Dilating a curved ball Let $B$ be a ball sitting inside a manifold $(M^n, g)$. Now, let us dilate the metric $g$ to $\lambda g$, $\lambda$ being a positive number going to $\infty$. It seems intuitively true that the dilated ball, called $\lambda B$ would "tend" to $\mathbb{R}^n$. Is this true, and is there any way to make this precise, that is, what is the proper notion of "tending" here?
Edit: By a ball, I mean something diffeomorphic to the ordinary Euclidean ball, but with a different metric. 
 A: One possible answer to this question is to use the notion of pointed Gromov-Hausdorff convergence. The relevant definitions are elementary, though somewhat involved, so I'll refer you to Wikipedia and not repeat them here, but just state the precise result.
Let $(M, g)$ be a Riemannian manifold, and fix a distinguished $p \in M$. For any $\lambda > 0$ we'll denote by $\lambda M$ the Riemannian manifold $(M, \lambda g)$. (For large $\lambda$, $M$ is very "dilated" as you suggest.) 
Then one can show that the sequence $(\lambda M, p)$ of pointed metric spaces converges (in the pointed Gromov-Hausdorff sense) to $\mathbb{R}^n$ as $\lambda \to \infty$.
(To prove this, I imagine I would use the Taylor expansion
$$
  g_{ij} = \delta_{ij} - \frac{1}{3} R_{iljk} x^l x^k + O(x^3)
$$
for the Riemannian metric in normal coordinates, but I haven't written down a careful proof so it's possible this idea is off the mark.)
Remark: One nice thing that comes out of this picture -- of "dilating" the manifold $M$ at the point $p$ in order to make it limit onto its tangent space -- is that one can get some notion of "tangent cone" for more general length metric spaces $X$ by attempting the same dilation procedure; if a limit exists, then one calls such a limit the tangent cone of $X$ at $p$.
