$a_{2n}=-\sqrt{2n}$; $a_{2n+1}$=$\sqrt{2n+1}-\sqrt{2n}$
Lower limit and infimum both negative infinity as $-\sqrt{2n}$ gets arbitrarily small.
Upper limit = zero ($a_{2n+1}$ goes to zero as $n-> \infty$), supremum = 1 (obtained for n = 0)
Is it right to my procedure?
Is there a more formal method?