Let $(M, g)$ be a Riemannian manifold and $\gamma :[0, \delta] \to M$ a $C^1$-curve, $\gamma(0) = y$ and $\dot\gamma(0) \neq 0$. Then we have
$$\lim_{t\to 0} \frac{d_M(y, \gamma(t))}{\int_0^t |\dot \gamma|} = 1.$$
The proof suggested below uses the Taylor expansion of metric in a normal coordinate.
I post this question and answer here since
it helps complete a proof I wrote here, and
this result has some independent interest (I think), and
Would someone please check if the answer is correct, I am bit unfamiliar with the big $O$ argument.
On the other hand, it would be great if someone points me to a reference where this is proved.
Let $\delta >0$ be small so that $\gamma$ lies completely in the normal coordinate centered at $y$. In this normal coordinate, we write
$$\gamma(0) = 0, \ \ \ g_{ij}(x) = \delta_{ij} + O(|x|^2).$$
Write $v = \dot\gamma(0)$. Then we have
$$ \gamma (t) = tv + O(t^2), \ \ \ \dot\gamma = v+ O(t)$$
$$\begin{split}
\int_0^t \|\dot\gamma (s)\|ds &= \int_0^t \| v + O(s)\| ds \\
&= \int_0^t \sqrt{\|v\|^2 + O(s)} ds \\
&= \int_0^t \left( \|v\| + O(s)\right) ds \\
&= t\| v\| + O(t^2) \\
&= \|tv\| + O(t^2)\\
&= |\gamma (t) – O(t^2)| + O(t^2) \\
&= |\gamma(t)| + O(t^2).
\end{split}$$
Now in normal coordinate, $|\gamma (t)| = d_M(y, \gamma(t)).$ Thus
$$\frac{\int_0^t \|\dot\gamma\|}{d_M(y, \gamma(t))} = 1 + \frac{O(t^2)}{t} \to 1$$
as $t\to 0$.