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 $n$?



Normalized packing number
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. 


