# Calculus III - Multivariate Continuity

1. Jan 22, 2012

### Jonmundsson

1. The problem statement, all variables and given/known data
Let
$\begin{equation*} f(x,y) = \begin{cases} \dfrac{x^3 - y^3}{x^2 + y^2}, \hspace{1.1em} (x, y) \neq (0,0) \\ 0, \hspace{4em} (x,y) = (0,0) \end{cases} \end{equation*}$

Is $f$ continuous at the point $(0,0)$? Are $f_x$ og $f_y$ continuous at the point $(0,0)$?
2. Relevant equations

Polar coords

3. The attempt at a solution

If you convert f to polar it's easy to see that it is continuous (since it doesn't depend on $\theta$. I'm just wondering if the derivative of the polar function is dependent on $\theta$ then it isn't continuous and therefore neither are $f_x$ and $f_y$

Steps:

$\displaystyle \lim _{(x,y) \to (0,0)} \dfrac{x^3 - y^3}{x^2 + y^2} = \lim _{r \to 0} \hspace{0.3em} r (cos^3 \theta - sin^3 \theta) = 0$

Define $g(r) = r (cos^3 \theta - sin^3 \theta)$ then $g'(r) = cos^3 \theta - sin^3 \theta$ and $\displaystyle \lim _{r \to 0} cos^3 \theta - sin^3 \theta$ doesn't exist.

Cheers.

2. Jan 22, 2012

### SammyS

Staff Emeritus
How about ∂f/∂x and ∂f/∂y? Are the continuous at (0, 0) ?

3. Jan 22, 2012

### Jonmundsson

They aren't. I'm asking whether it is a coincident that the derivative of g and the partials of f aren't continuous.