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In Limit of function by convergent sequence, there is given a 'sufficient condition' in the proof; here is the excerpt:

Suppose that for each sequence $\langle x_n\rangle$ of points of S such that $\forall n\in \mathbb N_{>0}:x_n \ne c$ and $\displaystyle \lim_{n\to \infty}x_n=c,$ it is true that:

$$\lim_{n\to \infty}f(x_n)=l$$

What we will try to do is assume that it is not true that $\displaystyle \lim_{x\to c}f(x)=l,$ and try to find a contradiction.

So, if it is not true that $\displaystyle \lim_{x\to c} f(x)=l,$ then:

$$\exists \epsilon > 0: \forall \delta > 0: \exists x \in S: 0 < \mathrm d_1 \left({x, c}\right) < \delta \land \mathrm d_2 \left({f \left({x}\right), l}\right) \ge \epsilon$$

For all $n\in \mathbb N>0,$ define:

$$S_n = \left\{{x \in S: 0 < \mathrm d_1 \left({x, c}\right) < \dfrac 1 n \land \mathrm d_2 \left({f \left({x}\right), l}\right) \ge \epsilon}\right\}$$

By hypothesis, $S_n$ is non-empty for all $n\in N>0.$

Using the axiom of countable choice, there exists a sequence $\langle x_n\rangle$ of points in $S$ such that $x_n\in S_n ~~\forall n\in N>0.$

Then:

$\forall n \in \mathbb N_{>0}: x_n \ne c$ and $\displaystyle \lim_{n \to \infty} x_n = c$

but it is not true that:

$$ \lim_{n \to \infty} f \left({x_n}\right) = l$$

So there is our contradiction, and so the result follows.

I could conceive upto the application of $\rm{ACC}_;$ but couldn't understand what thy did after that.

First, they defined $S_n$ then after applying $\rm{ACC},$ how could they conclude $\forall n \in \mathbb N_{>0}: x_n \ne c$ and $\displaystyle \lim_{n \to \infty} x_n = c\;?$

Also, why is this proof the 'sufficient' condition and not the former one given in there?

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Every element $x_n$ is an element of $S_n$, so you know that for this element, two things hold:

  1. $0<d_1(x_n,c)<\frac 1n$
  2. $d_2(d(x_n), l) \geq\epsilon$

Specifically, the first property tells you that for each $n$, $0<d_1(x_n, c)$, which also implies that $x_n\neq c$

The first property also tells you that $$\lim_{n\to\infty} x_n=c$$

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  • $\begingroup$ also, can you tell why is this derivation the 'sufficient condition' and not the other provided there? $\endgroup$ – user142971 Jun 24 '16 at 6:03
  • $\begingroup$ @MAFIA36790 I don't understand that part of the question. The proof is of the type "If $A$, then $B$", which is just another way of saying that $A$ is a sufficient condition for $B$... $\endgroup$ – 5xum Jun 24 '16 at 6:07
  • $\begingroup$ That's what I wanted. Thanks. $\endgroup$ – user142971 Jun 24 '16 at 8:38

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