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

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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, $F(\epsilon) > 0$.

  • $\begingroup$ 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. $\endgroup$
    – Raoul
    Commented Apr 14, 2020 at 13:39
  • $\begingroup$ @Raoul thank you, I fixed it. Does it look fine now? $\endgroup$ Commented Apr 14, 2020 at 13:43
  • $\begingroup$ 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. $\endgroup$
    – Raoul
    Commented Apr 14, 2020 at 13:48
  • $\begingroup$ I see. Then, I am confused about how to prove the converse part now. Thank you though @Raoul $\endgroup$ Commented Apr 14, 2020 at 13:50
  • $\begingroup$ Do you know the sequential characterization of limits? That is, how to show that a function has a limit using sequences? $\endgroup$
    – Raoul
    Commented Apr 14, 2020 at 13:57

1 Answer 1


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 $$ \lim_{x \to a, \: x \neq a} f(x) = L. $$

  • 1
    $\begingroup$ Thank you so much Raoul. It is much clearer to me now. $\endgroup$ Commented Apr 14, 2020 at 14:43

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