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By the sequential characterization of definition of limit, suppose $f:D\to \mathbb{R}$, where $D\subset \mathbb{R}$, and let $a\in D$. If we want to show $\lim_{x\to a}f(x)= L$ by sequential characterization, we take $\{x_n\} \in D\setminus \{a\}$ with $\{x_n\} \to a$, we have $f(x_n) \to L$.

Question: Assuming the sequential definition of limit, and prove the epsilon-delta formulation of limit. One such proof is by contradiction, i.e. $0<|x_n - a |< \delta$ we have $|f(x_n)-L|\ge \epsilon$ for $n \ge N$ and it contradicts with the hypothesis. For $\delta$, the analysis book that I am using says it is "standard practice" to take $\delta=\frac{1}{n}$. Why $\frac{1}{n}$? Is it because $\frac{1}{n}$ is less than any arbitrary small number if we take $n$ large enough by Archimides' Principle? The book does not mention (nor define) $a$ as limit point, and imply that for each $n \in \mathbb{N}$, there exists $\{x_n\} \in D\setminus \{a\}$ such that $0<|x_n - a |<\frac{1}{n}$. How do we know we can find such a sequence in $D\setminus \{a\}$?

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Your book is a mess! How can it speak of limits without any assumption on $a$ and $D$? The existence of the sequence $\{x_n\}$ is a consequence of the fact that $a$ must be an accumulation point for $D$. Otherwise, it is false, in general. –  Siminore Jul 4 '12 at 8:53
    
Why the downvote??? –  abatkai Jul 4 '12 at 8:57
    
I think limit point really makes a difference. I mean in Rudin it is pretty clear that the point $a$ in my question is definitely a limit point. –  Daniel Jul 4 '12 at 9:13

1 Answer 1

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Answering for the sake of completeness;

(A) there is no particular reason to use $1/n$, but it is arguably the simplest sequence converging to $0$. This is why it is found on every page of an introductory analysis book.

(B) As @Siminore said, it is impossible to discuss limit at a point that is not a limit point of the domain of definition.

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