# Homework Help: Help Doing an Epsilon Delta Proof

1. Oct 22, 2009

### hungryhippo

1. The problem statement, all variables and given/known data

given a function defined by

f(x,y) {= |xy|^a /(x^2+y^2-xy), if (x,y) cannot be (0,0)

and = 0, if (x,y) = (0,0)

Find all values of the real number a such that f is continuous everywhere

e= epsilon
d= delta

In order to prove this, I know I need to do an epsilon delta proof for a limit. I know that,

|(x1,x2)-(y1,y2)| < d
|x1-y1, x2-y2| < d
sqrt ( (x1-y1)^2+(x2-y2)^2 ) < d

since we know, that for one (x,y) = (0,0) the above is just
sqrt(x^2+y^2) < d

also, assuming e<0

we have
|f(x,y)-f(xo,yo)|< e

saying that f(xo,yo) corresponds to when xo = 0 and yo =0, we have
|f(x,y)|< e
||xy|^a /(x^2+y^2-xy)| < e

From here on, I don't know which path to take...I don't have any background on delta epsilon and this is my first time seeing it. So i'd really appreaciate your help

Thanks in advance :)

2. Oct 23, 2009

### foxjwill

As it stands right now, there's no a that satisfies this, since f will always be discontinuous at the point (x,y) which satisfies x^2+y^2=xy no matter what a you choose.

3. Oct 23, 2009

### HallsofIvy

Are you required to use "epsilon-delta"? I would recommend changing to polar coordinates, then taking the limit as r goes to 0. The advantage of polar coordinates here is that the distance from (0,0) depends only on r, not $\theta$. The function will be continuous at (0,0) if and only if the limit, as r goes to 0, does not depend on $\theta$.

foxjwill's statement is incorrect. The fact that the denominator goes to 0 does NOT show that the limit does not exist, as, for example, the limit of f(x,y)= (x^2+ y^2- xy)/(x^2+ y^2- xy) as (x, y) goes to 0. There are, in fact, an infinite number of values for a that will make this function continuous at (0, 0).