证明：（一）由x1=1/2,x(n+1)=(xn²+1)/2.可得x1=1/2,x2=5/8.∴x1＜x2.又2x(n+1)=xn²+1≥2xn.===>x(n+1)≥xn.∴{xn}是递增数列.（二）易知,0＜x1＜x2＜1.假设0＜xn＜1,===>0＜xn²＜1.===>1＜xn²+1＜2.===>1/2＜(xn²+1)/2＜1.===>x(n+1)＜1.∴数列{xn}有上界1.∴{xn}存在极限.可设极限为a,在递推式两边取极限得：2a=a²+1.===>a=1.即极限为1.