Is it true that $\frac{d}{dt}\int_S \mathbf{B} \cdot d \mathbf{a}$ goes to zero if the amperian loop delimiting $S$ contracts indefinitely? I suppose to have an ordinary magnetic field: in the answer I'm not interested to involve Dirac delta: the integral goes to zero. I want to focus on another point: an infinitesimal physical quantity can have a finite time derivative? Of course derivative of zero is zero, but this flux is never strictly zero, and this trouble me because the step
$$
\frac{d}{dt}\int_S \mathbf{B} \cdot d \mathbf{a} \to 0
$$ 
(when the surface connected to the amperian loop can be taken indefinitely small) is used when we exploit Maxwell equations to fix boundary conditions on the discontinuity between two media. Maybe I'm getting flustered in the slightest thing, but this confuse me and I can't get to the bottom of this problem. How could I see clearly this passage?
 A: You're right that a function can be "small" at a point but have a "large" derivative at that point.  But maybe the confusion is that you're imagining the surface $S$ shrinking in time, so that it's only "small" at one instant.  But the surface doesn't shrink in time - you're taking the limit where it's "small" at all times.  And if a function is always small, then its derivative is also always small, because $d/dt\ (\epsilon \Phi(t)) = \epsilon\, d\Phi/dt$, where $\Phi = \int_s {\bf B} \cdot d{\bf S}$.  So you don't even have to exchange the time derivative with the line integral if you don't want to - the smallness of the loop integral will make $\Phi(t)$ infinitesimal for all times, so it's time derivative will also be infinitesimal.
A: As long as $\mathbf{B}$ is a continuous (once-differentiable) function, when you look at small enough sizes, $\mathbf{B}$ has a Taylor series, the first term of which is a constant. As you let the loop size shrink, only the constant term matters. But then $\int_S \mathbf{B}\cdot d\mathbf{a}\rightarrow \mathbf{B}\int_S d\mathbf{a}\rightarrow 0$ since the area of the enclosed surface goes to 0.
A: The previous answers and comments inspired to me what follow: maybe this is the simplest (and so the best) way to see why $\frac{d}{dt} \int_S \mathbf{B} \cdot d \mathbf{a}$ vanishes when the loop contracts indefinitely.
I can simply consider a small surface that doesn't vary in time, but it is essential take into account the possibility that the magnetic field (and consequently the flux) does vary. Now, if surface $d\mathbf{a}$ is very small we can ignore the integral and simply write $\Phi = \mathbf{B} \cdot d \mathbf{a}$ (in case we are on a media discontinuity we will broke into $\mathbf{B}_1 \cdot d \mathbf{a}_1 + \mathbf{B}_2 \cdot d \mathbf{a}_2$, this has no effects on next reasoning) from wich 
$\dot{\Phi} = \dot{\mathbf{B}}\cdot d \mathbf{a} + \mathbf{B} \cdot d \dot{\mathbf{a}} = \dot{\mathbf{B}}\cdot d \mathbf{a}$ (remember that the surface is constant in time). Because of the smallness of $d\mathbf{a}$ this can be read $\dot{\Phi}(t) \cong 0$. So the answer is yes: it goes to zero because previous steps show that time derivative too is pushed to zero by the infinitesimal surface.
