Here are a few functions whose continuity, differentiability and existence of partial derivatives are to be checked at the origin. I have given the answers, but I would really appreciate it if someone could check it for me :) $$1. f(x,y)=\sin x\sin(x+y)\sin(x-y)$$ Continuous, differentiable, partial derivatives exist because $$\lim_{h\to 0}\frac{1}{h}[f(0+h,0)-f(0,0)] = \lim_{h\to 0}\frac{1}{h}[f(0, 0+h)-f(0,0)] $$

$$2. f(x,y)=\left\{\begin{matrix} \ \frac {xy}{x^2+y^2}; (x,y)\neq (0,0) \\0\; ; (x,y)=(0,0) \end{matrix}\right. $$ Discontinuous, not differentiable, partial derivatives exist (because partial derivatives are not continuous) $$3. f(x,y)=\sqrt{\left | xy \right |}$$ Continuous, not differentiable, partial derivatives defined $$4. f(x,y)=\left\{\begin{matrix} \ \frac {x^2-y^2}{x^2+y^2}; (x,y)\neq (0,0) \\0\;; (x,y)=(0,0) \end{matrix}\right. $$ Discontinuous, not differentiable, partial derivatives undefined $$\lim_{h\to 0}\frac{1}{h}[f(0+h,0)-f(0,0)] = \lim_{h\to 0}\frac{1}{h}[\frac{h^2}{h^2}-\frac{0}{0}]$$ and $$\lim_{h\to 0}\frac{1}{h}[f(0,0+h)-f(0,0)] = \lim_{h\to 0}\frac{1}{h}[\frac{-h^2}{h^2}-\frac{0}{0}]$$ and the two partial derivatives are not defined. $$5. f(x,y)=\left\{\begin{matrix} \ 1\;; xy=0 \\0\;; xy \neq 0 \end{matrix}\right. $$ Discontinuous, partial derivatives defined, not differentiable (for this, I don't really understand how to go about this particular one: My answers are based on the graph) $$6. f(x,y)=1-\sin \sqrt{x^2+y^2}$$ Continuous (because it is trigonometric), partial derivatives not defined, not differentiable. $$f_x=\lim_{h\to 0}\frac{1}{h}[1-\sin\sqrt{h^2}-1+\sin 0]= -1$$ $$f_y=\lim_{h\to 0}\frac{1}{h}[1-\sin\sqrt{h^2}-1+\sin 0] =-1$$ I am extremely new to these concepts so my reasoning can be extremely flawed. If you could check these answers and, if you think I am wrong, point out as to why I am wrong, I would be extremely thankful :) @Avitus: It has been edited :) Please have a look.


1 Answer 1


On number 4.

In polar coordinates the limit $\lim_{(x,y)\rightarrow (0,0)}f(x,y)$ is equal to

$$\lim_{\rho\rightarrow 0}\frac{\rho^2}{\rho^2}\frac{\cos^2\theta-\sin^2\theta}{\cos^2\theta+\sin^2\theta},$$

whose result is dependent of $\theta$. The function $f$ is not continuous at $(0,0)$.

You can arrive at the same result without polar coordinates and choosing different paths to $(0,0)$, leading to different limit values. For example, try to reach $(0,0)$ along the lines

$$y=x, $$



You obtain the limits $\lim_{x\rightarrow 0}f(x,x)=0$ and $\lim_{x\rightarrow (0,0)}f(x,0)=1$. $f$ is not continuous at $(0,0)$, then.

  • $\begingroup$ Thank you. Are the others fine? Also, are the partial derivatives defined? $\endgroup$
    – Artemisia
    Nov 21, 2013 at 8:30
  • $\begingroup$ I have made some changes in the answers. Could you check them once again? $\endgroup$
    – Artemisia
    Nov 21, 2013 at 9:52
  • $\begingroup$ I think that it would be useful if you take 1-2 cases and show your computations for the partial derivatives and, most of all, differentiability in a new question. There are many technicalities to be checked. $\endgroup$
    – Avitus
    Nov 21, 2013 at 10:55
  • $\begingroup$ I don't understand... cases as in along a chosen set of lines? I don't know how to do that... I just calculated the partial derivatives and checked if they made sense at the origin, and if they have an indeterminate form, then they are not defined. $\endgroup$
    – Artemisia
    Nov 21, 2013 at 10:59
  • $\begingroup$ Please, write another question with number 4., showing your computations. It would help a lot to understand where the problems may lie. $\endgroup$
    – Avitus
    Nov 21, 2013 at 11:11

You must log in to answer this question.

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