Limit of Sequence?? The sequence: $-2, 0, -1, 0, -1/2, 0, -1/4, 0, -1/8, 0, -1/16, 0\ldots$
Heyy. So I was a little confused about this question. How would this have an existing limit if it keeps alternating between a value and $0$? I know if you have a sequence of $1, -1, 1, -1\ldots$ that would not have a limit.

So why is the sequence above have a limit?

Thank you in advance for any help!
 A: Just remember the definition of a sequence limit in term of $\varepsilon$-$\delta$: $(a_n)_{n} \subset \mathbb R$ converges to $a \in \mathbb R$ if and only if for all $\varepsilon > 0$, there exists $N_\varepsilon \in \mathbb N$ such that for all $n \ge N_\varepsilon$ we have
$$|a - a_n| < \varepsilon.$$
In this case, $a = 0$ and
$$
a_n = 
\begin{cases}
-1/2^{n/2} & \text{if } n \text{ is even},\\
0 & \text{otherwise}.
\end{cases}
$$
Let $\varepsilon > 0$. For $N_\varepsilon = \lceil 2\log_2(1/\varepsilon)\rceil$, we have, $\forall n \ge N_\varepsilon$, that
$$\left|a - a_n\right| = \left|a_n\right| < \varepsilon.$$
A: 
$\left<-2, 0, -1, 0, -1/2, 0, -1/4, 0, -1/8, 0, -1/16, 0\ldots\right>$
How would this sequence have an existing limit if it keeps alternating between a value and $\mathbf0$?
I know that the sequence $\left<1, -1, 1, -1\ldots\right>$ does not have a limit.

To be clear, $\left<-2, 0, -1, 0, -1/2, 0, -1/4, 0, -1/8, 0, -1/16, 0\ldots\right>$ is not alternating between some (constant) value and $\mathbf0,$ unlike $\left<1, -1, 1, -1\ldots\right>,$ which is alternating between two constant values. This crucial difference between them means that the latter cannot have a limit, while the former may.
Rather, the first sequence is approaching $0$ from the left; this means that its limit is zero (regardless of whether it ever reaches $0$).
