I would like verification on a calculation that shows $f$ below is differentiable at $(0,0)$ but $f_x$ is not continuous at $(0,0)$.
$$f(x,y)=(x^2+y^2)\sin(\frac{1}{x^2+y^2})$$
when $(x,y)\neq (0,0)$ and $f(0,0)=0$.
So assuming for the moment that the function is differentiable at $(0,0)$ (it is), this shows that $f_x(0,0)$ exists. In fact, it is easy to see $f_x(0,0)=0$ by applying the definition of the derivative and using the squeeze theorem. Then by standard computation we get that away from the origin $$f_x(x,y)=2x\left(\sin\left(\frac{1}{x^2+y^2}\right)-\frac{\cos\left(\frac{1}{x^2+y^2}\right)}{x^2+y^2}\right)$$
Here is where I'm a little fuzzy.... I found that $\lim_{(x,0)\to (0,0)} f_x=0$ and so this isn't helpful. I need to find a path that makes the limit not equal to zero, right? What if I find a path where the limit doesn't exist? Is this enough? I have a result that says
$$\lim_{(x,y)\to(0,0)}g(x,y)=\lim_{r\to 0^+} g(r\cos\theta,r\sin\theta)$$
I guess I prefer working in these coordinates because then I don't have to worry about the path. I computed the following
$$\lim_{r\to 0^+} f_x(r,\theta)=2r\cos\theta\left(\sin(r^{-2})-r^{-2}\cos(r^{-2})\right)$$
Where $2r\cos\theta\sin(\frac{1}{r^2})\to0$ by the squeeze theorem and the second term's limit does not exist (right?) So I have two questions:
1) Does this prove that $f_x$ is not continuous since this limit does not exist? Or should I work in Cartesian coordinates and find a path that shows the limit depends on the path.
2) Can I always do this switch to polar coordinates (where the above centred equation holds true)? Does there need to be polar symmetry? I don't believe this function has polar symmetry.
