已知a b 是常数 lim（a根号（2n^2+n+1) -bn))=1 则a+b=

来源：学生作业帮编辑：神马作文网作业帮分类：数学作业时间：2024/10/10 06:15:17

∵当a与b中只有一个为零时,lim(n->∞)[a√(2n²+n+1)-bn]不存在
当a与b同时为零时,lim(n->∞)[a√(2n²+n+1)-bn]=0
又lim(n->∞)[a√(2n²+n+1)-bn]=1 (已知)
∴a与b都不等于零
∵lim(n->∞)[a√(2n²+n+1)-bn]=1 ==>lim(n->∞){[(2a²-b²)n²+a²(n+1)]/[a√(2n²+n+1)+bn]}=1
==>2a²-b²=0.(1)
==>lim(n->∞){[a²(n+1)]/[a√(2n²+n+1)+bn]}=1
==>lim(n->∞){[a²(1+1/n)]/[a√(2+1/n+1/n²)+b]}=1
==>a²/(√2a+b)=1.(2)
∴解方程组(1)与(2)得a=2√2,b=4
故a+b=2√2+4.

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