Volume of balls in homogeneous manifolds Let $X=G/H$ be a homogeneous manifold, where $G$ and $H$ are connected Lie groups and assume there is given a $G$-invariant Riemannian metric on $X$.
Let $B(R)$ be the closed ball of radius $R>0$ around the base point $eH$  and let $b(R)$ denote its volume.
Is it rue that
$$
\lim_{\varepsilon\to 0}\ \limsup_{R\to\infty}\ \frac{b(R+\varepsilon)}{b(R)}=1?\qquad(\#)
$$
The idea somehow being that volume growth is largest with constant negative curvature in which case it is exponential and thus satisfies our claim.
 A: 
The idea somehow being that volume growth is largest with constant negative curvature 

is essentially the content of the Bishop–Cheeger–Gromov comparison theorem.  See for instance Lemma 36 of Peter Petersen's Riemannian Geometry, 2ed.  It states, in his notation, that in any complete Riemannian manifold $(M,g)$ of dimension $n$ with Ricci curvature bounded below by $(n-1)k$, the function
$$r \mapsto \frac{\operatorname{vol} B(p,r)}{v(n,k,r)}$$
is nonincreasing, where $B(p,r)$ is the ball in $(M,g)$ centered at $p$ with radius $r$, and $v(n,k,r)$ is the volume of the ball of radius $r$ in the space form of dimension $n$ and constant sectional curvature $k$.  A homogeneous space certainly has bounded Ricci curvature, and the interesting case for us is when $k$ is negative, so that $v(n,k,r)$ is the volume in hyperbolic space, which as you say grows exponentially.
So this result tells us that
$$\frac{\operatorname{vol} B(p,r)}{v(n,k,r)} \ge \frac{\operatorname{vol} B(p,r+\epsilon)}{v(n,k,r+\epsilon)}$$
or in other words
$$1 \le \frac{\operatorname{vol} B(p,r+\epsilon)}{\operatorname{vol} B(p,r)} \le \frac{v(n,k,r+\epsilon)}{v(n,k,r)}.$$
As $r \to \infty$ the right side converges to something of the form $e^{c \epsilon}$, which in turn goes to $1$ as $\epsilon \to 0$.
