题可以化简为
lim{[1+x+½x²+o(x²)]-(1+x)}/⅓(-x²)
=lim[½x²+o(x²)]/(-⅓x²)
= -3/2

Answer 3

错了，第二步