Let $M\subseteq \mathbb{R}^{m}$ a differentiable hyperface, $p\in M$ and $$A_{p}=p+T_{p}M= \{p+v : v\in T_{p}M\}.$$ Define $\varphi: M\to \mathbb{R}$ by $\varphi (x)=\frac{d(x,A_{p})}{||x-p||}$, show that $\lim_{x\to p}\varphi (x)=0$.
Let $\varepsilon>0$, I want to find $\delta>0$ such that, if $||x-p||<\delta$, then $$\left|\frac{d(x, A_{p})}{||x-p||}\right|=\frac{d(x, A_{p})}{||x-p||}<\varepsilon \Leftrightarrow d(x, A_{p})<\varepsilon ||x-p||<\varepsilon \delta.$$ is it possible to bound that distance like this?. otherwise, could you give me an idea or suggestion?
Consider the projection $q$ of $x$ to $A_p$; then you have a right triangle $pqx$, and $\varphi(x)$ is the sine of the angle $xpq$. If you prove that the angle itself goes to $0$, then the sine will go to zero as well. Geometrically, the angle goes to zero because the tangent plane approximates the hypersurface to second order. A proof can be obtained as follows:
Rotate and translate the space so that $A_p$ is a coordinate hyperplane $z=0$; let $v$ be the "other" coordinates. Then $M$ is locally graph of function $z=f(v)$ with zero derivative at $p$; by multivariable Taylor theorem with bound on remainder you get that $|z|/|v|$ goes to zero; but that is the tangent of the angle in question.