lim [√(n^2+pn) -(qn+1)]=q

来源：百度知道 编辑：UC知道 时间：2024/06/20 04:01:14
求p的值
n→∞
A，2 B,-2 C,4 D,-4

If q <= 0,

lim_{n->+Inf}[(n^2 + pn)^(1/2) - (qn+1)]

= lim_{n->+Inf}[(n^2 + pn)^(1/2) + (-q)n -1]

= +Inf.

So, q > 0.

lim_{n->+Inf}[(n^2 + pn)^(1/2) - (qn+1)]

= lim_{n->+Inf}[(n^2 + pn) - (qn+1)^2]/}[(n^2 + pn)^(1/2) + (qn+1)]

= lim_{n->+Inf}[n^2(1 - q^2) + n(p - 2q) - 1]/[(n^2 + pn)^(1/2) + (qn+1)]

= lim_{n->+Inf}[n(1 - q^2) + (p - 2q) - 1/n]/[(1 + p/n)^(1/2) + (q+1/n)]

So, q = 1.

lim_{n->+Inf}[(n^2 + pn)^(1/2) - (qn+1)]

= lim_{n->+Inf}[(n^2 + pn)^(1/2) - (n+1)]

= lim_{n->+Inf}[(p - 2) - 1/n]/[(1 + p/n)^(1/2) + (1+1/n)]

= (p - 2)/[1 + 1]

= (p - 2)/2

= q

= 1,

So, p = 2q + 2 = 4.

我想应该是题目没表述清楚，应该是n->0的极限吧，这样的话这个极限为1，那么q=1