Proof of the sequential criterion for limits Let $f:D\to\mathbb R$ and let $c$ be an accumulation point of $D$. Then
$(i)\lim_{x\to c}f(x)=L$ iff $(ii)$ for every sequence $(s_n)$ in $D$ that converges to $c$ with $s_n\neq c$ the sequence $(f(s_n))$ converges to $L$
I'm okay with one direction.
To prove the other direction (taking the contrapositive statement): 
Suppose $L$ is not a limit of $f$ at $c$. Find a sequence $s_n$ in $D$ such that $s_n$ converges to $c$ but $(f(s_n))$ does not converge to $L$
Since $L$ is not a limit of $f$ at $c$, $\exists\epsilon>0$ such that $\forall\delta>0$ $\exists x\in D$ such that $0<|x-c|<\delta$ implies $|f(x)-L|\ge\epsilon$.
Now the book I'm reading, Steven Lay's "Analysis with an introduction to proof" goes on as follows:
" In particular, for each $n\in\mathbb N$, there exists $s_n\in D$ with
$0<|s_n-c|<1/n$ such that $|f(s_n)-L|\ge\epsilon$"
Thus exhibiting $(s_n)$ as the required sequence.
I'm not sure why is it required that $\delta$ must be related to $1/n$
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ok, I want to show that there exists a sequence $s_n$ that converges to $c$ such that $(f(s_n))$ does not converge to $L$
Let $s_n$ coverge to $c$. Then $\forall \delta>0 \exists N\in \mathbb N$ such that $n\ge N \to |s_n-c|<\delta$
Now I want to make this statement into:
$\forall \delta>0 \exists s_n \in D$ such that $|s_n-c|<\delta$
please detail how that happens.
 A: The reason for the $1/n$ is that he is trying to produce a sequence that, by construction, converges to $c$.  It would work just as well if you replaced $1/n$ with any other sequence of positive numbers that converge to $0$.  Say $1/log(n)$ or $0.5^n$.
As for the argument that you gave, it is a fine proof of a different statement.  Namely that if $L$ is the limit, then all sequences must converge to $L$.  But it says nothing about what happens if $L$ is not the limit.  (He's trying to prove that if $L$ is not the limit, then some sequence $s_n$ converges to $c$ but $f(s_n)$ does not converge to $L$.).
A: This is to see if I'm getting this straight, so please critique my answer.
To show $\lnot(ii)$, I want to find a sequence $s_n$ such that $s_n$ converges to c but $f(s_n)$ does not converge to L, given $\lnot(i)$.
$\lnot(i)$ states that $\exists\epsilon>0$ such that $\forall\delta>0$  $\exists x\in D$ such that $0<|x-c|<\delta$ implies $|f(x)-L|\ge \epsilon$
Let $s_n$ be a convergent sequence such that $\forall n \in \mathbb N$  $[s_n\in D]$.
Then $\forall\delta>0$  $\exists N\in \mathbb N$ such that $n \ge N$ implies 
$0<|s_n-c|<\delta$ (since $\forall n \in \mathbb N$ $s_n \neq c$).
Since the existence on $x \in D$ in $\lnot (i)$  is dependent on $\delta$
I now must construct the sequence so that $\lnot (i)$ holds for all $n$, rather than just $n \ge N$, which requires $\delta$ to be related to the index of the sequence.
Thus, by $\lnot (i)$, for $\delta_n=1/n$,
$\exists s_1$ such that $0<|s_1-c|<\delta_1=1$
$\exists s_2$ such that $0<|s_2-c|<\delta_2=1/2$
$\exists s_3$ such that $0<|s_3-c|<1/3$
...
$\exists s_n$ such that $0<|s_n-c|<1/n$
Since $\forall n \in \mathbb N$ $s_n \in D$, $\lnot (i)$ applies and $0<|s_n-c|<1/n=\delta_n$ implies $|f(s_n)-L| \ge \epsilon$
