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Having trouvle with a quick limit provlem.

  1. Jun 3, 2012 #1
    [itex]F(x,y)-> (1,1)[/itex]
    (x^2 + y^2) / (x^2+y^2-2)


    I am just having an issue as I don't know how to deal with the -2 in the denominator.

    Thank you
  2. jcsd
  3. Jun 3, 2012 #2


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    What is the problem? Are we to assume that the second line is the definition of F? Is the problem to determine a value of (x, y) such that F(x,y)= (1, 1)? x]

    But that can't be right because [/itex]F(x,y)= (x^2+ y^2)/(x^2+ y^2+1)[/itex] maps pairs of numbers, (x, y) to a single number, not a pair of numbers.

    So please got back, read the problem again, and tell us what it really says.
  4. Jun 3, 2012 #3
    One part of the problem is discovering if the limit exists or not, then you have to prove it.

    The discovery itself can be difficult. Here are some strategies for that. Try some curves like straight lines with slope m and parabolas that go through the point, so that it is a one dimensional limit along those curves. You may find that the limit doesn't exist this way. But If you find no issue here, this does not prove that the limit exists, for the limit relies on any curve, which is hard to consider all, so we let x,y go to x0,y0 in an arbitrary way (compare proofs of 2-d limits that do exist).

    The algebra was not telling me anything quickly, so I cheated and looked at the plot on wolfram alpha (on my phone yo!). That told me that the limit didn't exist, and to try the curve y=x, and that for x<1, the denominator is negative, while for x>1, the denominator is positive. Oh, and sometimes when the limit goes to infinity, we say the limit exists, but not here even since different paths give positive and negative infinity.
  5. Jun 3, 2012 #4
    You can simplify the problem a little bit by noting the symmetry that reduces it to the limit as [itex]\xi[/itex] goes to 2, where [itex]\xi[/itex] = x[itex]^{2} [/itex] + y[itex]^{2} [/itex]. It may be instructive to plot the function (taking care to limit the results to constrain the blow-up as [itex]\xi[/itex]-2 becomes < 1 (as in the attached Mathcad-derived image of a cross-section through the curve)).

    As you note, the the limits are different depending upon whether [itex]\xi[/itex] [itex]\stackrel{+}{\rightarrow}[/itex] 2 or [itex]\xi[/itex] [itex]\stackrel{-}{\rightarrow}[/itex] 2.

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    Last edited: Jun 3, 2012
  6. Jun 3, 2012 #5
    Thanks for the responses guys.

    The actualy question reads
    [itex] g(x,y)=(x^2+y^2)/(x^2+y^2-2) [/itex]
    The question asks if this function has a limit as (x,y) approaches (1,1)

    To me it doesn't mean anything if you look at the value slightly left of the function and the value slightly right of it. As this means you could be approaching the point from above and from below respectively.

    However I remember reading about that in my book, but I thought it only made a difference if it was at the Specific point. I'm not sure if I am completely getting it yet.
  7. Jun 4, 2012 #6
    OK, if the question is after terminological exactitude, then there is no limit as the one-sided limits have to be equal for The Limit to exist. (If you look at the Mathcad image you should find that Mathcad states The Limit is undefined, even though the one-sided limits are.)

    Last edited: Jun 4, 2012
  8. Jun 4, 2012 #7
    Ozone, I am not sure I understand exactly where you are having problems. Please explain further your thinking.
  9. Jun 4, 2012 #8
    I was just having some trouble dealing with this in 3 space, but if the limit has to be identical and either side of the point, then this is a good enough way for me to prove that it doesn't exist. I just wanted to make sure I knew how to sufficiently prove this on my upcoming test. Thank you everyone
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