设X1=a,X2=b,Xn+2=(Xn+1+Xn)/2,n+1,2,3,......求Xn的极限，要

设X1=a,X2=b,Xn+2=(Xn+1+Xn)/2,n+1,2,3,......求Xn的极限，要

证明：因为0<x1<3所以x(n+1)<=[xn+(3-xn)]/2=3/2
所以｛xn}有界
又x(n+1)=√[Xn(3-Xn)] >=√[Xn(3-3/2)] =√(3/2)xn>=xn
所以{xn}递增
单调有界数列必有极限，设x=limxn=limx(n+1),则
x=√x(3-x)解得x=3/2
所以limxn=3/2

Xn+2=(Xn+1+Xn) =(Xn+2Xn-1+Xn-2）/4 因为2Xn-1=Xn-2+Xn-3 所以(Xn+2Xn-1+Xn-2)/4 =(Xn+2Xn-2+Xn-3)/4 迭代得Xn+2=(Xn+2X2+X1)/4=(Xn+2b+a)/4 lim Xn=lim Xn+2=(lim Xn+2b+a)/4 解得lim Xn=(a+2b)/3

证明：因为0<x1<3所以x(n+1)<=[xn+(3-xn)]/2=3/2 所以｛xn}有界 又x(n+1)=√[xn(3-xn)] >=√[xn(3-3/2)] =√(3/2)xn>=xn 所以{xn}递增 单调有界数列必有极限，设x=limxn=limx(n+1),则 x=√x(3-x)解得x=3/2 所以limxn=3/2