I wonder if the following holds in an arbitrary Riemannian manifold $M$:


assume $x\in M$, $h\in T\_x M$, do we have for $u\in T\_x M$ exponentiable (if necessary of small enough norm) that:


$$\lim\_{t\to 0} \frac{d(\exp\_{\exp\_x u}t\tau(h), \exp\_x (u+th))}{t}=0$$
where $\tau(h)$ is the parallel transport of $h$ along $[0,1]\rightarrow M, t\mapsto \exp\_x tu$.


If it doesn't hold in general, then which geometrical conditions are sufficient for it to hold, and do we have at least in general that for any $x\in M$, $h\in T\_x M$:
$$\lim\_{t\to 0, u\to 0} \frac{d(\exp\_{\exp\_x u}t\tau(h), \exp\_x (u+th))}{t}=0$$



Tangential behavior of Riemannian exponential
The second identity is always true because both arguments of $d$ are smooth functions of $u$ and $t$ and they coincide when $u=0$.


The first one holds true for all $u$ and $h$ only if the metric is flat. Indeed, the l.h.s. is the length of the difference of the initial velocity vectors of two curves $t\mapsto \exp\_{\exp\_x u} t\tau(h)$ and $t\mapsto\exp\_x(u+th)$. The first velocity vector is $\tau(h)$, so its length equals $|h|$. The second one is the derivative of $\exp\_x$ at $u$ along $h$. If its length equals $|h|$ for all $u$ and $h$, then $\exp\_p$ is a Riemannian isometry (by definition), so the metric is flat.


