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Proving a limit for a multi-variable equaton

  1. Apr 3, 2009 #1
    1. The problem statement, all variables and given/known data

    Using the definition of a limit, prove that

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

    2. Relevant equations

    Now, I know that the limit of f(x, y) as (x, y) approaches (a, b) is L such that lim (x, y) --> (a, b) f(x, y) = L. Also, for every number epsilon > 0, there is a delta > 0 such that |f(x, y) - L| < epsilon.

    I believe the above is the definition of a limit of two variables.

    3. The attempt at a solution

    In a sense, f(x, y) --> L (two VALUES) as (x, y) --> (a, b) (two POINTS). By making the distance between points (x, y) and (a, b) extremely small (some value epsilon), we make the distance between f(x, y) and L (some value delta) subsequently small. For any interval [L - epsilon, L + epsilon], there is a subsequent plane with center (a, b) and radius delta > 0 satisfying this.

    What I want to do is use some very small value of epsilon to find a value of delta that satisfies the definition of a limit.
     
  2. jcsd
  3. Apr 3, 2009 #2

    HallsofIvy

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    Generally, for problems like this, the best thing to do is to change to polar coordinates: [itex]x= rcos(\theta)[/itex], [itex]y= rsin(\theta)[/itex] because then r alone measures how close to (0,0) we are. If the limit as r goes to 0 is independent of [itex]\theta[/itex], then the limit exists and is that value. If the limit as r goes to 0 depends on [itex]\theta[/itex] then the limit of the function does not exist.
     
  4. Apr 3, 2009 #3
    [tex]|\frac{x^2y^2}{x^2+2y^2}-0|=|\frac{x^2y^2}{x^2+2y^2}|=|\frac{x^2}{x^2+2y^2}y^2|=|\frac{x^2}{x^2+2y^2}|y^2<y^2<x^2+y^2<\delta^2=\epsilon[/tex]

    So for a choice of :

    [tex]\delta=\sqrt{\epsilon}[/tex] it would work out.


    [tex]x^2<x^2+2y^2=>\frac{1}{x^2+2y^2}<\frac{1}{x^2}=>\frac{x^2}{x^2+2y^2}<1[/tex]


    [tex]x^2+y^2<\delta^2[/tex]

    edit: Halls suggestion is correct and helpful no doubt. but you have more options now!
     
  5. Apr 16, 2009 #4
    Thank you, you guys were very helpful!
     
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