In my textbook, I found the following exercise:
Show that the following vector field is differentiable at $(0,0,0)$: $$G(x,y)=\left(\frac {x^3 + y^3}{\sqrt{x^2 + y^2}},\frac {1}{x^2 + y^2 + 1}\right)$$
However, I see that it is not even continuous at the origin, so it cannot be, in any way, differentiable. Is this right or am I missing something?
All help will be greatly appreciated.
Write $G=(G_1,G_2)$. $G$ is differentiable at $(0,0)$ if and only if $G_1$ and $G_2$ are differentiable there.
$G_2$ is obviously differentiable. For $G_1$, the partial derivatives are each $0$:
$$\frac{\partial G_1}{\partial x} (0,0) = \lim_{h \to 0} \frac{G_1(h+0,0) - G_1(0,0)}{h} = \lim_{h\to 0} \frac{\frac{h^3}{|h|}}{h} = \lim_{h \to 0} |h| = 0$$
One gets $\frac{\partial G_1}{\partial y}(0,0)$ by symmetry. Now recall that $G_1$ is differentiable at $(0,0)$ if and only if the following limit exists and equals $0$:
$$\lim_{(h,k)\to (0,0)} \frac{G_1(0+h,0+k) - G_1(0,0) - \frac{\partial G_1}{\partial x}(0,0) h - \frac{\partial G_1}{\partial y}(0,0)k}{\sqrt{h^2+ k^2}}$$
Convert to polar coordinates to see why this is in fact the case.