Normalized packing number Suppose $M$ is a $n$-dimensional closed Riemannian manifold. Let the packing number $N(t)$ be the maximum number of balls with radius $t$ in $M$ that are disjoint. I am wondering whether the following limit has some geometric meaning:
$$
\lim_{t\to 0}\frac{N(t)t^n}{Vol(M)}
$$
Does the limit depend only on dimension $n$?
 A: This packing problem is the limit as $s\to\infty$ of the problem of minimal energy point configurations under the Riesz potential $V=1/r^s$. Hardin and Saff show (see Theorem 2.1) that the minimum energy $E(A,N)$ of $N$ points on the $d$-dimensional manifold $A$ satisfies
$$\lim_{N\to\infty} \frac{E_s(A,N)}{N^{1+s/d}} = \frac{C_{s,d}}{\mathcal H_d(A)^{s/d}}\text,$$
where $C_{s,d}$ is a constant independent of $A$ and $\mathcal H_d$ is the $d$-dimensional Hausdorff measure. As $s\to\infty$, the energy is given by $E_s(A,N)=e^{-s (\log 2t(A,N)+o(1))}$, where $t(A,N)$ is the optimal packing radius of $N$ spheres on $A$. After a bit of work, you should be able to get the desired manifold-independent limit. It is also possible that the packing problem itself is treated in the literature separately, but I am only familiar with the cited reference.
A: If I am not mistaking,  your  number is the same for all manifolds, does not depend  on the metric  and on the manifold and coincides with the packing number of  the standard ball in the euclidean  $R^n$. 
Indeed,  two metrics $g$ and $g'$  on $M$
that are  $\epsilon$ close one to another
(in the sence that for any $i,j$ the difference between 
the components $|g_{ij}- g'_{ij}|$ is at most  epsilon) 
   have  approximately the same  numbers $N(t)$: 
$$
N_{g}((1- C\cdot \epsilon) \cdot  t)) \le N_{g'}(t)  \ \textrm{ and } \  N_{g}((1+C\cdot \epsilon) \cdot t) \ge N_{g'}(t) 
$$
where $C$ is a constant (depending on $(M, g)$). Since the number $N_g(t)$ growth asymptotically with order  $t^{-n}$, this implies that your number $\lim_{t\to 0} \frac{t^n N(t)}{Vol(M)}$ is locally a constant.  
This implies that  if we continuously 
 deform the  metric we do not change your number  and therefore one can think that the metric $g$ is such that it is flat everywhere except for the set of volume
 $1/K \cdot  Vol(M)$.  For this metric your number is close,  for big $K$, to the  similar number for the euclidean metric and making $K$ bigger and bigger gives us what we want.   
