If an is the nth value, then [itex]a_{n+1}= \sqrt{2+ a_n}[/itex]
IF the sequence {an} converges (you will need to prove that, perhaps by proving that it is an increasing sequence and has an upper bound), call the limit A.
Then we must have [itex]\lim_{n\rightarrow \infty}a_{n+1}= \sqrt{2+ \lim_{n\rightarrow \infty}a_n}[/itex] so [itex]A= \sqrt{2+ A}[/itex].