Question

大一微积分 求数列极限

设m和n均为正整数 求极限
1.lim x→1 x^m-1/x^n-1= 提示（因式分解）

2.lim x→0 （1+mx）^n-(1+nx)^m/x^2= 提示（用二项式定理）

Answer 1

1. x^m -1 = (x-1)* [x^(m-1) + x^(m-2) + ..... + x + 1],
x^n -1 = (x-1)* [x^(n-1) + x^(n-2) + ..... + x + 1]
lim (x→1) (x^m-1)/(x^n-1)
= lim (x→1) [x^(m-1) + x^(m-2) + ..... + x + 1] / [x^(n-1) + x^(n-2) + ..... + x + 1]
= m/n
2. (1+m x)^n = 1 + n * (mx) + n(n-1)/2 * (mx)² + o(x³)
(1+n x)^m = 1 + m * (nx) + m*(m-1)/2 * (nx)² + o(x³)
(1+m x)^n - (1+n x)^m = [n*(n-1) * m² - m*(m-1) * n² ]/2 * x² + + o(x³)
原式= lim (x→0) [n*(n-1) * m² - m*(m-1) * n² ]/2 + + o(x)
= [ n*(n-1) * m² - m*(m-1) * n² ]/2
= mn(n-m)

追问

2最后忘除了吧 第二道的第一步是什么公式？

Answer 2

1. lim(x->1)[(x^m-1)/(x^n-1)=lim(x->1){(x-1)(x^(m-1)+x^(m-2)+...+1)}/{(x-1)(x^(n-1)+x^(n-2)+...+1)}
=lim(x->1){(x^(m-1)+x^(m-2)+...+1)}/{(x^(n-1)+x^(n-2)+...+1)}
=m/n
2. lim(x->0)[(1+mx)^n-(1+nx)^m]/x^2
=lim(x->0)[(1+nmx+n(n-1)(mx)^2/2+o(x^3))-(1+mnx+m(m-1)(nx)^2/2+o(x^3))]/x^2
=lim(x->0)[n(n-1)/2-m(m-1)/2+o(x)]
[n(n-1)-m(m-1)]/2