Intuition behind derivative as the best linear approximation I'm trying to understand how derivative for functions $f:\mathbb{R}^n \rightarrow \mathbb{R}^m$ works, which is defined as "the best linear approximation". Intuitively, given $T,M\in\mathcal L(\mathbb{R}^n, \mathbb{R}^m)$ and $a\in\mathbb{R}^n$, the affine approximation at $a$,  $\ell(x) = f(a) + T(x-a)$ is better than the approximation $F(x) = f(a) + M(x-a)$ if $||f(x) - \ell(x)|| \rightarrow 0$ faster than $||f(x) - F(x)||$ does as $x\rightarrow a$, i.e,
$$\lim_{x\to a} \frac{||f(x) - (f(a) + T(x-a))||}{||f(x) - (f(a) + M(x-a))||} = 0$$
The intuition says that if $T\in\mathcal L(\mathbb{R}^n, \mathbb{R}^m)$ is such that for all $M\in\mathcal L(\mathbb{R}^n, \mathbb{R}^m)$ with $M\not=T$ the previous limit is zero (this means that $\ell$ is a better affine approximation than any other affine approximation), then $T$ should be the derivative of $f$ at $a$, for functions of one variable this is true, i.e.
$\textbf{Proposition.}$ Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a function, $a\in\mathbb{R}$ and $L\in\mathcal L(\mathbb{R})$, the following statements are equivalent:
(a) $f$ is differentiable at $a$ with derivative L:
$$\lim_{x\to a} \frac{|f(x) - (f(a) + L(x-a))|}{|x-a|} = 0$$
(b) For all $M\in\mathcal L(\mathbb{R})$ with $M\not= L$, we get:
$$\lim_{x\to a} \frac{|f(x) - (f(a) + L(x-a))|}{|f(x) - (f(a) + M(x-a))|} = 0$$
My question is: Is this true for functions in higher dimensions? i.e.
$\textbf{Proposition.}$ Let $f:\mathbb{R}^n \rightarrow \mathbb{R}^m$ be a function, $\hat{a}\in\mathbb{R}^n$ and $L\in\mathcal L(\mathbb{R}^n, \mathbb{R}^m)$, the following statements are equivalent:
(a) $f$ is differentiable at $\hat{a}$ with derivative L:
$$\lim_{x\to \hat{a}} \frac{||f(x) - (f(\hat{a}) + L(x-\hat{a}))||}{||x-\hat{a}||} = 0$$
(b) For all $M\in\mathcal L(\mathbb{R}^n, \mathbb{R}^m)$ with $M\not= L$, we get:
$$\lim_{x\to \hat{a}} \frac{||f(x) - (f(\hat{a}) + L(x-\hat{a}))||}{||f(x) - (f(\hat{a}) + M(x-\hat{a}))||} = 0$$
Is the above statement true?
 A: This does not work in higher dimensions.
When $n > 1$ you can construct $M$ that is identical to $L$ in all but one dimension.
Therefore there is an affine space $A$ that passes through $\hat a$
in which
$$ \frac{\lVert f(x) - (f(\hat{a}) + L(x-\hat{a}))\rVert}
        {\lVert f(x) - (f(\hat{a}) + M(x-\hat{a}))\rVert} = 1 $$
for all $x \in A.$
Since every punctured neighborhood of $\hat a$ contains points in $A,$ it is not possible that the ratio above have the limit $0.$
When $L$ is the derivative of $f$ at $\hat a,$
what makes $L$ the "best" linear approximation is that $M = L$ is
the unique linear function that satisfies
$$ \lim_{x\to a} \frac{\lVert f(x) - (f(\hat{a}) + M(x-\hat{a}))\rVert}
                      {\lVert x-a\rVert} = 0. $$
When $M \neq L$ the limit does not exist.
The limit
$$ \lim_{x\to a} 
   \frac{\lVert f(x) - (f(\hat{a}) + L(x-\hat{a}))\rVert}
        {\lVert f(x) - (f(\hat{a}) + M(x-\hat{a}))\rVert} $$
also does not exist when $M\neq L,$ but that would also be true even if $L$ is not the derivative of $f$ at $\hat a.$
