设y=[√(n^2+1)/(n+1)]^nlny=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
