Question: If the limit as $x$ approaches $3$ of $x^2$ is $9$, how do I use the epsilon delta definition to prove this? I was trying to work it out this way but could do figure out how.

Answer: 
Anything specifying a *range* on a continuous parameter does not typically reduce the dimensionality of the system. If there exist points 'in the middle of' all the ranges, they are locally unaware of the boundary, and hence locally the dimensionality is unaffected.

You have exactly one proper *constraint* on the motion, which is the total energy condition $\sum \mathbf p^2=E^2=\text{const.}$ There are $6N$ degrees of freedom. Hence your system is $6N-1$ dimensional.

The phase space volume is currently infinite since $r\_y,r\_z$ can take arbitrarily negative values. If you bound them below by zero then you have a $3N-1$ dimensional sphere of radius $E$ mutiplied by a $3N$ dimensional cube of side length $l$. The total phase space volume is the product of the volumes of these two objects, since they are independent. Note that by the volume of the sphere I mean the surface area in $3N$ dimensions, the intrinsic volume of the sphere as a $3N-1$ ddimensional manifold.

The units of the volume are $(\text{momentum})^{3N-1} (\text{length})^{3N}$
I think it'll be helpful to first think of 1 dimensional physical space which for your case of $N$ non interacting particles is actually $2N$ dimensions (position and momentum). In this space, the conditions read:
$$p\_i = c\_i,\ r\_i \in[0,l]$$
So the dimension of this space is simply $N$ (since $p$ is constant i.e. zero dimensional). Now let's move to $2D$ physical space ($4N$ phase space). This time we have:
$$p\_{i1}^2 + p\_{i2}^2 = c\_i^2,\ r\_i \in[0,l]$$
We only have on constraint, so this is a $4N-1$ dimensional space. What is this space? if we consider one particle and ignore one of the $r\_i$'s, this is a hollow cylinder in $3D$ space, embedded in a $4D$ space. Now let's return to physical $3D$ space, It should be obvious that the dimension of the space is $6N-1$.