from what i have learned till now at uni
a real sequence [math](x_n)[/math] is said have a limit [math]L \in \mathbb{R}[/math]
if the following is true. To each positive number a, corresponds an integer [math]n_0[/math], such that [math]|x_n-L|[/math] is less than a for all integers [math]n \ge n_0[/math]
obviously this condition satisfies if the sequence is 0,0,0,0,0,... and L is taken to be 0, for all n.
Also we can use the monotonic sequence theorem, which states that
let [math](x_n)[/math] be a sequence which is non decreasing for [math]n \ge N[/math]. If [math](x_n)[/math] is bounded above, then (x_n) converges, and the limit is the supremum of the s of the set [math]{x_{n}:n \in \mathbb{Z},n \ge N}[/math]. If (x_n) is not bounded above, then (x_n) diverges to [math]\infty[/math]
so from above the limit of sequence 0,0,0,0,0,.. is 0, as the sequence is non decreasing. and 0 is the lowest upper bound of the set {x_n}
the same theorem can also (but pointless) be applied to show that the limit of the sequence
(0.9,0.99,0.999,0.9999,...) is 1.