Geodesic distance and Euclidean distance of an embedded Riemannian manifold Let $(M,g)$ be a Riemannian manifold embedded into $R^n$ with $g$ induced by the Euclidean distance of $R^n$ (as a result this is an isometric embedding). How to prove that 
$$\lim_{x\to y} \frac{d_M(x,y)}{\|x-y\|_E}=1.$$
What I got so far: Let $r:I\to M$ be the geodesic curve connecting $x$ and $y$ and 
$d_M(x,y)=\int_0^1 \|r'(t)\|_g dt$ (is this correct?). But I don't know how to use the geodesity and isometric embedding to show the limit above.
 A: Note that by definition, $ d(x, y) \ge \| x-y\|_E$ for all $x, y\in M$, so 
$$ \frac{d(x, y)}{\|x-y\|_E} \ge 1, \ \ \ \forall x, y\in M, x\neq y.$$
So it suffices to show that 
$$\tag{1}\limsup_{y\to x} \frac{d(x,y)}{\|x-y\|_E} \le 1$$
We assume that $M$ has dimension $k <n$. Using a rotation and translation, we assume that $x = 0\in M$ and $T_xM = \mathbb R^k \times \{0\} \subset \mathbb R^k \times \mathbb R^{n-k}$. Since $M$ is embedded in $\mathbb R^n$, there is $R>0$ and an open set $U$ with $0\in U\subset \mathbb R^k$ so that 
$$ M\cap B_x(R) = \{ (z, u(z)) : z\in U\},$$
here $u: U \to \mathbb R^{n-k}$. Now let $y\in M\cap B_x(R)$. So $y = (z, u(z))$ for some $z$ and 
$$ \| x-y\|_E = \sqrt{\|z\|^2 + \| u(z)\|^2}\ge \|z\|.$$
Note that we have $u(0) = \nabla u(0) = 0$. On the other hand, let $\gamma(t) = (tz, u(tz))$. Then this is a curve in $M\cap B_x(R)$ joining $x$ to $y$. Thus 
\begin{align*}
 d(x, y) &\le \int_0^1 \| \gamma'(t)\|_g dt\\
& = \int_0^1 \sqrt{\|z\|^2 + |\langle z, \nabla u (tz)\rangle|^2}\,  dt
\end{align*}
where $\nabla u (tz)$ is the gradient of $u$ at the point $tz$ and $\langle \cdot, \cdot \rangle$ is the dot product on $\mathbb R^k$ (and we also used that $\gamma'(t) = (z, \langle z, \nabla u (tz)\rangle)$). Thus we have  
$$ \frac{d(x, y) }{\|x-y\|_E} \le \frac{1}{\|z\|} \int_0^1 \sqrt{\|z\|^2 + |\langle z, \nabla u (tz)\rangle|^2} \, dt = \int_0^1 \sqrt{1 + |\langle \nu, u(tz)\rangle|^2} \, dt ,$$
where $\nu = z/\|z\|$. Since $\nabla  u \to 0$ as $y\to x$, we conclude that 
$$ \int_0^1 \sqrt{1 + |\langle \nu, u(tz)\rangle|^2} \, dt  \to 1.$$
Thus (1) is shown. 
