# On the definition of limit of a sequence

I apologize in advance if this turns out to be a trivial question, but when we define the limit of a sequence $x_n$ (it it exists) as the number $a$ such that $\forall \delta > 0\ \exists r : n \ge r \Rightarrow | x_n - a | < \delta$, is there a particular reason why we use $< \delta$ instead of $\le\ \delta$?

I first thought of this when reading the proof of the uniqueness of the limit. And indeed that proof does not hold if we use the "less or equal" form. But then I failed at trying to construct a sequence that under that new definition, would have two distinct limits. So I'm thinking that must not be the reason. So why is it that the definition uses the $< \delta$ form? Is it just a matter of convention?

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The only reason is convention; it is quite easy to prove that the two formulations are equivalent. – arjafi Aug 4 '12 at 17:40

It makes no difference. The version $|x_n-a|\lt \delta$ is more traditional. Since it makes no difference, one might as well go along with tradition. By the way, what you wrote would be slightly easier to read if instead of $\delta$ you used everyone's default favourite little guy $\epsilon$.
Remark: For proving the uniqueness, it is true that the wording of the proof might change slightly. If $a$ and $b$ are distinct, let $\delta=\frac{|b-a|}{3}$. By the definition of limit, if $n$ is large enough then $|x_n-a|\le \delta$ and $|x_n-b|\le \delta$, which contradicts the Triangle Inequality. If we use the more common $\lt$ version of the definition of limit, we can get away with $\delta=\frac{|b-a|}{2}$. But the geometric idea does not change: we can't be simultaneously real close to $a$ and real close to $b$. The rest is minor detail.
There are many seemingly different definitions of 'limit'. The one which becomes most useful in analysis if you proceed in that discipline is the statement that, given any neighbourhood $U$ of the limit point, almost all the members of the sequence (which corresponds to: $\exists r: \forall n\ge r$ ) are contained in $U$. Usually one chooses neihgbourhoods to be open sets (otherwise you'd have to additionally request them to have nonempty interior, which is inconvenient), which corresponds to $<$ if you choose to work with intervals or balls. This is why one usually uses $<$, but $\le$ would work equally well in your case.