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Limits: Fxns of 2 variables

  1. Feb 29, 2008 #1
    Hello

    I've only recently begun studying (in math, anyway) functions of two variables, so forgive me if my terminology is vague or incorrect (or if I am completely misguided!).

    Suppose I want to find the limit of f(x,y) as (x,y) --> (0,0). Now, I know that the limit will exist (other conditions satisfied) only if f(x,y) --> A as (x,y) --> (0,0) from ALL directions.

    To prove that the limit = Q, would it thus be sufficient to be able to show a.) that for approaches along ALL lines y = cx (c some real number), the limiting value f(x,y) will be independent of the value of c? (and since all curves through the origin can be approximated in this vicinity by a tangent line through the origin, surely this implies that all possible directions are accounted for) And then b.) that if the limiting value can be found to be Q for approach along SOME line, then by a.), the limit must exist and = Q?

    Is this acceptable? Thanks for your help.
     
    Last edited: Feb 29, 2008
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  3. Feb 29, 2008 #2

    HallsofIvy

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    No, that is not sufficient. For example, if
    [tex]f(x,y)= \frac{x^2y}{x^4+ y^4}[/tex]
    for (x,y) not (0, 0), then the limit of f(x,y), as (x,y) goes to (0,0) along any straight line, is 0. But the limit of f(x,y), as (x,y) approaches (0,0), along the parabola, y= x2, is 1/2, so there are points arbitrarily close to (0, 0) such that f(x,y) is NOT close to 0 as well as point arbitrarily close to (0, 0) such that f(x,y) is NOT close to 1/2. That function has no limit at (0,0).

    (That example is from Salas, Hille, and Etgen's Calculus text.)

    Well, no, absolutely not! There are much simpler examples where approaching (0, 0) along two different lines give different values and so the limit itself does not exist.
    For example,
    [tex]f(x,y)= \frac{xy+ y^3}{x^2+ y^2}[/itex]
    obviously gives limit 0 as you approach (0, 0) along the x and y axes but limit 2/5 as you approach along the line y= x.
    (Again, that example is from Salas, Hille, and Etgen.)

    No, it is not acceptable. You can't prove that you get the same limit approach by every possible path- there are just too many possibilities.

    In my opinion, the best thing to do is to convert to polar coordinates. That way, the "closeness" to (0, 0) is determined entirely by the variable r- if you can show that the limit, as r goes to 0, is independent of [itex]\theta[/itex], then the limit exists and is that value. For example, suppose
    [tex]f(x,y)= \frac{3x^3}{x^2+ y^}[/tex]
    as long as (x, y) is not (0,0). We can look at as many lines or curves through (0, 0) as we wish and show that the limit is 0 but we can do all possible curves so that doesn't prove that the limit exists.

    Changing to polar coordinates, [itex]x= r cos(\theta)[/itex] and [itex]y= r sin(\theta)[/itex] so
    [tex]f(x,y)= f(r,\theta)=\frac{3r^3 cos^3(\theta)}{r^2cos^2(\theta)+ r^2sin^2(\theta)}= \frac{r^3 cos^3(\theta)}{r^2}= r cos(\theta)}[/tex]
    which goes to 0 as r goes to 0 no matter what [itex]\theta[/itex] is.

    That tells us that, close enough to (0, 0), the value of f(x,y) is close to 0 so the limit is 0.
     
  4. Feb 29, 2008 #3
    Concerning a), the set of all lines of the form y = cx does not include the line consisting of the y-axis so not all possible directions are accounted for. And why would the value of c not matter?

    Concerning b), just because for some line the limit is Q doesn't mean that for other lines the limit will be Q as well.
     
  5. Feb 29, 2008 #4
    Ok, I see where I went wrong. Thank you for your help! HallsOfIvy, the polar co-ordinate transformation seems like a very good approach. Thanks a lot!
     
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