For the function, $$f = \begin {cases} \frac{sin(x^3)}{x^2+y^2}, \text{if } (x,y) \neq (0,0) \\ 0, \text{if } (x,y) = (0,0) \end {cases}$$ compute $\frac{\partial f}{\partial x}(0,0)$ and show that $\frac{\partial f}{\partial x}$ is not continuous at $(0,0)$.

Now I have already calculated the partial derivative if $(x,y)\neq(0,0)$ to be $\frac{\partial}{\partial x}\left(\frac{\sin{\left(x^{3} \right)}}{x^{2} + y^{2}}\right)=\frac{x \left(3 x \left(x^{2} + y^{2}\right) \cos{\left(x^{3} \right)} - 2 \sin{\left(x^{3} \right)}\right)}{\left(x^{2} + y^{2}\right)^{2}}$ but would the partial derivative at $(0,0)$ be $0$?

And if so then how can I use it prove discontinuity at $(0,0)$ since I have not yet been taught how to calculate limits using $\epsilon \ and \ \delta$?

Any help would be appreciated!


1 Answer 1


You have$$\frac{\partial f}{\partial x}(0,0)=\lim_{h\to0}\frac{f(h,0)-f(0,0)}h=\lim_{h\to0}\frac{\sin(h^3)}{h^3}=1.$$On the other hand, note that, if $y\ne0$,$$\frac{\partial f}{\partial x}(2y,y)=\frac{12}{5} \cos \left(8 y^3\right)-\frac{4 \sin\left(8 y^3\right)}{25 y^3}$$and that therefore$$\lim_{y\to0}\frac{\partial f}{\partial x}(2y,y)=\frac{28}{25}\ne1.$$

  • 1
    $\begingroup$ I was thinking of going this route but how can I use this to prove that it is discontinuous? $\endgroup$ Mar 29, 2021 at 14:14
  • $\begingroup$ @CaporalFourrier I've added a paragraph to my answer. $\endgroup$ Mar 29, 2021 at 14:17
  • $\begingroup$ I get the edit but is this the only way; by using symmetry? $\endgroup$ Mar 29, 2021 at 14:18
  • $\begingroup$ Where did I use which symmetry? $\endgroup$ Mar 29, 2021 at 14:19
  • 1
    $\begingroup$ I am referring to this post: math.stackexchange.com/questions/3096620/… $\endgroup$ Mar 29, 2021 at 14:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .