
求极限 n趋向于无穷 lim((根号下n^2+1)/(n+1))^n
1个回答
展开全部
设y=[√(n^2+1)/(n+1)]^n
lny=nln[√(n^2+1)/(n+1)]=n[1/2ln(n^2+1)-ln(n+1)]
lim(n→∞)lny=lim[1/2ln(n^2+1)-ln(n+1)]/n^(-1)
=lim(n→∞)[n/(n^2+1)-(n+1)]/[-n^(-2)](洛必达法则)
=-lim(n→∞)(n-1)n^2/[(n^2+1)(n+1)]
=-1
所以lim(n→∞)y=1/e
lny=nln[√(n^2+1)/(n+1)]=n[1/2ln(n^2+1)-ln(n+1)]
lim(n→∞)lny=lim[1/2ln(n^2+1)-ln(n+1)]/n^(-1)
=lim(n→∞)[n/(n^2+1)-(n+1)]/[-n^(-2)](洛必达法则)
=-lim(n→∞)(n-1)n^2/[(n^2+1)(n+1)]
=-1
所以lim(n→∞)y=1/e
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询