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Math Help - Multivariate rational function

  1. #1
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    Multivariate rational function

    I'm working on a section on limits ofmultivariate functions. In looking through the examples, for f(x,y) = (4-xy) / (x^2 + 3y^2), it is said that this is a rational function and thus continuous, the next question however, f(x,y) = y^4 / (x^4 + 3y^4), goes into analysis of the limit. Isn't the latter one also a rational function?
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    Quote Originally Posted by theowne View Post
    I'm working on a section on limits ofmultivariate functions. In looking through the examples, for f(x,y) = (4-xy) / (x^2 + 3y^2), it is said that this is a rational function and thus continuous, the next question however, f(x,y) = y^4 / (x^4 + 3y^4), goes into analysis of the limit. Isn't the latter one also a rational function?
    The first is defined everywhere except at (0,0) where it increases without bound, the second is undefined or rather indeterminate at (0,0) and so the limit there has to be investigated (and if it exists if you define the function to take that value at (0,0) this extended function will be continuous).

    CB
    Last edited by CaptainBlack; October 8th 2009 at 02:08 PM.
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    Quote Originally Posted by theowne View Post
    I'm working on a section on limits ofmultivariate functions. In looking through the examples, for f(x,y) = (4-xy) / (x^2 + 3y^2), it is said that this is a rational function and thus continuous
    Surely that is not what it says! It does not follow that a function is continuous just because it is a rational function. What is true is that a rational function is continuous [b]wherever it is defined[/itex]. The fact is that this function is NOT continuous at (0, 0) because it is not defined there.

    , the next question however, f(x,y) = y^4 / (x^4 + 3y^4), goes into analysis of the limit. Isn't the latter one also a rational function?
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    Quote Originally Posted by CaptainBlack View Post
    The first is defined everywhere and so there is no problem, the second is undefined at (0,0) and so the limit there has to be investigated (and if it exists if you define the function to take that value at (0,0) this extended function will be continuous).

    CB
    Am I missing something? How is that first function, f(x,y) = \frac{4-xy}{x^2 + 3y^2}, defined at (0,0)? Its denominator is 0 there.
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    Quote Originally Posted by HallsofIvy View Post
    Am I missing something? How is that first function, f(x,y) = \frac{4-xy}{x^2 + 3y^2}, defined at (0,0)? Its denominator is 0 there.
    Oppss... misreading things again

    CB
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