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Is the function $f(x) = \lim_{k \to +\infty} \tan{kx}$ a right example?

But I do not know whether this is a function at first. Is this a function?

Could you give a correct real-valued function that is continuous at precisely one point?


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Your example is a function that's only defined at a single point, $x=0$; the limit doesn't exist for any other $x$. – joriki Aug 27 '11 at 7:27
up vote 24 down vote accepted

The function $$f(x)=\begin{cases}x\text{ if }x\in\mathbb{Q}\\0\text{ if }x\notin\mathbb{Q}\end{cases}$$ is continuous at $x=0$ but nowhere else.

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More generally, if $\phi(x)$ is any bounded and nowhere continuous function, let $f(x) = x \phi(x)$. – Nate Eldredge Aug 27 '11 at 14:15
Why is the above example (given by Zev) continuous at zero? – Kara Apr 11 '13 at 21:45
@Kara: Let $x_n$ be any sequence converging to zero, then $f(x_n)$ is also a sequence converging to zero. – Asaf Karagila Apr 11 '13 at 21:47
@Kara because $-|x| \leq f(x) \leq |x|$ for all $x$. – N. S. Apr 11 '13 at 21:55
@Kara: The definition of "preimage of open set is open" is for continuity everywhere, not for continuity at a point. In metric spaces we have simpler ways to verify continuity. – Asaf Karagila Apr 11 '13 at 22:07

Your example isn't going to work because it actually isn't defined for any nonzero $x$. If $x$ is nonzero, then $kx$ gets either positively or negatively infinite as $k \to \infty$, and $\tan kx$ is going to zoom around periodically and so will not have a limit.

The standard example of a function continuous at only 1 point is something like the one given by Zev Chonoles, where $f$ takes the value of some continuous function (in this case, $f(x)=x$) on a dense subset of the reals that has a dense complement (in this case, $\mathbb{Q}$), and another continuous function (in this case, $f(x)=0$) on the complement. If the two functions coincide at exactly one point, then you get continuity at that point; everywhere else, the function bounces around crazily between the two continuous functions it is cobbled together from, because it takes each one's value on a dense subset of $\mathbb{R}$.

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What a great explanation! :-) – Joseph O'Rourke Aug 27 '11 at 13:03

The function $$f(x)=\begin{cases}0\text{ if }x\in\mathbb{Q}\\x\text{ if }x\notin\mathbb{Q}\end{cases}$$ is continuous at $x=0$ but nowhere else.

This is more natural I belive...

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very nice answer – Bhaskara-III Feb 27 at 12:03
Thanks @Bhaskara-III – cello Feb 27 at 12:09

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