Proving linearity of derivative The derivative for a function $f : \mathbb{R}^n \to \mathbb{R}^m$ defined in some open set containing $x$ at $x$ is defined (at least in Rudin and other references), if it exists, to be the linear function $A : \mathbb{R}^n \to \mathbb{R}^m$ such that
\begin{equation}
\lim_{t \to 0} \left| \frac{f(x + v t) - f(x)}{t} - Av\right| = 0.
\end{equation}
Suppose we drop the bolded assumption that $A$ is linear; can we then prove, from the remaining definition, that if $f$ is differentiable at $x$ its derivative must be linear?
I suspect the answer is no (although I certainly understand why the derivative only makes intuitive sense for $A$ linear). It is certainly possible to prove under stricter assumptions (for example continuity of derivative viewed as a map $\mathbb{R}^n \to L(\mathbb{R}^n, \mathbb{R}^m)$), but I am looking for something given only the above and other conclusions (for example the uniqueness of $A$ if it exists) that can be made without assuming $A$ is linear. If it is impossible, please provide a reason or better yet some sort of specific counter-example.
Thank you in advance.
 A: Yes, you must assume $A$ is linear. The value of $A(v)$ is the bi-directional derivative of $f$, at the point $x$, in the direction of $v$ and $-v$. Try the polar function
$$f(r, \theta) = r\sin(2\theta)$$
at the origin. The graph of this function consists of a union of lines passing through the origin in $\mathbb{R}^3$, so these bi-directional derivatives should exist in each direction. Given a vector $v = (r_0, \theta_0)$, then
$$\lim_{t\rightarrow 0} \frac{f(0 + tv) - f(0)}{t} = \lim_{t\rightarrow 0} \frac{f(tr_0, \theta_0)}{t} = \lim_{t\rightarrow 0} \frac{tr_0\sin(2\theta_0)}{t} = r_0\sin(2\theta_0).$$
If we define $A(r, \theta) = r\sin(2\theta)$, then the limit in the question holds. Moreover, uniqueness of limits implies that this is the only possible choice for $A$ so that the limit holds, but $A$ is not linear. For example, expressing $A$ as a function of cartesian coordinates to make addition easier,
\begin{align}
A(1,0) &= 1 \sin\left(2 \cdot 0\right) = 0 \\
A(0,1) &= 1 \sin\left(2 \cdot \frac{\pi}{2}\right) = 0 \\
A(1,1) &= \sqrt{2} \sin\left(2 \cdot \frac{\pi}{4}\right) = \sqrt{2}.
\end{align}
Hope that helps!
