The first thing I would do is calculate a few terms: a1= 5, a2= 0.5(5+ 10/5)= 3.5, a3= 0.5(3.5+ 10/3.5))= 3.1785... . Hmm, looks like it is a decreasing sequence! If I could prove that (perhaps by induction) and prove that is has a lower bound (which is obvious: every number is positive) then, by the "monotone convergence property", the sequence converges.
It is clear that a2< a1. Assume that ak+1[/sup]< ak[/sup] for some k. Can you prove, using the recurrance relation, that then ak+2< ak?
(It occurs to me that, in order to deal with the "10/an" part, in which an is in the denominator, you will need a lower bound, other than 0, on the sequence. Do you see how to prove that, in fact, 3 is a lower bound on the sequence? Use induction again.)
By the way, once you have proved that the sequence converges, it is easy to actually find the limit: If the sequence converges to a, then, taking the limit on both sides of a(n+1)= 0.5*(a(n)+ 10/a(n)) you get lim(a(n+1))= 0.5 (lim(a(n))+ 10/lim(a(n))) which, since all of those limits are the same, is a= 0.5(a+ 10/a). Solve that equation for a.