Prove that a function $f:\mathbb{R^n}\to\mathbb{R^m}$ is differentiable at $p$ with $Df(p)=A\in M_{m\times n} \mathbb{(R)} \iff f(x)=f(p)+A(x-p)+r(x) $ when $r=o(x-p) $
Hi everyone. I am trying to brush up on my calculus and prove the definition of differentiability, which would have been a trivial proof yet I can't think of a formal way to prove it.
I am trying to show that $ \lim_{x\to p} \frac{||f(x)-f(p)-A(x-p)||}{||x-p||}=0 $ or $\lim_{x\to p} \frac{|f_i(x)-f_i(p)-\sum^n_{j=1}a^i_j(x_j-p_j)|}{||x-p||} =0$ for each $1 \le i \le m \ , \ 1 \le j \le n$, and play with the properties of the matrix norm and the partial derivatives to get the result, the intuition is very clear to me but I don't know how to show it on paper. I would be happy to get your help if you can, thank you :)
If $f$ is differentiable at $p$, then define $r(x)$ as $f(x)-f(p)-Df(p)(x-p)$. Then$$\lim_{x\to p}\frac{\lVert r(x)\rVert}{\lVert x-p\rVert}=0$$and therefore $r(x)=o(x-p)$.
And if $f(x)=f(p)-Df(p)(x-p)+r(x)$ with $r(x)=o(x-p)$, then$$\lim_{x\to p}\frac{\lVert f(x)-f(p)-Df(p)(x-p)\rVert}{\lVert x-p\rVert}=\lim_{x\to p}\frac{\lVert r(x)\rVert}{\lVert x-p\rVert}=0.$$