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Homework Help: Need to bound this from above

  1. Oct 13, 2005 #1
    Hi all,
    we've been doing multi-variable functions and one exercise involves (or at least in the way I've been solving it) the need to bound the following from above (x and y go to infinity):

    \left| \frac{x+y}{x^2 - xy + y^2}\right|

    What I have done so far:

    \left| \frac{x+y}{x^2 - xy + y^2}\right| = \frac{1}{\left|x+y\right|}\ \left|\frac{(x+y)^2}{x^2-xy+y^2}\right| \le \frac{1}{\left|x+y\right|}.K

    You know, I think I could prove that the second part is <= K for some K and using the theorem about the product of limited function and function going to zero I would have it.
    Anyway, I can't find that K...Could you help me please?
    Thank you.
    Last edited: Oct 13, 2005
  2. jcsd
  3. Oct 13, 2005 #2


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    Homework Helper

    I'm not sure I understand the problem. What I think you're saying is you need an upper bound for

    [tex]\left| \frac{x+y}{x^2 - xy + y^2}\right|[/tex]

    for x and y any real numbers. BUT THAT doesn't seem to be the case because the little thing diverges at 0 to infinity. So I'm not sure what kind of limit you're looking for.

    Anyway, here's some general advice that you may find useful:

    In this sort of problem, where there are several degrees of freedom contributing to a problem, you should always consider hunting around for a transformation that into new coordinates that makes the problem easier. What you want to do is to xform the problem into one where one of the coordinates can be trivially seen to maximize the function.

    This is one of those tricks that may only be learned by seeing a few examples worked out. The problem you have is one where the bottom looks somewhere between (x-y)^2 and (x+y)^2 while the top looks like (x+y). So I'd be looking at sums and differences of x and y.

  4. Oct 14, 2005 #3
    Thank you Carl,

    I'm sorry if my initial post isn't clear. I don't want to bound for any x and y real, both x and y go to infinity and in that case it can be bounded by some K from above...
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