Question Re. Simple epsilon proofs

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Here's an example that helps illustrate my question:

Prove: A sequence in R can have at most one limit.

Proof:

Assume a sequence {xn}n[tex]\in[/tex]N has two limits a and b.
By definition:

-For any [tex]\epsilon[/tex]>0, there exists an N[tex]\in[/tex]N such that n[tex]\geq[/tex]N implies that |xn-a| < [tex]\epsilon[/tex]/2.
-A similar argument can be made for the limit b.

Thus:

|a-b| [tex]\leq[/tex] |xn-a| - |xn-b| < [tex]\epsilon[/tex]/2 + [tex]\epsilon[/tex]/2 = [tex]\epsilon[/tex]

Thus a=b.


Now here's my question...in the step immediately following "By definition," the actual definition of a limit of a sequence shows that n[tex]\geq[/tex]N implies that |xn-a| < [tex]\epsilon[/tex], not that n[tex]\geq[/tex]N implies that |xn-a| < [tex]\epsilon[/tex]/2. So is it acceptable to put [tex]\epsilon[/tex] over any number when convenient in a proof? As in the above proof, it is convenient to put [tex]\epsilon[/tex]/2 instead of just [tex]\epsilon[/tex] so that in the final step the two add up to [tex]\epsilon[/tex] and show that a=b.

Thanks.

edit: sorry for the awkward formatting, if anything is unclear, let me know I'll explain it in words.
 
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Yes, if the limit exists, then you can replace epsilon with any positive number (so epsilon/2 is of course positive), because that's simply what epsilon typically denotes. Many people want the last step of a proof to be just epsilon, but often times it doesn't matter. In this case, it doesn't matter whether we have epsilon or 2*epsilon at the very end, since epsilon is arbitrary anyways. Note you have a typo in the leftmost inequality at the final step of the proof.
 
snipez90 said:
Yes, if the limit exists, then you can replace epsilon with any positive number (so epsilon/2 is of course positive), because that's simply what epsilon typically denotes. Many people want the last step of a proof to be just epsilon, but often times it doesn't matter. In this case, it doesn't matter whether we have epsilon or 2*epsilon at the very end, since epsilon is arbitrary anyways. Note you have a typo in the leftmost inequality at the final step of the proof.

Thanks for the reply, that's what I thought but wanted to hear it from someone who actually knew before I went on doing proofs on an assumption.