# Polar coordinates not smooth at origin?

Let $f(r,\theta)=(r,\theta)$ be polar coordinates on $D^2$, $r \in [0,1]$ and $\theta \in [0,2\pi)$.

A function $f:\mathbb R^m \to \mathbb R^n$ is said to be differentiable at a point $x_0$ if there exists a linear map $J: \mathbb R^m \to \mathbb R^n$ such that $$\lim_{h \to 0} {f(x_0 + h) - f(x_0) - J(x_0)h \over \|h\|} = 0$$ (see here)

I am now trying to verify that $f$ is not smooth at the origin. My work: $$J_{(0,0)} = Id$$ and $$\lim_{(\varepsilon, \delta) \to (0,0)} {(\varepsilon, \delta) - (0,0) - (\varepsilon, \delta) \over \|(\varepsilon, \delta)\|} = 0$$

Unofrtunately, this means that the coordinates $f$ is differentiable at the origin.

How to prove that $f$ is not smooth at $(0,0)$?

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You are trying to prove that the identity map, which is linear, is not smooth. This is a very bad conjecture, since any linear map is of class $C^\infty$. –  Siminore Oct 6 '13 at 8:26