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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}"

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    You haven't said what $X$ is. Is it a subset of $\mathbb R$?2012-07-15
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    Yes. It is a susbest of $\mathbb{R}$2012-07-15

4 Answers 4

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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.

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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 $a$.

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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 $x$.

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    How should we restict the sequence $x_n$ then?2012-07-15
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    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?2012-07-15
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    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.2012-07-15
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    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?2012-07-15
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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 $a$.

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    @JonasMeyer: Yes, that is a much better counterexample.2012-07-15
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    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 sequence2012-07-15
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    As all of the answers above show; you can't without additional constraints.2012-07-15
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    I agree. What are these are the additional constraints? Uniform continuity is one example. What can we say about the sequence (positive, monotonic) etc.2012-07-15
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    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.2012-07-15
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    So this proof could work (with modifications of course) for one sided limits and limits at infinity?2012-07-15
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    @Nameless: No, it is false in those situations, too. E.g. let $f:\mathbb R\setminus\{0\}\to\mathbb R$ be defined by $f(x)=\sin\left(\frac{1}{x}\right)$, $x_n=\frac{1}{\pi n}$ to see that it doesn't work for one-sided limits, and $f(x)=\sin(x)$, $x_n=\pi n$ to see that it doesn't work for limits at infinity.2012-07-15
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    @JonasMeyer Let me change the equestion: A rigorous calculus book that I have been reading defines the logarithmic function $\log(x)$ as the integral of $1/x$. For the proof of $\lim_{x\to \infty} \log(x)=+\infty$ he easily proves that $\lim_{n\to \infty} \log(2^n)=+\infty$ and writes "by the continuity of $\log(x)$ we have te result". But why?? That's why I asked this question. What's so special about that case?2012-07-15
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    @Nameless: Instead of trying to change this question, you could ask a new question where you make clear what your new specific question is. (It is not continuity, but monotonicity of the log in that case. Monotone functions have (finite or infinite) one-sided limits and limits at infinity, but that is a separate question from what you asked.)2012-07-15
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    @JonasMeyer Very well. If I don't come up with any answer I will ask it tomorrow then.2012-07-15