Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

2 variable limit

  1. Feb 22, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove: [tex]lim_{(x,y)\rightarrow (0.0)} \frac{x^4 y}{x^4 + y^2} = 0[/tex]

    2. Relevant equations

    x = rcos(theta)
    y= rsin(theta)

    3. The attempt at a solution

    since (x,y) are at the origin, I could use polar coordinates

    [tex]lim_{(x,y)\rightarrow (0.0)} \frac{x^4 y}{x^4 + y^2} = lim_{(r)\rightarrow 0} \frac{r^4 cos^4 \vartheta r sin \vartheta}{r^4 cos^4 \vartheta + r^2 sin^2 \vartheta} =
    lim_{r \rightarrow 0} \frac{r^2 cos^4 \vartheta sin \vartheta}{r^2 cos^4 \vartheta + sin^2 \vartheta}[/tex]

    I got that far, but I'm not sure how to get rid of the r^2 in the denominator.
    Last edited: Feb 22, 2009
  2. jcsd
  3. Feb 22, 2009 #2
    I will try to avoid giving you an answer, but a hint is:
    The instructor/book probably wants you reorganize the problem in terms of a single parameter that unconditionally goes to zero as (x,y)->(0,0), and prove that the other parameter doesn't affect the result. In other words; you have to find a limit and also prove that the limit is independent of the particular path to (0,0); i.e. is unique.

  4. Feb 22, 2009 #3
    Oh yes, and you made a calculational mistake in your polar conversion.
  5. Feb 22, 2009 #4
    Sorry, I wrote the question wrong, but the convertion is right (or should be). I also replaces (x,y) -> (0,0) with r -> 0.

    But I'm still not sure what to do. I could divide both the numerator and denominator by r^2, but then I would have sin (theta) / r^2 but that doesnt exist.
  6. Feb 22, 2009 #5
    You still have an algebra mistake; you won't get a unique answer until it's fixed.
    Once fixed the problem is interesting, and a little messy.
    You need to convert the denominator so that you can estimate it as r->0 .
    You can do it the systematic way (which is messy), or by considering cases (which looks like hand waving but isn't). Neither seems neat.
    If you come up with a neat answer why not post it?
    I presume that you are not using LHopital's rule? That is only neat way I can see.
    I always choose the generic approach; so I am prejudiced.

  7. Feb 22, 2009 #6


    Staff: Mentor

    Instead of converting to polar coordinates, I suggest looking at the limit along a few paths. IOW, as the point (x, y) moves along the graphs of a few functions.

    Along the curve y = x^4:
    (x^4 * y)/(x^4 + y^2) = x^8/(x^4 + x^8) = x^4/(1 + x^4) [itex]\rightarrow[/itex] 0 as x [itex]\rightarrow[/itex] 0. This suggests, but isn't proof, that maybe the limit is 0.

    Try finding the limit along some other paths, using simple power functions. If the limit is still zero along the paths you try, that doesn't prove anything, but if you can find one path for which this limit is not zero, then you can say with confidence that the limit doesn't exist.
  8. Feb 22, 2009 #7


    User Avatar
    Science Advisor

    Do you need to "get rid of the r^2 in the denominator"? As r goes to 0, the numerator goes to 0, no matter what [itex]\theta[/itex] is. The denominator goes to [itex]sin(\theta)[/itex] as r goes to 0 so the fraction goes to 0, no matter what [itex]\theta[/itex] is.
    (If [itex]sin(\theta)[/itex] is 0, [itex]cos(\theta)= 1[/itex] so the fraction is [itex]0/r^2[/itex] which is 0 for all r.)
  9. Feb 22, 2009 #8
    would that actually work on a test (as in I'll get close to full marks at least)?

    As for L'Hopital's rule, my prof said it doesnt work in this case since it's 2 variables, or would it work once I used polar coordinates?
  10. Feb 22, 2009 #9


    User Avatar
    Science Advisor

    The point of putting the function into polar coordinates is that r itself measures the distance from the point to (0,0). If the limit as r goes to 0 is independent of [itex]\theta[/itex], then that is the limit of the function.
    Last edited by a moderator: Feb 23, 2009
  11. Feb 22, 2009 #10
    Okay, the flaw: the numerator has to have r^5, not r^4. The r term from r* sin() in the numerator inadvertently got dropped.
    Having said that, HallsofIvy is right even with only r^4; I didn't do the case evaluation carefully enough.
    You won't ace the test unless you demonstrate explicitly that for any epsilon you can state a delta such the | expr(r=delta)|<epsilon for all theta. There are functions such that this will fail; even if expr(r=0) is well defined.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook