# Trouble in Proving the Sequential Criterion for Limits at Infinity

Let $$A\subseteq R,$$ let $$f : A\to R,$$ and suppose that $$(a,\infty)\subseteq A$$ for some $$a \in R.$$ Then the following statements are equivalent:

(i) $$L=\lim_{x\to \infty}f(x)$$

(ii) For every sequence $$(x_n)$$ in $$A \cap (a,\infty)=(a,\infty)$$ such that $$\lim x_n=\infty$$, the sequence $$f(x_n)$$ converges to $$L.$$

I tried proving this in the following way:

First we try to show, $$(i)\implies (ii).$$ If, $$\lim_{x\to\infty}f(x)=L$$ then this means,

For every $$\epsilon\gt 0$$ there exists a $$K>a$$ such that if $$x\geq K,$$ we have $$|f(x)-L|\lt \epsilon.$$

Now, if $$(x_n)$$ is a sequence in $$A \cap (a,\infty)=(a,\infty)$$ such that $$\lim x_n=\infty$$ then, $$\exists M\in\Bbb N$$ such that $$x\gt K(\gt a),$$ when $$n\geq M.$$ But if, $$x_n>K$$ we have, $$|f(x_n)-L|\lt \epsilon.$$ To sum, we have shown that, if $$n\geq M$$, then, $$|f(x_n)-L|\lt \epsilon.$$ But $$\epsilon \gt 0$$ and $$(x_n)$$ was arbitrary and this implies, that for every sequence $$(x_n)$$ in $$A \cap (a,\infty)=(a,\infty)$$ such that $$\lim x_n=\infty$$, the sequence $$f(x_n)$$ converges to $$L.$$

However, while to show the converse, I think I make my argument faulty:

Next, we try to show, that $$(ii)\implies (i).$$

If for every sequence $$(x_n)$$ in $$A \cap (a,\infty)=(a,\infty)$$ such that $$\lim x_n=\infty$$, the sequence $$f(x_n)$$ converges to $$L,$$ then we fix, an arbitrary $$\epsilon\gt 0.$$

Now, we denote, each sequence as $$X_i=(x_n^{(i)})$$, where $$i\in \Bbb N$$ and $$X_i\in A \cap (a,\infty)=(a,\infty)$$. If $$\lim X_i=\infty$$ for a particular $$i\in Bbb N$$ then $$\exists M_i\in \Bbb N$$ such that, $$|f(x_n^{(i)}-L|\lt \epsilon$$ for all $$n\geq M.$$ For each, $$X_i$$ we obtain, an $$M_i$$ and if, $$M=\max(M_i)_{i\in \Bbb N},$$ we may take, $$x\geq M$$ and then, $$|f(x)-L|\lt\epsilon.$$ Hence, we claim that, $$L=\lim_{x\to \infty}f(x).$$

I doubt my proof about the converse statement as, $$\max\{\text{infinite set}\}$$ has no meaning at all. We can take $$\max$$ element in a finite set and not in an infinite one. Secondly, in the part, where I wrote, "$$x\geq M$$ and then, $$|f(x)-L|\lt\epsilon$$" might be faulty as well. This is because, if $$x\geq M$$ then, $$|f(x)-L|\lt\epsilon$$ will be true, only if, $$x$$ is in some properly diverging sequence $$X_i$$ tending to $$\infty.$$ But, how can we gurantee, that $$x$$ will be in such a sequence $$X_i$$ just like that?

For these two reasons, I feel my argument breaks down.

All in all, I don't understand how to show, $$(ii)\implies (i)$$ ?

• For $(ii) \implies (i)$, try the contrapositive. Commented Aug 26, 2023 at 7:29
• @terran Contrapositive of what ? $(i)\to (ii)$ ? Commented Aug 26, 2023 at 7:38
• The contrapositive of $(ii) \implies (i)$ is $\text{not } (i) \implies \text{not } (ii)$. Commented Aug 26, 2023 at 7:41
• Yes, that's the approach I meant! :-) Commented Aug 26, 2023 at 16:08
• @terran Thanks a ton for confirming ! Commented Aug 26, 2023 at 16:12

We can prove that $$(ii)\implies (i)$$ in a much simpler way. We show this, by proving that the contrapositive of the statement i.e $$\neg (i)\implies \neg (ii)$$ is true.

The contrapositive translates to the below statement:

If $$\lim_{x\to\infty}f(x)\neq L$$ then $$\exists (x_n)\in A\cap (a,\infty)$$ satisfying, $$\lim x_n=\infty$$ such that, $$\lim f(x_n)\neq L\tag 1$$

This is true, for if $$\lim_{x\to\infty}f(x)\neq L$$ then, $$\exists \epsilon\gt 0$$ such that, $$\forall K_i\gt a$$ $$\exists x'_i\gt K_i$$ we have, $$|f(x'_i)-L|\geq \epsilon.$$ We consider the sequence $$(x_i')\in A\cap (a,\infty)$$ and note that, $$\lim x_i' =\infty.$$

This can readily be verified if, $$\alpha\in \Bbb R$$ then, $$\exists K_{i_0}\in\Bbb N$$ such that $$K_{i_0}\gt \alpha$$ then, $$x'_{i_0}>K_{i_0}>\alpha$$ and hence, $$\forall i\geq i_0$$ we have, $$x'_i\gt K_i\gt K_{i_0}\gt\alpha$$ so that $$x'_i\gt \alpha$$ for all $$i\geq i_0.$$ Hence, $$\lim x'_i=\infty$$ follows.

But we note that, $$|f(x'_i)-L|\geq \epsilon$$ for all $$i\in \Bbb N.$$ So, the required sequence in $$(1)$$ is $$(x_i').$$ So, $$(1)$$ is indeed true and so, $$(ii)\implies (i).$$

The portion, $$(i)\implies (ii)$$ in OP seems correct.

As hinted in the comments, you can show that if $$\lim_{x \to \infty} f (x) \ne L$$, then there is some sequence $$(x_n)$$ (where $$x_n > a$$ for each $$n$$) with $$\lim_{n \to \infty} x_n = \infty$$ such that $$(f (x_n))$$ does not converge to $$L$$.

Since $$\lim_{x \to \infty} f (x) \ne L$$, then there is some open ball $$C = B (L, \varepsilon)$$ such that $$f^{-1} (C)$$ does not contain any set of the form $$A \cap (r, \infty)$$, where $$r \in \mathbb{R}$$. Since $$(2 |a|, \infty) \subseteq (a, \infty) \subseteq A$$, this implies there is some element $$x_1 > 2 |a| \ge 0$$ of $$A$$ such that $$f (x_1) \notin C$$. Now, since $$(2 x_1, \infty) \subseteq (a, \infty) \subseteq A$$, there is some $$x_2 > 2 x_1$$ contained in $$A$$ such that $$f (x_2) \notin C$$. Continue inductively so that we get a sequence of points of $$A$$

$$x_1 < x_2 < \dots$$

each of which is positive and none of whose images is contained in $$C$$. We claim that this sequence satisfies the desired properties.

It is easy to see that $$(f (x_n))$$ does not converge to $$L$$, since the neighborhood $$C$$ of $$L$$ contains no points of this sequence (by design).

To show that $$\lim_{n \to \infty} x_n = \infty$$, assume for the sake of contradiction that $$x_n < s$$ for each $$n$$ for some $$s \in A$$. Note that $$s$$ must be positive since each $$x_n$$ is positive. By design, $$x_{n + 1} > 2 x_n$$ for each $$n$$, so that $$x_n > x_1 (2)^{n - 1}$$. So, we have that $$2^{n - 1} < \frac{x_n}{x_1} < \frac{s}{x_1} \implies n < 1 + \log_2 \left( \frac{s}{x_1} \right)$$ for each positive integer $$n$$, clearly a contradiction.