已知lim(n的平方+1/n+1+an-b)=1,求a,b的值
问题描述：
已知lim(n的平方+1/n+1+an-b)=1,求a,b的值
答
lim(n^2+1/n+1+an-b)
=lim(n^2+1/n+1+(an-b)(n+1)/n+1)
=lim[(1+a)n^2+(a-b)n+(1-b)]/ (n+1)
又 lim(n^2+1/n+1+an-b)=1,可知：
[(1+a)n^2+(a-b)n+(1-b)]/ (n+1)中[(1+a)n^2的系数1+a=0,即a=-1
故：lim(n^2+1/n+1+an-b)
=lim[(1+a)n^2+(a-b)n+(1-b)]/ (n+1)
=lim[(a-b)n+(1-b)]/ (n+1)=1
则有：a-b=1,即b=a-1=-2
故：a=-1,b=-2