# Defining the continuity and differentiability of multi variate functio

1. Jun 6, 2013

### trap101

Let f: R2-->R be defined by f(x,y) = xy2/(x2+y2 if (x,y) ≠ 0, f(0,0) = 0

a) is f continuous on R2?
b) is f differentiable on R2?
c) Show that all the dirctional derivatives of f at (0.0 exist and compute them

Attempt:

a) I had an idea to show that multivariate functions are continuous, but now I think it's faulty. I did this:

xy2/(x2+y2 ≤ |xy2|/|(x2+y2| < |xy2|

now taking the lim (x,y)-->(0,0) of |xy2| is 0 as such the function is also continuous.

But looking at a similar example from my notes where the only difference in the function was:

xy2/(x2+y6

We had shown that this was not continuous. So now I'm not so sure if the method I used was valid.

b) I'm a little stumped on showing differentiability in this setting, it doesn't seem to involve using the definition directly because of all the possible paths. So what other option may I have?

This is all in terms of discussing the trouble spot. i.e approaching (0,0)

2. Jun 6, 2013

### voko

Find the limit when $x_n = y_n = \frac 1 n$, and when $x_n = y^2_n = \frac 1 n$.

3. Jun 6, 2013

### trap101

I dont follow. By doing that i get different solutions taking n-->0.

4. Jun 6, 2013

### Zondrina

So you have :

$f(x,y) = \frac{xy^2}{x^2 + y^2}$ if $(x,y) ≠ (0,0)$
$f(x,y) = 0$ if $(x,y) = (0,0)$

For part a) where you're asked if $f$ is continuous, this should ring bells about the $ε - δ$ definition if you want to show it rigorously. Notice you can already tell that the function is continuous everywhere ( since it is a quotient of polynomials and polynomials are continuous ) except at (0,0). So using the definition with the information we're given :

$\forall ε > 0, \exists δ > 0 \space | \space 0 < |(x,y) - (0,0)| < δ \Rightarrow |\frac{xy^2}{x^2 + y^2} - 0| < ε$

So now you want to massage the expression into a suitable form :

$|f(x,y) - L| = \frac{|x|y^2}{x^2 + y^2}$

Now, using the fact that $|x| < δ$ and $|y| < δ$ you should be able to continue.

5. Jun 6, 2013

### voko

n was supposed to be integer, going to infinity. Does that make (x, y) go to zero?