求lim[x^(n+1)-(n+1)x+n]/(x-1)^2x-->1=lim(t->0)[[1+(n+1)t+(n+1)n/2t^2+o(t^2)]-(n+1)-(n+1)t+n]/t^2？不懂

Question

求lim[ x^(n+1)-(n+1)x+n]/(x-1)^2 x-->1
=lim(t->0) [ [ 1 + (n+1)t + (n+1)n/2t^2 + o(t^2)] -(n+1)-(n+1)t + n]/t^2？不懂

Answer 1

①令：x = 1+t (t->0)
lim(x->1) [ x^(n+1)-(n+1)x+n]/(x-1)^2
=lim(t->0) [ (1+t)^(n+1)-(n+1)(1+t) + n]/t^2
【二项式展开, o(t^2) 表示t^2的高阶无穷小量：（n+1)n(n-1)t^3/3! + ... + t^(n+1) 】
=lim(t->0) [ [ 1 + (n+1)t + (n+1)n/2t^2 + o(t^2)] -(n+1)-(n+1)t + n]/t^2
=(n+1)n/2
② 罗必塔法则:
lim(x->1) [ x^(n+1)-(n+1)x+n]/(x-1)^2
=lim(x->1) [(n+1)x^n -(n+1)]/2(x-1)
=lim(x->1) (n+1)nx^(n-1)/2
=n(n+1)/2