$a_{2n}=-\sqrt{2n}$; $a_{2n+1}$=$\sqrt{2n+1}-\sqrt{2n}$
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?
The supremum is a maximum and, as André said, it is obtained at $\,m=1\,:\,\,a_1=\sqrt{3}-\sqrt{2}\,$ , since the function $\,f(x):=\sqrt{2x+1}-\sqrt{2x}\,$ is monotone descending.
The infimum certainly is $\,-\infty\,$, as $\,-\sqrt{2n}\to -\infty\,$ , and since this is also the limit of a subseq. of the seq. this is also the $\,\displaystyle{\underline{\lim}_{n\to\infty}a_n}\,$ .
Finally, $\,\displaystyle{\overline{\lim_{n\to\infty}}a_n=\lim_{n\to\infty}\left(\sqrt{2n+1}-\sqrt{2n}\right)}=\lim_{n\to\infty}\frac{1}{\sqrt{2n+1}+\sqrt{2n}}=0$