Help Doing an Epsilon Delta Proof

[hungryhippo]

Homework Statement

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 :)

 

[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.

 

[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).