lim n^2(k/n-1/n+1-1/n+2-.-1/n+k)(其中K是与N无关的常数)

k/n -1 /(n+1) - 1/(n+2)-……-1/(n+k)
= [1/n - 1/(n+1)] + [1/n - 1/(n+2)] + [1/n-1/(n+3)] + .+ [1/n - 1/(n+k)]
= 1/[n(n+1)] + 2 / [n (n+2)] + 3 /[n(n+3)] + .+ k / [n(n+k)]
原式 = lim(n->∞) n² { 1/[n(n+1)] + 2 / [n (n+2)] + 3 /[n(n+3)] + .+ k / [n(n+k)] }
= 1 + 2 + 3 + .+ k
= k(k+1)/2

证明：
关键在于缩放.
用定义证明既是需要证,对任意小的正数ε,一定存在某个正整数M,使得n＞M时,有
|n²/3^n - 0| ＜ ε
记 an=n²/3^n,则有
a(n+1)/an
= [(n+1)²/3^(n+1)] / [n²/3^n]
= (n+1)²/3n²
令 a(n+1)/an ＜ 1/2
有 (n+1)² ＜ 3n² ,得 n ≥ 5
∴ 当 n≥5时 ,an ≤ a5 · (1/2)^(n-5)
∴ 当n＞M时,
|n²/3^n - 0| ＜ ε
n²/3^n ＜ ε

=lim n^2（(1/n+1-1/n）+（1/n+2-1/n)······一直到1/n+5然后每一项开始计算把n^2再乘进去可得极限为-15

lim (n^2-n+3) / (1-3n^2)
上下同时除以n^2
=lim (n^2-n+3)/n^2 / (1-3n^2)/n^2
=lim (1-(1/n)+(3/n^2)) / (-3+(1/n^2))
因为1/n趋于0,1/n^2趋于0
故
=(1-0+0) / (-3+0)
=-1/3
有不懂欢迎追问

lim n^2+5n+4/n+1
n→-1
=lim (n+1)(n+4)/n+1
n→-1
=lim (n+4)
n→-1
=3

你没交代清,若只是n-->无穷,则：
lim n^2(1/n-1/(n+1)+1/n-1/(n+2)+1/n-1/(n+3)+...+1/n-1/(n+k))
化简一下为：
lim n/(n+1)+2n/(n+2)+...kn/(n+k)
=1+2+...+k
=k(k+1)/2

用等价无穷小代换:
(1-cosx)~x^2/2,x趋于0,
则1-cos1/n~(1/n)^2/2
则原式=limn^2[(1/n)^2/2]
=1/2