根據(jù)題意,設(shè)此數(shù)列為an,an>0則a1=根號(hào)2,a(n+1)=根號(hào)下(2+an),即[a(n+1)]^2=2+an
易得a2>a1
[a(n+1)]^2-(an)^2=[a(n+1)+an]*[a(n+1)-an]=an-a(n-1)
根據(jù)數(shù)學(xué)歸納法得a(n+1)>an即an為遞增數(shù)列
下面證明an<2
a1=根號(hào)2<2
假設(shè)ak<2,a(k+1)=根號(hào)下(2+ak)<根號(hào)（2+2）=2
有數(shù)學(xué)歸納法可得,an<2
綜上,由極限存在準(zhǔn)則得an存在極限.
解得,an的極限為2.