要用到洛比達法則.
lim [ln(1+x)-x+(1/2)x^2]/x^n
=lim [1/(1+x) - 1 +x] /[n·x^(n-1)]
=lim [1 - (1+x) +x·(1+x)] /[n·x^(n-1)·(1+x)]
=lim [x²] /[n·(x^n + x^(n-1) )]
=lim [2x] /[a·x^(n-1) + b·x^(n-2) )]
=lim 2 /[a1·x^(n-2) + b1·x^(n-3) )]
其中a1,b1都是常數(shù).
若分母不為0也不為∞,則x^(n-3)=1.
則n-3=0
n=3.
所以當x趨于0時,ln(1+x)-x+(1/2)x^2無窮小的階數(shù)為3