What is the limit of $f(\frac{(x,y)}{\vert (x,y) \vert})$?

Solution 1:

Let $(x, y) = (r\cos \theta, r\sin \theta), r\geq0$. Then $f(x,y)=rg(\cos \theta, \sin \theta)$.

So, we have $$ \frac{f(x,y)-f(0,0)}{r} = g(\cos \theta, \sin \theta). $$

Taking $r \to 0$, we get $D_\theta f(0,0) = g(\cos \theta, \sin \theta)$, where $D_\theta$ is the directional derivative along a vector $(\cos \theta, \sin \theta)$.

Take $\theta =0, \pi$ and note that $g(1,0)=g(-1,0)=0$, we conclude $f_x(0,0)=0$. Likewise, by taking $\theta = \pi/2, 3\pi/2$, we conclude $f_y(0,0)=0$.

So, if $f$ is differentiable at $(0,0)$, $D_\theta f(0,0)$ must be $0$ for any $\theta$. However, this is not the case unless $g \equiv 0$ since $D_\theta f(0,0) = g(\cos \theta, \sin \theta)$.