Let $M$ be a manifold compactly embedded in $\Bbb R^m$. For any $x\in M$, $T_xM$ can be identified with subspace of $T_x\Bbb R^m$ and we can talk about $x+tv$ for $v\in T_xM$.
As $M$ is compact, for a small neighborhood $M_{\delta}=\{y\in\Bbb R^m|\ \text{dist}(y,M)<\delta\}$ we can define $$ \pi_M:M_{\delta}\to M $$ to be the projection onto the nearest point in $M$.
How do we show that $$ \lim_{t\to 0} \frac 1t \text{dist}(x+tv,M)=\lim_{t\to 0} \frac 1t |(x+tv) -\pi_M(x+tv)| = 0 $$ for any $x\in M$, $v\in T_xM$?
Intuitively, this is saying that a tangent vector is really tangent to the submanifold. My background in differential geometry is not very strong so I don't really know how to prove this. Any help is appreciated
Let $\gamma:(-\epsilon,\epsilon)\to M$ be a smooth curve with $\gamma(0)=x$ and $\gamma'(0)=v$ (the existence of such a $\gamma$ is the definition of what it means for $v$ to be in $T_xM$, or at least one such definition). Then by definition of the derivative, $$\lim_{t\to 0}\frac{\gamma(t)-x}{t}=v$$ which implies $$\lim_{t\to 0}\frac{|(x+tv)-\gamma(t)|}{t}=0.$$ Since $\gamma(t)\in M$, $|(x+tv)-\gamma(t)|\geq |(x+tv)-\pi_M(x+tv)|$, which implies your limit is $0$ as well.