# Continuity, differentiability and existence of partial derivatives

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.

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

$$y=x,$$

and

$$y=0.$$

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.

• Thank you. Are the others fine? Also, are the partial derivatives defined? Nov 21, 2013 at 8:30
• I have made some changes in the answers. Could you check them once again? Nov 21, 2013 at 9:52
• 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. Nov 21, 2013 at 10:55
• 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. Nov 21, 2013 at 10:59
• Please, write another question with number 4., showing your computations. It would help a lot to understand where the problems may lie. Nov 21, 2013 at 11:11