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Multivariable epsilon delta proofs

  1. Oct 18, 2012 #1
    1. The problem statement, all variables and given/known data

    lim (x,y) -> (0,0) xy/sqrt(x^2+y^2) = 0


    3. The attempt at a solution

    my understanding of my actual goal here is kind of poor

    given ε>0 there exist ∂>0 s.t. 0 < sqrt(x^2 + y^2) < ∂ then 0<|f(x,y) - L| < ε

    | xy/sqrt(x^2 + y^2) - 0 | < ε

    (xy * sqrt(x^2 + y^2)) / (x^2 + y^2) < ε


    |xy|/|(x^2 + y^2)| * sqrt(x^2 + y^2) < ε

    so some number always less than one(i think?) times ∂ < ε. I can't say i know where im going with this at all. oh god please help.
     
  2. jcsd
  3. Oct 18, 2012 #2

    tiny-tim

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    hi oreosama! :smile:

    (have a delta: δ and try using the X2 button just above the Reply box :wink:)
    the trouble with that is that you're leaving the xy out of it … you need to use your δ for xy also

    hint: either convert to polar coordinates, or divide top and bottom by xy :wink:

    (in fact, try both!)
     
  4. Oct 18, 2012 #3
    all the demonstrations my professor used didnt involve polar so I've been trying to avoid using that

    so,

    sqrt(x2+y2)/((x2+y2)/xy)

    this doesnt help much, I dont understand my goal with manipulating this stuff at all. I see (x2+y2) which maybe you could call δ2 giving you

    xy/δ simplified but I still have no idea what to do with the xy or if this is the right direction at all
     
  5. Oct 18, 2012 #4

    tiny-tim

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    = δ/√((x/y)2 + (y/x)2) :wink:
     
  6. Oct 18, 2012 #5
    δ/√((x2/y2 + y2/x2))

    δ/√((x4+y4)/(x2y2))

    experimenting with throwing the x2y2 in the radical was a nice adventure but once again i dont see what I can do :confused:
     
  7. Oct 18, 2012 #6

    tiny-tim

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    won't either x/y or y/x have to be ≥ 1 ? :smile:
     
  8. Oct 18, 2012 #7
    I see why that is, but don't know how to apply the information.
     
  9. Oct 18, 2012 #8

    SammyS

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    Shouldn't those be x2/y2 , y2/x2, or have absolute values, since x/y and y/x could be negative?
     
  10. Oct 18, 2012 #9

    Zondrina

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    Switching to polars here would not be useful, dividing by xy would be more work than needed.

    Use the fact that |x| ≤ (x2 + y2)1/2 and |y| ≤ (x2 + y2)1/2

    Now what do you see happening?
     
  11. Oct 18, 2012 #10
    as with the last one, I can see that it's a true statement but don't know what I can do about it.
     
  12. Oct 18, 2012 #11

    Zondrina

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    Using what I said :

    |x||y| ≤ x2 + y2

    So how would you continue your chain of inequalities.

    EDIT : In fact, using this information we can create as small of a neighborhood as we want because |x||y| ≤ a(x2 + y2) where a[itex]\in[/itex]ℝ
     
  13. Oct 18, 2012 #12
    √(|x||y|) ≤ √(x2 + y2) would give me a term to replace with δ


    i dont know what my goal is. shotgunning algebra manipulation until magic happens is not giving me any knowledge. what do I do?. is my goal to not have any more x and y terms? because once again I don't see them all gone..
     
  14. Oct 18, 2012 #13

    Zondrina

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    You're probably assuming that you're looking for some δ in terms of ε, while in this case you simplify down to :

    δ ≤ ε

    Now, what this means is we can make the radius of δ as small as we would like as long as we don't get bigger than ε! So choosing δ = ε or δ = ε/2 or δ = ε/a with a>1 being a real number would all work. Your particular choice does not matter in this case, so you can make your radius as small as you like, but it can be no larger than ε. So δ=2ε would not work, δ=aε would not work with a>1 as before.

    Do you sort of see what I'm trying to say here?
     
  15. Oct 18, 2012 #14

    tiny-tim

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    put δ = ε …

    will that be less than ε ? :smile:
     
  16. Oct 18, 2012 #15
    i dont even know what youre asking me to do


    im expecting at some point to simplify down to something like that yes. it seems like once the manipulation gets to the point where you have to infer that a relationship between the variables is less than something i break down. never seen it used before and I still dont know how to break these things down.
     
  17. Oct 18, 2012 #16

    Zondrina

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    Usually what you want to try and do is to get the term |f(x,y) - L| into the form 0 < |x-a|,|y-b| < δ and then find your δ in terms of ε.

    These take lots of practice to know when to apply the triangle inequality, or to know how big or small something can be. Practice, practice and more practice and eventually these will become second nature.
     
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