Find a function $f: \mathbb{R} \to \mathbb{R}$ that is continuous at precisely one point?

Solution 1:

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.

Solution 2:

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 $\mathbb{R}$.