A map whose infinitesimal distortion tends to $1$ preserve lengths of paths? Let $X,Y$ be length spaces, and suppose $f:X \to Y$ satisfies $$ (*) \, \, \lim_{y \to x} \frac{d^Y(f(x),f(y))}{d^X(x,y)}=1$$ for all $x \in X$. 
Question: Is it true that $f$ preserves lengths of paths?
A naive attempt to prove this hits a problem since we do not know anything about the rates of convergence around different points in $X$.
We can approximate the length of a path $\alpha:[0,1] \to X$ via $L(\alpha) \approx \sum_i d^X(\alpha(t_i),\alpha(t_{i+1}))$ with partitions $\{t_i\}$ which become finer. 
A plausible heuristic would be that when $t_i \approx t_{i+1} \Rightarrow \alpha(t_i) \approx \alpha(t_{i+1})$, hence $d^X(\alpha(t_i),\alpha(t_{i+1})) \approx d^Y\big(f(\alpha(t_i)),f(\alpha(t_{i+1}))\big) $. The problem is that the $t_i's$ are changing when we make the partitions finer. (As far as I could see, even trying to use a "structured mechanism of construction" (like putting $t_i^n=\frac{i}{n}$) does not imply the desired result.
 A: Take a path $\alpha:[0,1]\to X$. Fix $\epsilon>0$ and consider the set $A_\epsilon$ of points $x\in \alpha([0,1])$ such that there exist $x_0,\ldots,x_n\in \alpha([0,1])$ with $x_0=\alpha(0)$ and $x_n=x$ such that for every $i=0,\ldots,n-1$
$$\left|\frac{d^Y(f(x_i),f(x_{i+1}))}{d^X(x_i,x_{i+1})}-1\right|<\epsilon\qquad (*)$$
i.e. there is a discrete path with low distorsion connecting $\alpha(0)$ to $x$. 
It is easy to prove that $A_\epsilon$ is nonempty and both open and closed in $\alpha([0,1])$, therefore by connectedness coincides with $\alpha([0,1])$.
Now given a curve $\gamma:[0,1]\to Z$ with $Z$ metric space and a partition $\underline t=\{t_i\}_{i=1}^m$ with $0\leq t_\leq \ldots\leq t_m\leq 1$, let
$$L(\gamma,\underline t):=\sum_{i=1}^{m-1} d(\gamma(t_i),d(t_{i+1}))$$
so that
$$L(\gamma)=\sup_{\underline t\in \mathcal{P}(0,1)} L(\gamma,\underline t).$$
Consider now a path $\alpha:[0,1]\to X$, and fix $\epsilon >0$. Then:

For any partition $\{t_i\}_{i=1}^m$ there is a refinement $\{t_i'\}_{i=1}^n$ with 
  $$(1-\epsilon)L(\alpha,\underline t)\leq (1-\epsilon)L(\alpha,\underline t')\leq L(f(\alpha),\underline t')$$
  and
  $$L(f(\alpha),\underline t)\leq L(f(\alpha),\underline t')\leq (1+\epsilon)L(\alpha,\underline t').$$

To prove this just apply the previous result to each subpath $\alpha|_{[t_i,t_{i+1}]}$ (and use the triangle inequality).
In particular $(1-\epsilon)L(\alpha,\underline t)\leq L(f(\alpha))$
and $L(f(\alpha),\underline t)\leq (1+\epsilon)L(\alpha)$, therefore first passing to the supremum over all partitions and then using that $\epsilon $ was arbitrary we obtain that the lengths are equal.
A: (1) If $t$ is arc length of $c:[a,b]\rightarrow X$, then $$
\lim_{\varepsilon\rightarrow 0}\ \frac{d_X(c(t),c(t+\varepsilon
)}{|\varepsilon |} =1
$$ for almost $t$. Define $A$ to be set of all such $t$.
(2) For $x$, there is $\delta=\delta (x)>0$ s.t.
$$  d_X(x,y)<\delta \Rightarrow
d_Y(f(x),f(y))  \leq (1+\epsilon )\cdot d_X ( x, y) $$
Hence if $ \gamma (t) :=(f\circ c)(t)$, define a strictly increasing
sequence $t_n\in [a,b],\ t_0=a$ s.t.
$$ d_Y(\gamma (t_i),\gamma (t_{i+1} )) \leq (1+
\epsilon ) d_X ( c(t_i),c(t_{i+1}) ) $$
If $t_i\rightarrow t$, then we have a sequence $t_i$ again starting at $t_0=t$. For convenience we use $t_i$.
Hence $$ d_Y( \gamma (t_0),\gamma (t_n))\leq \sum_i d_Y (\gamma (t_i),
\gamma (t_{i+1}) )$$ $$\leq \sum_i(1+ \epsilon )\ d_X( c(t_i),c(t_{i+1}) ) \leq
(1+\varepsilon ) (t_n-t_0) $$
So $\gamma$ is an $1$-Lipschitz curve.
(3) Hence $$ v(t):=\lim_\varepsilon\ \frac{d_Y(\gamma (t),\gamma
(t+\varepsilon )) }{\varepsilon }  $$ exists for almost $t$. Define
$A'$ to be a set of all such $t$. And
$$ {\rm length}\ \gamma =\int_a^b v(t)\ dt=b-a$$
since $$ t\in A\cap A'\Rightarrow v(t)=\lim_\varepsilon\
\frac{d_Y(\gamma (t),\gamma (t+\varepsilon )) }{
d_X(c(t),c(t+\varepsilon )) } \frac{ d_X(c(t),c(t+ \varepsilon
))}{\varepsilon } =1 $$
Hence, $f$ is a length preserving map.
A: In order to use the Del's argument above, we need to be careful in how we define $ A_{\epsilon}$:
$ A_{\epsilon}$ should be defined as the set of points $x=\alpha(t)\in \alpha([0,1])$ such that there exist a finite sequence $x_0=\alpha(0),x_1=\alpha(t_1),\ldots,x_{n-1}=\alpha(t_{n-1}),x_n=x=  \alpha(t)$ such that 
$$ 0 <t_1 < \dots < t_{n-1} <t,$$ and for every $i=0,\ldots,n-1$
$$\left|\frac{d^Y(f(x_i),f(x_{i+1}))}{d^X(x_i,x_{i+1})}-1\right|<\epsilon\qquad (*)$$
i.e. there is a discrete "forward" path with low distortion connecting $\alpha(0)$ to $x$.  
We need the "forward" property for use in partitions later, by definition, length of a curve is deinfed as supremum over such "forward" partitions. 
Here is an attempt to prove $A_{\epsilon}$ is closed:
Let $x=\alpha(t) \in \bar A_{\epsilon}$.
Let $y_n \to x,y_n \in A_{\epsilon}$. say $y_n=\alpha(t_n)$. If there is a subsequence $t_{n_k}$ of the $t_n$ such that $t_{n_k} < t$, then it's easy.
(We use continuity of $\alpha$ at $t$ together with assumption on the distrotion).
The harder part is to show we can "go backward": Suppose $t_n >t$. I do not see a way to show $\alpha(t) \in A_{\epsilon}$ in this case.

Edit:
Here is a way to finish this: Define $B_{\epsilon}=\{t \in [0,1] \, | \, \alpha(s) \in A_{\epsilon} \, \text{ for all } \, s \le t \}$.
In other words, $B_{\epsilon}$ is the set of all times $t$ such that for every time $s \le t$ there is a discrete forward path with low distortion connecting $\alpha(0)$ to $\alpha(s)$.  
$B_{\epsilon}$ is non empty, and is easily seen to be open and closed, hence $B_{\epsilon}=[0,1]$, and we are done.
