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Proving this limit

  1. Mar 28, 2008 #1
    1. The problem statement, all variables and given/known data

    How can I show that the limit exists (or doesn't exist) for this function and prove it? I can't think of a function that will sandwich it to show it's 0 or a way to set x and y to make the limit not equal to zero! (oh and I'm trying to do all this without the use of delta epsilon methods!)

    Thank you for all your help!



    2. Relevant equations

    lim as (x,y) goes to (0,0) for the function:
    [(X^2)(Y^2)] / [X^4 + Y^2]

    3. The attempt at a solution

    I've tried looking at the limit by setting x and then y to zero and moving along each axis, which gives a limit of zero. So does using y=mx and approaching from a straight line. I've tried a couple of non-linear substitutions for y or x but it doesn't seem to get me anywhere. I guess that everything points to a limit of zero, but my problem is that you can only PROVE something does not have a limit by using the above methods and to PROVE that something has a specific limit then I believe that you require the sandwich rule and use 0 as the lower bound. Can anyone please help me with a function to use in the sandwich rule for this question?

    Thanks for your help guys! :)
     
  2. jcsd
  3. Mar 28, 2008 #2

    tiny-tim

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    … try polar coordinates …

    Hi CathyC! :smile:

    Hint: with x^2 and y^2, and being interested in (0,0), the obvious trick would be to try putting x = r.costheta, y = rsintheta.

    Does that help? :smile:
     
  4. Mar 28, 2008 #3
    Can you give me a bigger hint then that? I've always been terrible at trig. I tried looking at sin(2theta)=2(sintheta)(costheta) and playing around with that but it doesn't seem to get me anywhere as theta can be anything.
     
  5. Mar 28, 2008 #4

    tiny-tim

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    Hi CathyC! :smile:

    Forget polar coordinates (they do work, but …).

    I just did a little logical thinking, and came up with:

    divide top and bottom by (x^2)(y^2), to give:

    1/[1/(x^2) + x^2/y^2];

    and then the bottom … ? :smile:
     
  6. Mar 28, 2008 #5
    Sorry, I still can't work it out :'( took some time to play around with it, but I still can't see it. It takes me a long time to see the obvious with trig.
     
  7. Mar 28, 2008 #6

    HallsofIvy

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    There is not a whole lot of trig involved here- just algebra. There is certainly no reason to worry about trig identities like sin(2[itex]\theta[/itex]). You have
    [tex]\frac{X^2Y^2}{X^4 + Y^2}[/tex]

    If [itex]X= r cos(\theta)[/itex] and [itex]Y= r sin(\theta)[/itex], what is X2? What is Y2? What is X4?

    The crucial point here is that (X,Y) going to (0,0) means that r goes to 0- no matter what [itex]\theta[/itex] is. Is the limit, as r goes to 0, independent of [itex]\theta[/itex]?
     
    Last edited: Mar 28, 2008
  8. Mar 28, 2008 #7

    tiny-tim

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    Hi CathyC! :smile:

    ok, 1/[1/(x^2) + x^2/y^2];

    we want to prove that this -> 0,

    Now, the bottom is the sum of two squares, so they're obviously both positive.

    So 1/[1/(x^2) + x^2/y^2] < 1/[1/(x^2)].

    So, for any epsilon, choose (x^2 + y^2) < epsilon;

    then |1/[1/(x^2) + x^2/y^2]| = 1/[1/(x^2) + x^2/y^2] < 1/[1/(x^2)] = x^2 < epsilon. :smile:

    (in trig, that would be 1/[1/r^2cos^) + cot^] < 1/[1/r^2cos^)] = r^2cos^ < r^2)
     
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