Why may we restrict the value of $\epsilon$ in limit proofs? I am reading Steven R Lay's "Analysis with an introduction to proof". The theorem is that if $(s_n)$ is a positive sequence and $lim~(\frac{s_n}{s_{n+1}}) = L < 1$, then $lim~s_n = 0$.
The proof begins by noting that, since $L < 1$, there exists a number $c$ such that $L<c<1$. Then, let $\epsilon = c-L$. Since the sequence is positive, we have that $|\frac{s_n}{s_{n+1}}| = \frac{s_n}{s_{n+1}}$, and therefore since by hypothesis the limit $\frac{s_n}{s_{n+1}}$ exists,
$$\frac{s_n}{s_{n+1}} < \epsilon + L = c$$.
Then, because $\frac{s_n}{s_{n+1}}$ is convergent, it is bounded [above], say by $M$. It follows that for all $n>(N+1)$,
$$ 0< s_n < s_{n-1}c < s_{n-2}c^2 < ... < s_k c^k$$
Let $M = \frac{s_k}{c^k}$, we obtain $0 < s_n < Mc^n$ for all $n>k = N+1$. Since $0<c<1$, $lim~c^n = 0$, and thus $lim~s_n = 0$.
But the definition of the existence of a limit is that the limit is true for all $\epsilon > 0$, so how can we restrict $\epsilon$ by assuming that $\epsilon =c-L$?, as we then have not proven the limit is true for all $\epsilon > 0$?
 A: Here, the author of the proof chose a real $r$ such that is equal to $r=c-L$. Replace $\epsilon$ name with $r$ in the proof and you'll be convinced.
You have been misled with a reflex consisting in associating $\epsilon$ to the $\epsilon-\delta$ definition. This is not what should be done here. By the way, the choice of $\epsilon$ naming by the author is not judicious.
A: Are you worried that the proof only covers small values of $\epsilon$?
This is sufficient since it is usually the small values that are challenging.  For example, if you can prove the limit for $\epsilon \le 1$ then then you can use your answer for $\epsilon = 1$ for anything bigger.  The $\delta$ does not need to be the best or most efficient answer in any sense.  Better than good enough is good enough.
A: The idea here isn't to take an arbitrary $\epsilon$ and show that some sequence converges. That is what you do in 99% of exercises, so it's an understandable misunderstanding.
However, in this case it goes in reverse: We freely choose an $\epsilon$ that suits our needs, and use the assumption that $\frac{s_n}{s_{n-1}}$ converges along with that $\epsilon$ to prove that $\frac{s_n}{s_{n+1}}$ is eventually below $c$.
