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Math Help - Multivariable limit problem

  1. #1
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    Multivariable limit problem

    If

    <br />
limit<br />
(x,y,z) -> (1,-1,1) for<br /> <br />
(xy + yz + zx)/(1 + xyz)

    When I plug in values, I get -1/0 which is undefined. My question is, how do I prove that the limit does not exist? I can't set y=mx because y and x do not necessarily follow the same path to the point (1,-1,1) and using polar coordinate substitution does not seem to work here. I tried setting z=x but that still gives me a 0 in the denominator because of the -1 for the y. Finally, multiplying the equation by various values of 1 (such as multiplying y squared to both the numerator and denominator) don't solve the undefined issue.
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  2. #2
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    Quote Originally Posted by Lord Darkin View Post
    If

    <br />
limit<br />
(x,y,z) -> (1,-1,1) for<br /> <br />
(xy + yz + zx)/(1 + xyz)

    When I plug in values, I get -1/0 which is undefined. My question is, how do I prove that the limit does not exist? I can't set y=mx because y and x do not necessarily follow the same path to the point (1,-1,1) and using polar coordinate substitution does not seem to work here. I tried setting z=x but that still gives me a 0 in the denominator because of the -1 for the y. Finally, multiplying the equation by various values of 1 (such as multiplying y squared to both the numerator and denominator) don't solve the undefined issue.
    You could try seeing what happens if you take a sequence of points converging to (1,–1,1), for example \bigl(1,-1,1+1/n\bigr). Do the values of the function at those points converge to a limit as n\to\infty?
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  3. #3
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    Ah, if that were the case then the limit would be -1 ...

    But then I have an undefined value and a real value ... does that mean the limit does not exist? Or do I have to keep trying different methods to get to that point and hope that those limits don't equal -1?
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  4. #4
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    If the limit along one path to the limit point is different from the limit along a different path to the limit point, then the limit does not exist. On the other hand, if the limits along any path to the limit point are all the same, finite number, then the limit does exist.
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  5. #5
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    I tried 2 methods: 1 using what opalg gave me, and the other with n going to infinity for (1, -1 + 1/n, 1) which yielded -1/0 so I guess that settles it. The limit D.N.E. because -1/0 =/= -1.
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  6. #6
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    If the limit along any path is infinite, the limit does not exist. But your reasoning works as well.
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