Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Assertion: If $f:X\setminus\left\{a\right\}\to \mathbb{R}$ is continuous and there exists a sequence $(x_n):\mathbb{N}\to X\setminus\left\{a\right\}$ such as that $x_n\to a$ and $f(x_n)\to \ell$ prove that $\lim_{x\to a}f(x)=\ell$

I have three questions: 1) Is the assertion correct? If not, please provide counter-examples. In that case can the assertion become correct if we require that $f$ is monotonic, differentiable etc.?

2)Is my proof correct? If not, please pinpoint the problem and give a hint to the right direcition. Personally, what makes me doubt it are the choices of $N$ and $\delta$ since they depend on another

3)If the proof is correct, then is there a way to shorten it?

My Proof:

Let $\epsilon>0$. Since $f(x_n)\to \ell$ \begin{equation} \exists N_1\in \mathbb{N}:n\ge N_1\Rightarrow \left|f(x_n)-\ell\right|<\frac{\epsilon}{2}\end{equation} Thus, $\left|f(x_{N_1})-\ell\right|<\frac{\epsilon}{2}$ and by the continuity of $f$ at $x_{N_1}$, \begin{equation} \exists \delta_1>0:\left|x-x_{N_1}\right|<\delta_1\Rightarrow \left|f(x)-f(x_{N_1})\right|<\frac{\epsilon}{2} \end{equation} Since $x_n\to a$, \begin{equation} \exists N_2\in \mathbb{N}:n\ge N_2\Rightarrow \left|x_n-a\right|<\delta_1\end{equation} Thus, $\left|x_{N_2}-a\right|<\delta_1$ and by letting $N=\max\left\{N_1,N_2\right\}$, \begin{gather} 0<\left|x-a\right|<\delta_1\Rightarrow \left|x-x_N+x_N-a\right|<\delta_1\Rightarrow \left|x-x_N\right|-\left|x_N-a\right|<\delta_1\\ 0<\left|x-a\right|<\delta_1\Rightarrow \left|x-x_N\right|<\delta_1+\left|x_N-a\right| \end{gather} By the continuity of $f$ at $x_N$, \begin{equation} \exists \delta_3>0:0<\left|x-x_N\right|<\delta_3\Rightarrow \left|f(x)-f(x_N)\right|<\frac{\epsilon}{2} \end{equation} Thus, letting $\delta=\max\left\{\delta_1+\left|x_N-a\right|,\delta_3\right\}>0$ we have that, \begin{gather} 0<\left|x-a\right|<\delta\Rightarrow \left|x-x_N\right|<\delta\Rightarrow \left|f(x)-\ell+\ell-f(x_N)\right|<\frac{\epsilon}{2}\Rightarrow \left|f(x)-\ell\right|-\left|f(x_N)-\ell\right|<\frac{\epsilon}{2}\\ 0<\left|x-a\right|<\delta\Rightarrow\left|f(x)-\ell\right|<\left|f(x_N)-\ell\right|+\frac{\epsilon}{2}<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon \end{gather} We conclude that $\lim_{x\to a}f(x)=\ell$

Thank you in advance

EDIT: The proof is false. One of the mistakes is in this part:

"Thus, letting $\delta=\max\left\{\delta_1+\left|x_N-a\right|,\delta_3\right\}>0$ we have that, \begin{gather} 0<\left|x-a\right|<\delta{\color{Red} \Rightarrow} \left|x-x_N\right|<\delta{\color{Red} \Rightarrow} \left|f(x)-\ell+\ell-f(x_N)\right|<\frac{\epsilon}{2}\end{gather}"

share|cite|improve this question
You haven't said what $X$ is. Is it a subset of $\mathbb R$? – Jonas Meyer Jul 15 '12 at 19:13
Yes. It is a susbest of $\mathbb{R}$ – Nameless Jul 15 '12 at 19:33
up vote 3 down vote accepted

Your assertion is wrong. A counterexample is for instance given by the sign function, $sgn : \mathbb R \rightarrow \mathbb R$. The sign function is continuous on $\mathbb R\backslash \{0\}$, but $$ \lim_{n\rightarrow \infty} sgn(1/n) = 1, $$

and $$ \lim_{n\rightarrow \infty} sgn(-1/n) = -1. $$ Here $(1/n)$ and $(-1/n)$ are both sequences that converge to zero, but the sequences $(sgn(1/n))$ and $sgn(-1/n)$ are very much different.

The mistake in your proof is that the distance between an arbitrary point $x$ that is close to $a$ and members of the sequence does not become arbitrary small, so you don't have something like for all $\delta_3$ there is an $N_3$ such that $$ \vert x-x_n\vert≤\delta_3, ~~\text{for } n≥N_3. $$ But your proof would need something like this.

In our counterexample with the function $sgn$ this more or less means that if the sequence is given by $-(1/n)$ then I know something about $sgn(x)$ for negative $x$, but I can not say anything about the function values for positive $x$.

share|cite|improve this answer
How should we restict the sequence $x_n$ then? – Nameless Jul 15 '12 at 19:26
Don't you mean $\left|x-x_n\right|\le \delta_3$ for $n\ge N_3$? I never implied that. In which line exactly is the fault located? – Nameless Jul 15 '12 at 19:32
heuristically speaking, you would need something like a "spacefilling" sequence for your assertion to be true. Also you say in the equation that follows the sentence "By the continuity of f ..." that all $x$ for which $\vert x-x_N\vert ≤ \delta_3$ those $x$ also suffice $\vert f(x)-f(x_n)\vert < \epsilon/2$. But you forgot to think about what $x$ suffice $\vert x-x_N\vert ≤ \delta_3$ in the first place. – h.h.543 Jul 15 '12 at 19:43
Does it matter? We don't want to prove the existence of such $x$ but rather the implication ($\Rightarrow$) for the limit. Isn't that right? – Nameless Jul 15 '12 at 19:50

Another example: Let $f:\mathbb R\setminus\{0\}\to\mathbb R$ be defined by $f(x)=\mathrm{sign}(x)$, $x_n=\frac{1}{n}$. This example is also monotone and differentiable. However, it is not uniformly continuous, and a uniformly continuous function will have a limit at $a$.

share|cite|improve this answer

You need to have $f(x_n) \to l$ for all sequences $x_n \to a$, not just one sequence.

For example, let $a=(0,0)$ with $f(x,y) = \frac{x y}{x^2+y^2}$. This is continuous on $\mathbb{R}^2 \setminus \{a\}$, and the sequence $x_n=(\frac{1}{n},0) \to a$, with $f(x_n) \to 0$ (excuse abuse of notation), but $f$ is not continuous at $a$.

share|cite|improve this answer
@JonasMeyer: Yes, that is a much better counterexample. – copper.hat Jul 15 '12 at 19:18
I can prove that this is true if we have the ''all sequences'' part. The question is how can we determine a limit with just one sequence – Nameless Jul 15 '12 at 19:25
As all of the answers above show; you can't without additional constraints. – copper.hat Jul 15 '12 at 19:27
I agree. What are these are the additional constraints? Uniform continuity is one example. What can we say about the sequence (positive, monotonic) etc. – Nameless Jul 15 '12 at 19:40
If you are in $\mathbb{R}$, you just need the left and right limits to have the same finite value. Then you can define $f$ to have the appropriate value there. – copper.hat Jul 15 '12 at 19:45

There are some important examples in complex analysis, say $D$ is the unit disk in $\mathbb C = \mathbb R^2$ (so not in $\mathbb R$ as in this question). Some examples of functions, analytic and hence continuous in $D$ are studied, where radial limits exist, but not tangential limits. These will be counterexamples to what you ask in that case.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.