Show that $\sqrt{|xy|}$ is discontinuous at $(0,0)$?

Question

How to show that $\sqrt{|xy|}$ is discontinuous at $0$. My professor has told me to break the function in piece wise and use $h>0, k>0$, $h<0, k<0$ and then solve it but I don't get him.

Answer 1

If the question is

Prove that $f(x,y)=\sqrt{\left|xy\right|}$ is not differentiable at $(0,0)$.

We have $$\frac{\partial f}{\partial x}(0,0)=\lim_{h\to 0}\frac{f(h,0)-f(0,0)}{h}=\lim_{h\to 0}\frac{0}{h}=\lim_{h\to 0}0=0.$$ By simmetry, $\dfrac{\partial f}{\partial y}(0,0)=0$. If $f$ is differentiable at $(0,0)$ necessarily its differential $\lambda$ is $$\lambda \begin{pmatrix}{h}\\{k}\end{pmatrix}=\begin{pmatrix}{\dfrac{\partial f}{\partial x}(0,0)}&{\dfrac{\partial f}{\partial y}(0,0)}\end{pmatrix}\begin{pmatrix}{h}\\{k}\end{pmatrix}=0$$ Now, $$\frac{|f[(0,0)+(h,k)]-f(0,0)-\lambda (h,k)|}{\left\|(h,k)\right|}=\frac{\sqrt{|hk|}}{\sqrt{h^2+k^2}}$$ $$\underbrace{=}_{\text{polar coordinates}}\frac{r\sqrt{\left|\sin \theta\cos \theta\right|}}{r}=\sqrt{\left|\sin \theta\cos \theta\right|}$$ that is, the limit $$\lim_{(h,k)\to (0,0)}\frac{|f[(0,0)+(h,k)]-f(0,0)-\lambda (h,k)|}{\left\|(h,k)\right\|}$$ does not exist, so $f$ is not differentiable at $(0,0).$

Answer 2

Let's assume $x,y≥0$ so that $f(x,y)=\sqrt{xy}$. Taking the gradient, we get $$\nabla f(x,y)=\langle{\,\frac{1}{2}\sqrt{\frac{y}{x}},\;\frac{1}{2}\sqrt{\frac{x}{y}}}\,\rangle $$

Suppose that $x=y=t$. Then $$\nabla f=\langle{\,\frac{1}{2}\sqrt{\frac{y}{x}},\;\frac{1}{2}\sqrt{\frac{x}{y}}}\,\rangle=\langle{\,\frac{1}{2},\;\frac{1}{2}}\,\rangle. $$ Suppose again that $x=4y=t$. Then $$\nabla f=\langle{\,\frac{1}{2}\sqrt{\frac{y}{x}},\;\frac{1}{2}\sqrt{\frac{x}{y}}}\,\rangle=\langle{\,\frac{1}{4},\;1}\,\rangle. $$

Since these are the same everywhere along the lines $x=4y$ and $x=y$, they must take the same value at $x=y=0$ – but they don't ($(1/2,1/2) \neq (1/4,1)$). Thus $f$ is non-differentiable at $(x,y)=(0,0)$.