Let $(M,g)$ be a Riemannian manifold, $p \in M$. You will often hear people say "$(T_pM, g(p))$ is the infinitesimal geometry of $M$ at $p$". I would like to upgrade `infinitesimal' to 'local' in some suitably weak sense (if possible). Of course, $M$ will not be flat in general so we can't hope for an open subset $U \subset M$ to be isometric to a subset of $T_pM$. I would like the following weaker statement to be true:
For $K$ arbitrarily close to 1, there are open subsets $p \in U_K \subset M$ and $V_K \subset T_pM$ and a $K$-bilipschitz homeomorphism $f:U_K \to V_K$.
I have a vague sense that this should follow from the continuity of the metric. I've tried to choose a coordinate chart $U \subset \mathbb{R}^n$ small enough so that the two metrics (ie $g$ and $g(p)$) are nearly identical, but I don't see how to use this information to bound $\frac{d_{g}(x,y)}{d_{g(p)}(x,y)}$.
At this stage I am not even sure if the statement is true.
I think that it is true. Let $f:=\exp_p: U\subseteq T_p M \to V\subseteq M$ the exponential map, $V = \exp_p(U)$. Wlog. let $\exp_p$ on $U$ a diffeomorphism. Then it is $df_0 = \text{id}_{T_p M}$. Let $K>1$ arbitrary. Then there is some $\varepsilon > 0$ with $B_{\varepsilon}(0) \subseteq U \subseteq T_p M$ and $$\max_{|X_v|=1, v \in \overline{B_{\varepsilon/2}(0)}} |df(X_v)| \leq K$$ since $df$ is continuous and $df_0 = \text{id}$. So, for $v,w \in B_{\varepsilon/2}(0) \subseteq T_p M$ $$d(f(v),f(w)) \leq L(f\circ \gamma) = \int_{0}^{1} |(f\circ \gamma)'|dt \leq \max_{|X_v|=1, v \in \overline{B_{\varepsilon/2}(0)}} |df(X_v)|\int_{0}^1 |\gamma'|dt \leq K L(\gamma)$$ for all curves $\gamma$ in $B_{\varepsilon/2}(0) \subseteq T_p M$ connecting $v$ with $w$. Therefore, after taking $\inf$, $d(f(v),f(w)) \leq K d(v,w)$. The same for the inverse of $f$.