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Quick limit question

  1. Jan 28, 2014 #1
    I want to show that a function x(a,b) is bounded by some value l, knowing that the extremes of the function occur when a = b = k is approached. The function blows up here.

    Being a function of two variables, is it valid proof to set a EQUAL TO k, and then take the limit of this new x(b) as b -> k, getting l as the result?
     
  2. jcsd
  3. Jan 28, 2014 #2

    jgens

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    Not knowing the specific situation here my hunch is no. In the general case you definitely need to take limits over all paths (a,b)->(k,k) instead of just checking a specific one.
     
  4. Jan 28, 2014 #3
    I was under the impression that all of the limits would be the same if the limit exists?
     
  5. Jan 28, 2014 #4

    jgens

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    Ah! If you know a priori that the limit exists, then yes fixing a specific path works.
     
  6. Jan 28, 2014 #5
    Well, I guess I don't, really. I guess I am just making the assumption because the function looked very simple. I have forgotten all about the fact that this is a "path" limit.

    Basically, this is the function (k is constant):

    x(a,b) = (a-b)/((ba/k^2) - 1)

    Where a,b are in (-k, k).
    I needed to show that x is in (-k,k) also. Doing algebraic manipulations with inequalities didn't lead me to the result, so I remarked that the functions extremes are when ba = k^2 and so a = k and b = k or a = -k and b = -k, then I proceeded to do as I explained in the OP to show what those values were. I never really considered the possibility that the function's limit did not exist because it seemed so simple (and I think the "almost" symmetry of the function (switching a and b only switches sign) also played a part in thinking I could fix a path.)

    I think I need to review some things, though.
     
  7. Jan 28, 2014 #6

    jgens

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    It is easy to see the limit does not exist in that case. For simplicity I will assume k=1 and consider the paths where 1) a=1 and b→1 and 2) a→1 and b=1. In the first case we have x(a,1) = (a-1)/(a-1) = 1 so the limit along this path returns the value 1. In the second case we have x(1,b) = (1-b)/(b-1) = -1 so the limit along this path returns the value -1. If the limit existed, then these two would agree, so evidently the limit does not exist.
     
  8. Jan 28, 2014 #7
    Does there exist some path where the limit is not k or -k?
     
  9. Jan 28, 2014 #8

    jgens

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    In this case there is such a path. Go along the line a=b as (a,b)→(k,k).
     
  10. Jan 28, 2014 #9
    Alright, it looks like I need to find some other method to show that the range is in (-k, k) then.

    Thank you for all the help, do you have any suggestions?

    Edit: Please keep any suggestion brief, I would like the joy of solving it.
     
    Last edited: Jan 28, 2014
  11. Jan 29, 2014 #10

    jgens

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    Well I think this function is actually unbounded around (k,k). So hopefully that helps put you on the right track.
     
  12. Jan 30, 2014 #11

    jgens

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    Oops! Totally missed that you have a restricted domain -k < a,b < k. In this case your function is indeed bounded and you can prove this essentially by arguing cases. Sorry about that.
     
    Last edited: Jan 30, 2014
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