# Continuity of a multivariable absolute value function

The function is as follows: $$f(x,y)=\sqrt{\left |xy \right |}$$ I have to check whether it is continuous, differentiable and has defined partial derivatives at $(0,0)$. My attempt is as follows:

1. Function is discontinuous at the origin.

2. Not differentiable at the origin because of the pointy peak (haha); and

3. Partial derivatives are as follows: $f_x=\frac{\sqrt y}{2\sqrt x}$ and $f_y=\frac{\sqrt x}{2\sqrt y}$ and are not defined at origin.

Is my reasoning correct?

• Why is the function discontinuous at the origin? Nov 21, 2013 at 15:19
• Because modulus functions are discontinuous at the origin...? :) Nov 21, 2013 at 15:24
• That's not true. Maybe you are confusing yourself with the fact that the function $f:\mathbb{R} \to \mathbb{R}$, $f(x)=|x|$ is not differentiable at the origin... Nov 21, 2013 at 15:28
• Ah maybe... so the function is continuous? Hmmm. And differentiable? Nov 21, 2013 at 15:30
• Be careful with partial derivatives, your formulae are wrong! Nov 21, 2013 at 16:03

To show that the function is continuous at the origin, you may use the polar coordinates: putting $x=\rho \cos \theta$ and $y=\rho \sin \theta$ with $\rho >0$ and $\theta \in [0, 2 \pi)$, you have that $$\sqrt{|xy|}=\sqrt{\rho^2 |\cos \theta \sin \theta|}\le \rho \to 0$$if $\rho \to 0$. About the differentiability, I think that you are right (anyway it is better if you do a background check with the definition).
Since $\sqrt{|x|} \to 0$ as $x \to 0$, it follows immediately that $$\sqrt{|xy|} = \sqrt{|x|} \sqrt{|y|} \to 0$$ as $|x|+|y| \to 0$. Hence $f$ is continuous at the origin. Since $f=0$ on both coordinate axes, the two partial derivatives do exist at the origin, and are both zero. Can you study the issue of differentiability by yourself? Hint: check that the directional derivative along the direction $\mathbf{v}=(1,1)$ does not exist at the origin.