Question

若lim（(a+2b)n^2+2n+1）/（bn+3）=1/2 则实数a+b为

Answer 1

∵lim(n->∞){[(a+2b)n²+2n+1]/(bn+3)}=1/2

∴必有a+2b=0.(1) (若a+2b≠0,则lim(n->∞){[(a+2b)n²+2n+1]/(bn+3)}不存在)

==>lim(n->∞)[(2n+1)/(bn+3)]=1/2

==>lim(n->∞)[(2+1/n)/(b+3/n)]=1/2 (分子分母同除n)

==>(2+0)/(b+0)=1/2

==>2/b=1/2

==>b=4.(2)

故 由(1)式和(2)式得 a+b=-b=-4.