Prove $\lim_{x\to a}f(x)$ exists iff for every $\epsilon>0$ there exists a $\delta>0$ s.t. for all $x_1,\,x_2\in D\cap (B(a,\delta)\setminus\{a\})$

Can I please get feedback on my proof below? Thank you.

$$\def\a{{\bf a}} \def\b{{\bf b}} \def\f{{\bf f}} \def\r{{\bf r}} \def\x{{\bf x}} \def\y{{\bf y}} \def\R{{\mathbb R}} \def\Z{{\mathbb Z}} \def\N{{\mathbb N}}$$

Let $$\f\colon D\to \R^m$$, $$D\subseteq\R^n$$, and let $$\a$$ be a cluster point of $$D$$. Prove $$\displaystyle{\lim_{\x\to\a}\f(\x)}$$ exists if and only if for every $$\epsilon>0$$ there exists a $$\delta>0$$ such that for all $$\x_1,\,\x_2\in D\cap (B(\a,\delta)\setminus\{\a\})$$, $$\|\f(\x_1) - \f(\x_2)\|<\epsilon$$. Essentially, we are using a continuous version of the Cauchy Critierion.

$$\textbf{Solution:}$$ Suppose $$\displaystyle{\lim_{\x\to \a}\f(\x)}$$ exists, and let $$\epsilon >0$$. Then, there exists $$\delta > 0$$ such that $$||\f(\x)-\f(\a)|| <\frac{\epsilon}{2}$$ for all $$\x \in (B(\a,\delta)\setminus\{\a\}) \cap D$$. Let $$\x_1,\,\x_2\in (B(\a,\delta)\setminus\{\a\})\cap D$$. Then, $$||\f(\x_1)-\f(\a)|| <\frac{\epsilon}{2}$$ and $$||\f(\x_2)-\f(\a)|| <\frac{\epsilon}{2}$$. Therefore, by triangle inequality, $$||\f(\x_1)-\f(\x_2)|| \le ||\f(\x_1)-\f(\a)|| + ||\f(\x_2)-\f(\a)|| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$$

Conversely, we must show there do not exist two sequences $$\x_n \to \a$$ and $$\y_n \to \a$$, with $$\f(\x_n)\to L_1$$ and $$\f(\y_n) \to L_2$$ and $$L_1 \neq L_2$$, because if they do exist, then the limit can not exist.

Let $$L_2>L_1$$. Then, we can choose $$\epsilon>0$$ such that for any $$x\in B(L_2;\epsilon)=(L_2-\epsilon,L_2+\epsilon)$$ and for any $$y\in B(L_1;\epsilon)=(L_1-\epsilon,L_1+\epsilon)$$, we have $$|x-y|>\epsilon$$. This follows since $$|\x-\y|\geq \x-\y>(L_2-\epsilon)-(L_1+\epsilon)=(L_2-L_1)-2\epsilon$$, which is greater than $$\epsilon$$ if $$\epsilon<\frac{L_2-L_1}{3}$$.

Choose such an $$\epsilon$$. By hypothesis, there exists $$\delta_1>0$$ such that if $$\x,\y \in B(\a;\delta_1)\backslash\{\a\}$$, then $$|\f(\x)-\f(\y)|<\epsilon$$. Since $$\x_n \to \a$$ and $$\f(\x_n)\to L_1$$, there exists $$N_{\x}$$ such that for all $$n\geq N_{\x}$$, we have $$\x_n \in B(\a;\delta)\backslash\{\a\}$$ and $$\f(\x_n)\in B(L_1;\epsilon)$$. Similarly, there exists $$N_{\y}$$ such that for all $$n\geq N_{\y}$$, we have $$\y_n \in B(\a;\delta)\backslash\{\a\}$$ and $$\f(\y_n) \in B(L_2;\epsilon)$$. Take any $$n_0 \geq \max\{N_x,N_{\y}\}$$. Then $$\x_{n_0},\y_{n_0}\in B(\a;\delta_1)\backslash\{\a\}$$, but $$|\f(\x_{n_0})-\f(\y_{n_0})|\geq \f(\x_{n_0})-\f(\y_{n_0})>(L_2-L_1)-2\epsilon>\epsilon,$$ a contradiction.

In the higher dimension case, the argument follows that $$\epsilon>0$$ can be chosen small so that $$d(B(L_1;\epsilon),\, B(L_2;\epsilon))>\epsilon$$.We wish to have $$\epsilon$$ sufficiently small such that $$d(B(L_1, \epsilon), B(L_2,\epsilon)) > \epsilon$$. So, let $$F(\epsilon) = d(B(L_1, \epsilon), B(L_2,\epsilon)) - \epsilon$$. Then $$F(0) = ||L_2-L_1|| > 0$$. Hence, for $$\epsilon$$ sufficiently small, $$F(\epsilon) > 0$$.

• Your premise for the converse is not correct. Having no limit does not mean that there exists $x_n, y_n \to a$ with $f(x_n)$ and $f(y_n)$ having different limits (think $f(x) = 1/x$ at $0$). But you can definitely use sequences, and use the characterization of limits with sequences: take $x_n \to a$; show that $(f(x_n))$ is convergent (because Cauchy)\$; and finally check that the limit does not depend on the chosen sequence. Commented Apr 14, 2020 at 13:39
• @Raoul thank you, I fixed it. Does it look fine now? Commented Apr 14, 2020 at 13:43
• I am sorry, but I do not see the difference. You are still beginning the second part of the argument with "Conversely, we must show there aren't two sequences...", which is not a right way to start. Commented Apr 14, 2020 at 13:48
• I see. Then, I am confused about how to prove the converse part now. Thank you though @Raoul Commented Apr 14, 2020 at 13:50
• Do you know the sequential characterization of limits? That is, how to show that a function has a limit using sequences? Commented Apr 14, 2020 at 13:57

We want to check the sequential characterization of limits: there is a $$L \in \mathbb{R}^m$$ such that, for any sequence $$(x_n)$$ with $$x_n \to a$$ and $$x_n \neq a$$ for all $$n$$, we have $$f(x_n) \to L$$.

So consider such a sequence $$(x_n)$$. Take $$\epsilon, \delta > 0$$ as in the assumption. Since $$x_n \to a$$, for $$m,n$$ large enough, we have $$x_n, x_m \in B(a,\delta) \backslash \{a\}$$, and thus $$\| f(x_n) - f(x_m)\| < \epsilon.$$ Therefore $$(f(x_n))$$ is Cauchy, so it converges, say to $$L$$.

Now, if $$(y_n)$$ is another sequence such that $$y_n \to a$$ and $$y_n \neq a$$ for all $$n$$, then also $$(f(y_n))$$ converges, say to $$L'$$. But take $$\epsilon, \delta > 0$$ as in the assumption. For $$n$$ large enough, we have the following.

• First, $$x_n, y_n \in B(a,\delta) \backslash \{a\}$$, and thus $$\| f(x_n) - f(y_n)\| < \epsilon$$.
• Additionally, $$\|f(x_n) - L \| < \epsilon$$.
• Finally, $$\|f(y_n) - L' \| < \epsilon$$.

By triangle inequality, this gives $$\| L - L' \| < 3 \epsilon$$. As this is true for any $$\epsilon$$, this means that $$L = L'$$. Finally, this shows that for any sequence $$(x_n)$$ with $$x_n \to a$$ and $$x_n \neq a$$ for all $$n$$, we have $$f(x_n) \to L$$, which means that $$\lim_{x \to a, \: x \neq a} f(x) = L.$$

• Thank you so much Raoul. It is much clearer to me now. Commented Apr 14, 2020 at 14:43