lim [n+1-√(n²+n)] as n->∞= lim [n-√(n²+n)]+1= lim [n-√(n²+n)][n+√(n²+n)]/[n+√(n²+n)]+1,分子有理化= 1-lim n/[n+√(n²+n)]= 1-lim 1/[1+√(1+1/n)]= 1-1/[1+√(1+0)]= 1-1...