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theorem: let $ f:\mathbb{A} \to \mathbb{R} $ where $ \mathbb{A} \subseteq \mathbb{R} $ and let $c$ be a cluster point of $\mathbb{A}$ ,then: if for every sequence $(x_{n})$ in $\mathbb{A}$ that converges to c such that $x_{n} \not=c $ for all $ n \in \mathbb{N}$, the sequence $(f(x_{n}))$ converges to L, then this implies $\lim_{x \to c} f(x)=L$

my teacher's proof:

assume that $\lim_{x\to c} f(x)\not= L$

this means $\exists \epsilon \in \mathbb{R^+},\forall \delta \in \mathbb{R^+}, $ such that $|f(x_{\delta})-L|>\epsilon $ for atleast one $x_{\delta}$ in the $\delta$ neighbourhood of $c$ (denoting with subscript $ \delta $ implying this $x$ depends on $ \delta $)

my teacher here implied that there exists a sequence $(x_{n})$ such that this sequence converges to c, and contains this particular $x_{\delta}$ and from the statement in the theorem " if for every sequence $(x_{n})$ in $\mathbb{A}$ that converges to c such that $x_{n} \not=c $ for all $ n \in \mathbb{N}$, the sequence $(f(x_{n}))$ converges to L "

says that the sequence $ f(x_{n})$ converges to L (from the initial statement of the theorem), and then says, from the assumption that $ |f(x_{n})-L|>\epsilon $ whenever $x_n$ is in the delta neighbourhood of c.

from here a contradiction arises, and he goes on to say that the assumption that $\lim_{x\to c} f(x)\not= L$ is wrong.

My belief is that this proof is wrong, or at best incomplete, since one cannot say that $x_{\delta}$ readily belongs to the sequence in question. and even if one constructs a sequence using all the $ x_{\delta}$s, it's not necessary that the resulting sequence $((x_{\delta})_{n})$ converges to c. kindly check this proof

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    $\begingroup$ Hint: consider the sequence $\delta_n = 1/n$. This should allow you to construct a sequence $(x_n) := (x_{\delta_n})$ that converges to $c$ and that satisfies $|f(x_n) - L| > \epsilon$ for every $n \in \mathbb{N}$ $\endgroup$
    – david_sap
    Feb 20, 2023 at 10:53
  • $\begingroup$ dear @david, I have seen this proof in bartle and sherbert, considering delta to be 1/n, and constructing a sequence using this, and this is my go-to proof, which i agree with. my problem is solely with my teacher's proof, who takes this for any delta and argues that this sequence converges to c. $\endgroup$ Feb 20, 2023 at 10:59
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    $\begingroup$ I'd say that the provided version doesn't really make sense: when it says "there exists a sequence $(x_n)$ such that this sequence [...] contains this particular $x_\delta$", there is no "particular" $\delta$ here: you get a $\delta$ for every $\epsilon$, but you didn't fix any value for $\epsilon$. $\endgroup$
    – david_sap
    Feb 20, 2023 at 11:05

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For some $\varepsilon > 0$, by taking $\delta = 1/n$ for $n \geq 1$, there exists $x_n$ in the $\delta$-neighborhood of $c$ such that $|f(x_n)-L|>\varepsilon$ (that you denoted $x_\delta$). This constructs a sequence $(x_n)_{n \geq 0}$ that, by definition, converges to $c$, and such that $(f(x_n))_{n \geq 0}$ cannot converge to $L$, resulting in a contradiction.

It is by considering $\delta$ smaller and smaller ($\delta=1/n$ or any positive sequence converging to zero) that you are able to construct a sequence of points that contradicts the assumption.

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  • $\begingroup$ I agree with this proof @Jpole, but could you verify my teachers assumption that this sequence converges for any delta, not just for the special case where delta=1/n $\endgroup$ Feb 20, 2023 at 11:01

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