并不复杂呀
x->0时
lim(x-arctanx)/ln(1+x^3)
=lim[1-1/(1+x^2)]/[3x^2/(1+x^3)]
=lim[x^2/(1+x^2)]/[3x^2/(1+x^3)]
=lim(1+x^3)/3(1+x^2)=1/3
或者用麦克劳林展开式arctanx＝x-1/3*x^3+1/5*x^5-1/7*x^7+1/9*x^9
x->0时,ln(1+x^3)是x^3的等价无穷小所以
x->0时
lim(x-arctanx)/ln(1+x^3)
=lim(1/3x^3)/x^3
=1/3