Suppose that when n=k, xk

Then when n=k+ 1, x (k+ 1) = √ (2+xk) < √(2+2)=2 holds.

So for any n, xn

Because x(n+ 1)=√(2+xn)>0, so 0.

√(2/2^2+ 1/2)= 1

So x (n+ 1) >: Xn, that is, {xn} monotonically increases.

To sum up, {xn} is monotonically bounded, that is, {xn} limit exists.

Let the limit of {xn} be a, and then find the limit on both sides of x (n+ 1)=√(2+xn).

A=√(2+A)

A^2-A-2=0

(A-2)(A+ 1)=0

A=2 or-1 (truncated)

So the limit of {xn} is 2.