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Help Doing an Epsilon Delta Proof

  1. Oct 22, 2009 #1
    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 :eek:

    Thanks in advance :) :confused:
     
  2. jcsd
  3. Oct 23, 2009 #2
    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.
     
  4. Oct 23, 2009 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    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 [itex]\theta[/itex]. The function will be continuous at (0,0) if and only if the limit, as r goes to 0, does not depend on [itex]\theta[/itex].

    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).
     
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