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Limit quest

  1. Apr 24, 2005 #1
    I'm searching for a function f(x,y), such that

    1) [tex] \lim_{t\rightarrow 0} f(at,bt)=E \quad\forall(a,b)\ne(0,0) [/tex]
    2) [tex] \exists a,b|\lim_{t\rightarrow 0}f(at,bt^2)=F\ne E [/tex]
  2. jcsd
  3. Apr 24, 2005 #2


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    Sounds like homework. Anyways, what thoughts have you had so far?
  4. Apr 24, 2005 #3
    I think it's impossible, but I wanted to have other people opinion. I will tell you why :

    take f(x,y) and just use polar coordinates : x=r*cos(theta), y=r*sint(theta). Just put : t=r, cos(theta)=a, sin(theta)=b. Then from 1), the limit as (x,y)->(0,0) exists and is E. Hence, it should be independent on how you tend towards (0,0).

    Since b) is a special case of the limit (x,y)->(0,0), then it should be equal to E.

    But I'm not sure if this reasoning is correct.
  5. Apr 24, 2005 #4


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    You have it backwards -- if you show the limit doesn't depend on how you approach 0, then you can conclude the limit exists.

    All you've shown is that if the limit exists, it is E.
  6. Apr 24, 2005 #5
    I don't think I know much about this type of math, but the restriction (a, b) != (0,0) only applies to 1). In 2), (a, b) can equal (0, 0), and you can use that.
  7. Apr 24, 2005 #6


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    You don't have to "cheat" like that, though. In fact, I think you can change part 2 to say "For all a and b", but the problem then becomes more difficult.

    Anyways, you try converting part 2 to polar coordinates too?
  8. Apr 25, 2005 #7
    Yes Bicycletree is right : (a,b) should be different than (0,0) in b...

    In fact the question can be simply stated in words : if you show that approaching (0,0) on every straight line leads to the same limit, then can we deduce this is the limit approaching whatever way you want (on curves)...I think I could just say : a curve, when becoming near to 0 can be approximated by a straight line ?
  9. Apr 25, 2005 #8
    I don't want to answer your question exactly but I want you to consider the following polar function [tex] f(r, \theta) = r/\theta [/tex] where [tex]0 < \theta \le 2\pi[/tex]. Certainly if you fix a theta the limit as r approaches 0 is 0. But if you take the path [tex]g(t) = (1/t,1/t) [/tex]. Then [tex]f\circ g(t) =1[/tex] for t>1. But the path is spiraling towards the origin since r is approaching 0. This is a cooked example that won't satisfy your question but first try creating a cooked example that will satisfy your question (something piecewise... this is easy) then if you are feeling adventurous you can come up with some a little less cooked looking (harder).

    good luck
  10. Apr 25, 2005 #9

    matt grime

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    But once more you're assuming that f must be continuous (at 0,0) to show that the result is false, surely that should tell you where to look for a counter example?

    The common way to do this kind of question is to make f(x,y) a function of the form g(x,y)/h(x,y) where you can in the first case pull out factor of t so that we're left with f(ta,tb)=tf(a,b) but if we put in (at,bt^2) we cancel all the factors of t so that f(at,bt^2)=f(a,b) and the limit is independent of t (but such that f is not constant).

    Examples that don't quite work here:

    f(x,y) = x^2/y

    f(ta,tb)=ta^2/b which tends to zero as t tends to zero

    f(ta,t^2b)= a^2/b

    of course this doesn't quite work since b could be zero. I leave it to you to sidestep this problem.
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