# Does $\lim\limits_{x \rightarrow c} f(x)$ exist if the sequence $\{ f(x_n)\}_{n=1}^\infty$ is Cauchy?

I'm struggling a little with this question:

Let c be a cluster point of $$A ⊂ \mathbb{R}$$, and $$f : A → \mathbb{R}$$ be a function. Suppose for every sequence $$\{x_n \}$$ in A, such that $$\lim x_n = c$$, the sequence $$\{ f (x_n )\}_{n=1}^{\infty}$$ is Cauchy. Prove that $$\lim_{x \rightarrow c} f(x)$$ exists.

I also have the following lemma, which should probably help a lot:

Let $$S ⊂ \mathbb{R}$$ and $$c$$ be a cluster point of $$S$$. Let $$f:S→ \mathbb{R}$$ be a function. Then $$f(x)→L$$ as $$x→c$$ if and only if for every sequence $${x_n}$$ of numbers such that $$x_n ∈ S - \{c \}$$ for all n, and such that $$\lim x_n = c$$, we have that the sequence $${ f (x_n)}$$ converges to L.

I feel like I'm so close but I'm missing something incredibly obvious.

Following the hint from Robert Shore, this is my attempt:

We try to prove that all sequences $${ f (x_n)}$$ converge to a unique limit $$L$$. Assume for sake of contradiction that there are two sequences $$\{x_{n_1}\} , \{x_{n_2}\}$$ where $$\lim \{x_{n_1}\} = \lim \{x_{n_2}\} = c$$ but $$\lim \{f(x_{n_1})\} = L_1 \neq \lim \{f(x_{n_2})\} = L_2$$. Then from the epsilon-delta definition of limits, we have: $$\forall \epsilon > 0, \exists M_1, \forall n \geq M_1 \implies |f(x_{n_1}) - L_1| < \epsilon/2$$ $$\forall \epsilon > 0, \exists M_2, \forall n \geq M_2 \implies |f(x_{n_2}) - L_2| < \epsilon/2$$ Then $$|L_1 - L_2| = |f(x_{n_1}) - L_1 - (f(x_{n_2}) - L_2 )|$$ $$\leq |f(x_{n_1}) - L_1| + |f(x_{n_2}) - L_2|$$ $$< \epsilon/2 + \epsilon/2 =\epsilon$$ Since $$|L_1 - L_2| < \epsilon$$ for all $$\epsilon > 0, |L_1-L_2| = 0$$ and so $$L_1 = L_2$$

I somehow feel like this proof is going in the wrong direction - I'm not at all using the fact that $$\lim x_n = c$$

• By $R$ do you mean the real numbers $\Bbb R$? Or is $R$ some more general space? – Robert Shore Mar 5 at 22:14
• R is indeed the real numbers. I'll make this edit. – JaP Mar 5 at 22:15
• The proof is ok, except you have to choose an index $N$ such that the two inequalities you wrote are satisfied and then add and subtract $f(x_N)$. You're using that the limit of the two sequences is $c$ both because you're saying that the limits of the sequences $f(x_{n_i}})$ exist and in applying the Lemma you cited afterwards to claim that the limit of the function exists. – Federico Mar 5 at 23:07

Here's an outline of the proof. Since $$\Bbb R$$ is complete, every Cauchy sequence converges. Therefore, for every sequence $$\{x_n \}$$ converging to $$c$$, we have that $$\{f(x_n) \}$$ converges. All you need to prove is that all such sequences $$\{f(x_n) \}$$ converge to the same limit. (Let us know if you need help with that.) Once you know that, call that unique limit $$L$$ and then apply the Lemma.
• If two sequences $\{f(x_n) \}$ and $\{f(y_n) \}$ go to two different limits $L_1$ and $L_2$, consider what happens to the sequence obtained when you "interleave" the two sequences: $\{f(x_1), f(y_1), f(x_2), f(y_2), \ldots \}.$ It has to be Cauchy (because since $x_n$ and $y_n$ both converge to $c$, the interleaved sequence $\{x_1, y_1, x_2, y_2, \ldots \}$ also has to converge to $c$), but the difference between $L_1$ and $L_2$ ensures that it can't be. – Robert Shore Mar 5 at 23:10