lim(x→∞)(1-kx)的x分之一次幂，k∈z；
lim(x→∞)(x+1／x-1）的x次幂；
lim(x→∞){1／a(a+d) + 1／(a+d)(a+2d) ``` +1／[a+(n-1)d][a+nd]},a,d为不等于零的常数

lim(x→∞)(1-kx)的x分之一次幂，k∈z；
lim(x→∞)(1-kx)^(1/x)
=lim(x→∞){[(1-kx)^(1/-kx)]^(-kx)}^(1/x)
=lim(x→∞)e^[(-kx)*(1/x)]
=lim(x→∞)e^(-k)
=1/e^k

lim(x→∞)(x+1／x-1）的x次幂；
=lim(x→∞)[(x-1+2)/(x-1)]^x
=lim(x→∞)[1+2/(x-1)]^x
=lim(x→∞)【{[1+2/(x-1)]^(x-1/2)}^2/(x-1)】^x
=lim(x→∞)e^[2x/(x-1)]
=e^2

lim(n→∞){1／a(a+d) + 1／(a+d)(a+2d) ``` +1／[a+(n-1)d][a+nd]},a,d为不等于零的常数
=lim(n→∞){(1/d)[(1/a)-(1/a+d)]+(1/d)[(1/a+d)-(1/a+2d)]+……+[(1/d)(1/[a+(n-1)d]-1/[a+nd]}
=lim(n→∞)(1/d)*[(1/a)-(1/a+d)+(1/a+d)-(1/a+2d)+……+1/[a+(n-1)d]-1/(a+nd)]
=lim(n→∞)(1/d)*[(1/a)-1/(a+nd)]
=lim(n→∞)(1/d)*[nd/a*(a+nd)]
=lim(n→∞)n/[a*(a+nd)]
=lim(n→∞)1/{a*[(a/n)+d]}
=1/[a*(0+d)]
=1/(ad)

前面两道题目，利用的是重要极限：lim(x→∞)(1+x)^(1/x)=e
它们都属于这个重要极限的变形！
第三题的关键是拆项。考虑到分式的分母中两项之差均为d，将每一个分式拆成两个分式的差，然后就是消去同样的项。

看来楼上的解答，我只能是无语，还有点想笑。。。(ˇˍˇ）