lim[√(n^2+pn)-(qn+1)]=q求p的值n→∞A，2B,-2C,4D,-4

Question

lim [√(n^2+pn) -(qn+1)]=q
求p的值
n→∞
A，2 B,-2 C,4 D,-4

Answer 1

If q +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.