I'm having some trouble understanding the second half of this proof that proves the following theorem:
Let f : A → R, where A ⊂ R, and suppose that c ∈ R is an accumulation point of A. Then limx→c f(x) = L if and only if limn→∞ f(xn) = L. for every sequence (xn) in A with xn 6= c for all n ∈ N such that limn→∞ xn = c.
The first half of the proof:
First assume that the limit exists and is equal to L. Suppose that (xn) is
any sequence in A with xn 6= c that converges to c, and let > 0 be given. From
Definition 6.1, there exists δ > 0 such that |f(x) − L| < whenever 0 < |x − c| < δ,
and since xn → c there exists N ∈ N such that 0 < |xn − c| < δ for all n > N. It
follows that |f(xn) − L| < whenever n > N, so f(xn) → L as n → ∞.
The first half of this proof I follow; we simply show that for a sequence, x_n, a delta exists when the sequence converges to c, which implies an epsilon must exist to show the limit of f(x_n) = L.
What I can't understand is the following (picture used as I don't know how to format is properly):
The basis of it is proving the converse of the above proof when the limit does not exist. I'm specifically stuck on where they pulled the inequality 0 < |x_n - c| < 1/n from: this doesn't seem to be a rule anywhere that I have found and I haven't the faintest how they got that result.
If anyone could give some tips on what's happening here it would be incredibly appreciated.