TITLE: Differentiable at a point
QUESTION [11 upvotes]: My roommates and I have an argument you guys can help to settle (peace is at stake, don't let us down!) In undergrad calculus courses, one usually explains what it means for a function to be differentiable at a point x, and then differentiable in a domain. Then the focus is entirely on this latter notion. My question is:

Has the notion of differentiability at a point any interest?

That is, I'm looking for a theorem which is valid for a function regular at some point, but which needs significantly less regularity in a neighborhood of this point, or a good reason for which such a theorem doesn't exist. 
Of course, this question is very flexible, and any insight is welcome.

REPLY [6 votes]: There are functions that are only differentiable at one point, here is an example:
Consider the function $d:\mathbb{R}\to\mathbb{R}$ defined by $d(x)=0$ if $x$ is rational and $d(x)=1$ if $x$ is irrational. This function is not differentiable anywhere, since it is not continuous, however, it is surely bounded. Now look at $f(x)=x^2\cdot d(x)$.
Note that for a rational $x$ we have $f(x)=0$ and while for irrational $x$ we have $f(x)=x^2$. 
Next since there are rational numbers arbitrary close to any irrational $x$ our function is not continuous at irrationals. The same argument holds for rational $x$'s whenever $x\ne0$. 
At $x=0$ we have, for $h$ rational and $\ne0$,
$$\frac{f(x+h)-f(x)}{h}= \frac{f(0+h)-f(0)}{h}=\frac{f(h)-f(0)}{h}=\frac{0}{h}=0$$
while for irrational $h$ we get 
$$\frac{f(x+h)-f(x)}{h} = \frac{f(h)-f(0)}{h}=\frac{h^2}{h}=h$$
which tends to $0$ as $h\to0$, in particular $f$ is differentiable at $0$ and $f'(0)=0$.