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Limit in R^2

  1. Aug 21, 2007 #1
    I'm sort of new to the calculation of limits of functions defined in R^2( or R^n, n>=2)
    exp. the limit of sqrt{hk/(h^2+k^2)} as (h,k) --> (0,0)
    I think it should be 1/sqrt(2) if we let h=k. but i'm not sure if this is the right approach.
    what about other limits in the form "0/0" as x={x1,x2...xn}-->0 (x is an element of R^n)?
    just make x1=x2=...=xn?

  2. jcsd
  3. Aug 22, 2007 #2
    What happens if you use polar coordinates?
  4. Aug 22, 2007 #3
    Limits require the idea of distance. Well defined limits are those for which the answer is the same regardless of the direction of approach. In [tex]\mathbb{R}^n[/tex] there's the usual distance function as the sqrt of the sum of squares (also known as the 2-norm). So following AiRAVATA's suggestion, let [tex]h=r \cos \theta, k = r \sin \theta[/tex] and take the limit r->0. If this doesn't depend on theta, then you've got a well defined limit.
  5. Aug 22, 2007 #4
    A little comment on this. It is not sufficent to first fix theta, then take limit r->0, and be happy if the result was independent of theta.
  6. Aug 22, 2007 #5
    Indeed -- I meant that theta must be left free, and the limit must be found to be independent of theta. Sorry if I was unclear.
  7. Aug 22, 2007 #6
    well, if let [tex]h=r \cos \theta, k = r \sin \theta[/tex], [tex]lim_{(h,k)->(0,0)} \sqrt{\frac{hk}{h^2+k^2}}}[/tex] can be simplified to [tex]lim_{r->0}\sqrt{\cos \theta\sin \theta}[/tex] which is dependent on theta however, how should I proceed then?
  8. Aug 23, 2007 #7
    Then you have a problem. The limit is not well defined.
  9. Aug 23, 2007 #8


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    The problem is that limits in R2 must give the same thing when the limit point (here (0,0)) is approached along any path. Most Calculus texts give examples (often in the exercises) of functions that have the same limit as (0,0) is approached along any straight line, but a different limit approaching (0,0) along a parabola.

    In this case, it is simplest to do it the opposite way:
    [tex]\frac{hk}{h^2+ k^2}[/tex]
    if h= k, becomes
    [tex]\frac{h^2}{2h^2}= 1/2[/tex]
    for all h and so has limit, as (0,0) is approached along the line h= k, of 1/2.
    However, if we take h= 0, it becomes
    [tex]\frac{k*0}{k^2}= 0[/tex]
    for all k and so has limit, as (0,0) is approached along the line h=0, of 0.

    Since those two limits are different, the limit itself does not exist.

    If the limit does not exist, it is sufficient to show that limits along two different curves are different.

    Of course, if the limit does exist, then showing that you get the same thing along any number of curves is not sufficient. Usually, then, the simplest way to find the limit is to do as was originally suggested: convert to polar coordinates so that the distance from the origin depends on the single variable r. If the limit as r goes to 0 is independent of [itex]\theta[/itex] then that value is the limit of the function.
    Last edited by a moderator: Aug 23, 2007
  10. Aug 23, 2007 #9
    Thank you. HallsofIvy
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