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Math Help - Evaluating the limit of f(x,y)

  1. #1
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    Evaluating the limit of f(x,y)

    Evaluating the limit of f(x,y)-codecogseqn.png

    I have to evaluate the limit of the function above, if it exists. My text-book says that the limit to the function doesn't exist, but I really understand how.

    Here's how I solved it:


    Evaluating the limit of f(x,y)-codecogseqn-6-.png

    Evaluating the limit of f(x,y)-codecogseqn-2-.png

    Evaluating the limit of f(x,y)-codecogseqn-3-.png



    It wouldn't be the first time something was misprinted in my textbook, but I just want to make sure.

    Thanks in advance,

    Giest
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  2. #2
    Senior Member jakncoke's Avatar
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    Re: Evaluating the limit of f(x,y)

    yes u are correct.
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  3. #3
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    Re: Evaluating the limit of f(x,y)

    Contrary to what jakncoke says, your text book is correct, you are not. In order that a limit, of a function of two variables, exist, there must be some region around the point in which the function exists and has value sufficiently close to that value. Here, no matter how close to (3, 3) you are, there exist points, on the line y= x, for which f(x, y) does not exist.

    Remember that, even in functions of one variable, the function f(x)= \frac{x^2- 4}{x- 2} is NOT the same as g(x)= x+2. The first is not defined at x= 2 while g(2)= 4. Fortunately, the two functions differ only at x= 2 and so have the same limit there.
    In two dimensions, the problem is more complicated. f(x,y)= \frac{x^3- y^3}{x^2- y^2} and g(x,y)= \frac{x^2+xy+ y^2}{x+ y} are equal except where x= y. And that line includes (3, 3) as well as points arbitrarily close to (3, 3).
    Last edited by HallsofIvy; March 7th 2013 at 09:02 AM.
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  4. #4
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    Re: Evaluating the limit of f(x,y)

    The formal definition of limit as x approaches a, (no matter where x and a are, R, R2, etc) requires that a be a cluster point of the domain of the function. That is, every neighborhood of a contains points of the domain other than a. Without this requirement limits wouldn't even be unique.

    So now your specific situation. When writing a constraint under a limit, such as x = y, this is just shorthand for specifying the domain of the function is the constraint. So if you really meant x = y, the domain of the given function is empty and so (3,3) is certainly not a cluster point. Hence the limit does not exist. On the other hand, your proof where you cancelled the factor x-y seems to indicate the x = y is a typo. If for example you want the limit with constraint x unequal to y, then (3,3) is certainly a cluster point of the domain and your proof is correct.
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