A New Definition of Derivative

After translating and subtracting a linear function from $f$, we can assume that $a = 0$, that $f(0) = 0$ and that $A = 0$. So we're assuming that $f(x) = \varepsilon(x)||x||$ for some vector-valued function $\varepsilon(x)$ with $\varepsilon(x) \to 0$ as $x \to 0$. We must show that there is a matrix-valued function $\varphi(x)$ with $\varphi(x) \to 0$ as $x \to 0$ and $f(x) = \varphi(x) \cdot x$.

To achieve this, for all $x \ne 0$ we define $\varphi(x) \cdot h = \langle \frac{x}{||x||},h \rangle \varepsilon(x)$. We have $||\varphi(x)|| = ||\varepsilon(x)||$, where by $||\varphi(x)||$ I mean the operator-norm of $\varphi(x)$, so it is clear that $\varphi(x)$ satisfies our requirements.

Define the mapping $\psi$ as

$$\psi(x) = {f(x)-f(a) - A(x-a)\over |x-a|^2} (x-a)\cdot$$

then if $x\ne a$ you have that $x\ne a$ you have that $f(x)-f(a) - A(x-a) = \psi(x) (x-a)$. And you have that

$$||\psi(x)|| = {||f(x)-f(a) - A(x-a)||\over||x-a||}$$

So you have that $||\psi(x)||\to 0$ as $x\to a$. Now we have

$$f(x)-f(a) = A(x-a) + \psi(x) (x-a) = (A-\psi(x))(x-a)$$

Now we have that since $||\psi(x)||\to 0$ that $\varphi(x) = A-\psi(x)$ is continuous at $a$.