Is the Biot Savart Law applicable only for continuous currents? There is a widely known formula for the magnetic field due to a moving charged particle.
$$\frac{\mu_0}{4\pi} q \frac{\vec{v}\times\vec{r}}{r^3}  $$
The usual derivation is as follows.
$$ dB = \frac{\mu_0}{4\pi} i \frac{\vec{dl}\times\vec{r}}{r^3}$$ (Biot Savart Law) 
And then 
$$ i = \frac{dq}{dt}$$
so $$ i\vec{dl} = \frac{dq}{dt}\vec{dl} = dq\frac{\vec{dl}}{dt} = dq\vec{v} $$
Finally, 
$$ dB = \frac{\mu_0}{4\pi} dq \frac{\vec{v}\times\vec{r}}{r^3} $$
which on integration gives the above formula.
However, my teacher says that this formula is not correct since Biot Savart Law itself is applicable only for continuous flows, whereas a charged particle constitutes a discrete current. Is that true? If yes, is there any similar formula for the field due to a moving charged particle? Please show the derivation too in that case.
Edit:
Griffiths himself writes at one point in his book that this equation is "simply wrong". In a footnote, he also writes that it is wrong in principle wheras it is true for non-relativistic speeds, and later on in his book, he goes on to prove that. (Example 10.4) What my confusion is that this "true for non-relativistic speeds" is also true for Coulomb's law. Why isn't that law also "simply wrong" then ?
Thanks.
 A: The problem I have with your (phrasing of your) teacher's statement is the concept "discrete current". What does that even mean? 
When you look at the usual Biot-Savart law with continuous current, you consider an infinitesimally small line segment. As the size of the line segment becomes smaller, so does the amount of charge that you consider - until in the limit, you consider an amount of charge $dq$ that tends to zero as $dl$ tends to zero.
The only thing different when you have discrete particles is that the charge never tends to zero - it tends to a finite value. But that in no way invalidates the rest of the analysis.
As @leastaction@ said, if you consider the charge "lumpy" (a delta function) rather than continuous, the equations are virtually unchanged.
A: Perhaps this will help.  Consider this from the point of view that Electrostatics is about the E and B fields being constant in time.  The Coulombs expression for the E field is for static charge.  Coulombs 1/r^2 E-field is accurate some of the time because static charge (in statistical bulk) is possible.  A single moving electron (unlike the current in a long wire) definitely creates a changing B field at any position with respect to time, regardless of speed v, therefore it is not a "statics" situation for creating a magnetic field (for magnetism, statics requires a constant extended current such that the B field is a constant in time). This was my interpretation of Griffith's statement "simply wrong", in that for a moving point charge the expression will never be exact (though as you said, it is a good approx. for $v<<c$).  I hope this adds value to the discussion.
