# Proof verification: $(x_n)\rightarrow c$ $\implies$ $f(x_n) \rightarrow L$ iff $\lim_{x \rightarrow c} = L$

I am trying to develop a non-topological proof for the Sequential Criterion for Functional Limits. Can someone please check the accuracy of my proof? Thanks!

Sequential Criterion for Functional Limits: Let $$f: A\to\mathbb{R}$$. Given $$c$$ is a cluster point in $$A$$. Prove that the following statements are equivalent:

(a) $$\lim_{x \rightarrow c} = L$$.

(b) For all sequences $$(x_n) \subseteq A$$ with $$x_n \neq c$$ and $$(x_n)\rightarrow c$$ $$\implies$$ $$f(x_n) \rightarrow L$$

Proof. Suppose $$\lim_{x \to c} = L$$. Then, $$\forall \epsilon > 0, \exists \delta > 0,$$ s.t.$$0 < \mid x-c \mid < \delta \implies \mid f(x)-L \mid < \epsilon$$. Now, let $$(x_n) \subseteq A$$ with $$x_n \neq c$$ and $$(x_n)\rightarrow c$$. Pick $$\delta = \epsilon$$. Then, $$\forall \epsilon >0, \exists N \in \mathbb{N}$$ s.t. $$n \geq N \implies 0< \mid x_n - c \mid < \delta$$ which implies $$\mid f(x_n) - L \mid < \epsilon.$$

(Contrapositive) Conversely, suppose $$\lim_{x \to c} \neq L$$. Then, $$\exists \epsilon_0 > 0$$ s.t. $$\forall \delta > 0$$ it is that $$0 < \left| x-c \right| < \delta \implies \left| f(x) - L \right| \geq \epsilon_0$$. Note that this proposition is true for all $$x \in$$ dom $$f$$. Now, let $$(x_n) \subseteq A$$ be arbitrary with $$x_n \neq c$$. Then, based on our argument so far, we are justified in concluding that $$\forall n$$, it is that $$0< \left| x_n - c \right| < \delta \implies \left| f(x_n) - L \right| \geq \epsilon_0$$ are we are done.

• Missin $f(x)$ in the limit. – leo May 9 at 5:41

The first part of your proof is fine.

As for the converse, try proving its contrapositive.

• Thanks for the suggestion! I've edited my original proof to accommodate your suggestion. Is the "converse" direction of this proof correct or is it missing some details? – Ricky_Nelson May 6 at 20:52