∵x2=
1
2x1,且xn=
1
2(xn−1+xn−2)令n=3,
得 x3=
1
2(x2+x1)=
3
4x1,令n=4,
得 x4=
1
2(x2+x3)=
5
8x1,
∴x2−x1=−
1
2x1,x3−x2=
1
4x1,x4−x3=−
1
8x1,…,xn− xn−1 =(−
1
2)n−1,
于是xn=x1+(x2-x1)+…+(xn-xn-1)=x1+
−
1
2x1[1−(−
1
2)n−2]
1+
1
2
∴
lim
n→∞xn=x1−
1
3x1=2,x1=3.
故答案为3.