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Limit as (x,y,z)->0, I missed it on the exam but why?

  1. Oct 26, 2005 #1
    Hello everyone, This problem is bugging me because the professor showed me this is how you do it, the day before the exam, so what do i do? well I do it the way he told me, and I totally miss it. Here is the problem:

    I'm going to let f(x) be the function to make it easier to read:
    f(x) = [xy+yz+zx]/[x^2+y^2+z^2];

    Lim f(x);
    (x,y,z)->(0,0,0)

    So I let
    Lim f(x);
    (x,y,z)->(t,3t,0)
    and i got:
    (xy)/(x^2+y^2) = (t)(3t)/(t^2+(3t)^2) = 3t^2/10^2 = 3/10;

    Lim f(x);
    (x,y,z)->(0,3t,t)

    yz/(y^2+z^2) = (3t)(t)/((3t)^2+t^2)) = 3/10;

    Lim f(x);
    (x,y,z)->(t,0,3t)

    zx/(x^2+z^2) = 3/10;

    So I said the limit exists because they all go to 3/10, and yet he marked it wrong. Any ideas why this is wrong? Thanks. He's saying the limit doesn't exist.
     
  2. jcsd
  3. Oct 26, 2005 #2

    TD

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    Unfortunately, it's not sufficient that you find 3 (or "n") different paths to let (x,y,z) go to (0,0,0) with the same limit to conclude that limit is correct. In reverse though, it is sufficient to find two paths which yield a different result to conclude that the limit doesn't exist.

    As long as you're calculating it by approaching it differently and you keep finding the same values - the limit either exists (and is equal to that value), but you have to prove that then, or you haven't found a 'good' path yet to show it doesn't exist.

    Have you tried approaching it by approaching from one of the axis alone? E.g. (t,0,0)
     
    Last edited: Oct 26, 2005
  4. Oct 26, 2005 #3

    HallsofIvy

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    I guarantee that was not the way he told you! He may very well have shown how by getting different answers along different lines you can show that a limit does not exist but I feel certain he never said that getting the same limit proves that the limit does exist!
    There are, by the way, examples, in just about any Calculus book, in which you get the same limit approaching the origin along any straight line, but a different limit approaching along a parabola- so even showing that you get the same thing along any straight line does not guarantee you will have a limit.

    It would have been interesting if your professor had given you a problem where the limit did exist and was, say, 3/10 - and you argued that, since taking the limit along 3 different lines all gave 3/10, the limit was 3/10.
    A good professor would have marked that wrong- right answer, wrong reasoning- and it's the reasoning that is important! The best way to prove that a limit does exist, in problem with the limit at (0,0,0), is to convert to spherical coordinates. That way a single variable, [itex]\rho[/itex], measures the distance from the origin. If the limit as [itex]\rho[/itex] exists and is independent of the other variables, then the limit of the function exists and is equal to that limit.
     
    Last edited: Oct 26, 2005
  5. Oct 26, 2005 #4
    I have a question on that limit question too, is it legal to convert is to polar and let z=0, it still doesnt exist that way, but are you allowed to convert and approach on z=0?
     
  6. Oct 26, 2005 #5

    HallsofIvy

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    Approach "on z= 0"? Do you mean choose only paths in the xy-plane? Yes, that's "legal" but still doesn't prove that approaching along other paths won't give you a different answer. Or did you mean approach along the z-axis (letting x and y= 0)? Same answer!
     
  7. Oct 26, 2005 #6
    I mean take f(x) = [xy+yz+zx]/[x^2+y^2+z^2] then say consider z=0 so
    f(x) becomes

    f(x) = [xy]/[x^2+y^2]

    now convert that to polar and it is easy to see lim dne

    is this legal?
     
    Last edited: Oct 26, 2005
  8. Oct 26, 2005 #7

    Tide

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    No. That will not work. Your particular function has the interesting property that approaching the origin from any direction in the z = 0 plane gives the same result. But the limit does not exist! Follow Halls' advice and use spherical coordinates - all will become crystal clear! :)
     
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