Examples of Functions Alright so I am trying to find examples of functions that are differentiable at a point, but not continuous there.
Also a function continuous at no point; a function continuous only at one point.
Any ideas?
 A: Whereas I certainly agree with Apunam's answer to the first question, that a function differentiable at a point is continuous there, I must respectfully beg to differ with his second and third answers:
The function he defines, $f(x) = \sin(n \pi /2)$ for $n \in \Bbb N$ and otherwise $0$, is certainly not continuous at the points $n \pi / 2$ for odd $n$, but, being $0$ everywhere else (and I think we should note here that $\sin (n \pi /2) = 0$ for $n$ even), is in fact continuous on all of the open intervals falling between the points $n \pi /2$, $n$ odd.  So it is in fact only discontinuous on the countable, discrete set $\{ n \pi /2, n \; \text{odd} \}$; not really very discontinuous, especially when seen in the light of a function such as $\rho(x)$, where
$\rho(x) = 0, \; x \in \Bbb Q, \; \text{ the rationals}, \tag{1}$
$\rho(x) = 1, \; x \in \Bbb R - \Bbb Q, \; \text{the irrationals}. \tag{2}$
Or, if you like, you can consider $\sigma(x) = 1 - \rho(x)$, which reverses the roles of the rational and irrational numbers.  Neither $\rho$ nor $\sigma$ are continuous, anywhere.  Looking for example at $\rho(x)$, we see that for any $x_0$, whether $x_0 \in \Bbb Q$ or not, there are  values of $x$ arbitrarily close to $x_0$ such that $\vert \rho(x) - \rho(x_0) \vert = 1$; this property of course rests upon the fact that both $\Bbb Q$ and $\Bbb R - \Bbb Q$ are dense in $\Bbb R$.  And the corresponding assertion holds for $\sigma(x)$, again with the roles of the rationals and irrationals swiched.  $\rho$ and $\sigma$ are thus examples of functions which are truly discontinuous everywhere.
As for the third question, using $\rho$ or $\sigma$ we may define a function $g(x)$ which is not merely continuous, but in fact differentiable, at exactly one point, and not even continuous everywhere else.  For $a \in \Bbb R$, set $g(x) = (x - a)^2 \rho(x)$; I claim $g(x)$ is indeed differentiable at $x = a$, and that $g'(a) = 0$.  For clearly $g(a) = 0$; thus for $x \ne a$
$\vert \dfrac{g(x) - g(a)}{x - a}\vert = \vert \dfrac{g(x)}{x - a} \vert = \vert \dfrac{(x - a)^2\rho(x)}{x - a} \vert = \vert (x - a)\rho(x) \vert \le \vert x - a \vert \to 0 \; \text{as} \; x \to a, \tag{3}$
showing both that $g'(a)$ exists and that $g'(a) = 0$.  However, for any $b \ne a$, choosing $\delta < \kappa \vert b - a \vert$ for any positive $\kappa < 1$, we see that for $\vert x - b \vert < \delta$, we have 
$\vert x - a \vert = \vert (b - x) - (b - a) \vert \ge \vert \vert b - a \vert - \vert b - x \vert \vert = \vert b - a \vert - \vert x - b \vert$
$> \vert b - a \vert - \delta > (1 - \kappa)\vert b - a \vert > 0, \tag{4}$
since $\vert x - b \vert < \delta < \kappa \vert a - b \vert < \vert a - b \vert$.  We thus have $g(x) = 0$ when $x \in \Bbb Q$ but $\vert g(x) \vert > (1 - \kappa)^2(b - a)^2$ when $x \in \Bbb R - \Bbb Q$; there is no $\epsilon$, $0 < \epsilon < (1 - \kappa)^2(b - a)^2$, such that $\vert g(x) - g(b) \vert < \epsilon$ when $\vert x - b \vert \ < \delta$, no matter how small $\delta$ may be; such a functon $g(x)$ cannot possibly be continuous at $x = b$.  Again we have used the density of both $\Bbb Q$ and $\Bbb R - \Bbb Q$ in $\Bbb R$.
Other choices for $g(x)$ are possible; for example, I think it likely the above argument, with slight modifications, could be applied to $g(x) = \vert x - a \vert \rho(x)$, and $g(x)$ would be continuous only at $a$, but not differentiable there.  Such considerations also apply, I suspect, to the function $\Bbb Q(x)$ introduced by Reckless Reckoner in his comment, but I leave further deliberations on these matters to my readers for the present time.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
A: 
I am trying to find examples of functions that are differentiable at a point, but not continuous there.   

There is not any. Every function which is differentiable at a certain point  is also continues at that point.
Proof:
Let $f′(ξ)$ exists.
We have:
$$f(x)−f(ξ) = \dfrac{f(x)−f(ξ)}{x−ξ}⋅(x−ξ)                      
\to f′(ξ)⋅0  $$
as $x→ξ$
Thus:
$$f(x)→f(ξ) \ \ as\ \  x→ξ$$ Hence $f(x)$ is continues at $x=ξ$
or in other words:
$$\lim_{x→ξ} f(x)=f(ξ) $$   

Edit: Rest of the answer was incorrect so I have deleted it.
