lim(根号下（n^2+pn))-(qn+1))=q,则p的值为多少〉

Question

lim(根号下（n^2+pn))-(qn+1))=q,则p 的值为多少〉

Answer 1

分子有理化
lim(n→+∞)[√(n^2+pn)-(qn+1)]
=lim(n→+∞)[n^2+pn-(qn+1)^2]/[√(n^2+pn)+(qn+1)]
=lim(n→+∞)[(1-q^2)n^2+(p-2q)n-1]/[√(n^2+pn)+(qn+1)]
=lim(n→+∞)[(1-q^2)n+(p-2q)-1/n]/[√(1+p/n)+(q+1/n)]
由于lim(n→+∞)[√(1+p/n)+(q+1/n)]=1+q为常数
则(1-q^2)=0 q=1或-1
=(p-2q)/(1+q)=q
p-2q=q^2+q
p=q^2+3q=4或-2