# Check my teacher's proof for sequential criterion of limits of functions

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

• 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 $n \in \mathbb{N}$ Feb 20, 2023 at 10:53
• 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. Feb 20, 2023 at 10:59
• 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 $\epsilon$. Feb 20, 2023 at 11:05

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.