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Evaluate ##lim_{(x,y)\to(0,0)\frac{xy}{y-x^3}}##

  1. Oct 10, 2015 #1
    1. The problem statement, all variables and given/known data
    Evaluate ##lim_{(x,y)\to(0,0)\frac{xy}{y-x^3}}##

    2. Relevant equations


    3. The attempt at a solution
    I've tried move towards the origin along the path ##\gamma_{1} (t) = (t,2t^3)##, which leads to ##2t=0##. Then I've tried to use ##\gamma_{2} (t) = (0,t)## which leads to ##\frac{0}{t}##, then applying the L'Hospital's rule this limit is 0 too (can I always do it when I have a single-variable limit?).

    I've suspected that the limit exists and its value is 0, then I've tried to use polar coordinates:

    ##lim_{(x,y)\to(0,0)\frac{xy}{y-x^3}}##

    ##lim_{r \to 0^{+}} \frac{r^2 cos(k)sin(k)}{r(sin(k)-r^2 cos^3 (k))}##

    But I've got stuck there because I can't figure how to show that ##\frac{cos(k)sin(k)}{sin(k)-r^2 cos^3 (k)}## is bounded.


    A further question: It is related to the question on L'Hospital's rule, above. In regards to the limit

    ##lim_{(x,y) \to (0,0)} \frac{xy(x-y)}{x^4 + y^4}##
    The path ##\gamma_{A} (t) = (t,0)## leads to 0.
    ##\gamma_{B} (t) = (t,2t)## leads to ##\frac{-2t^3}{17t^4}=\frac{-2}{17t}##. Can I tend t to zero, conclude that the limit goes to the minus infinity and because of that say that the original limit doesn't exist? Maybe it seems a silly question, but until now I haven't seen something like this happening. It is always something bounded multiplied by zero.
     
  2. jcsd
  3. Oct 10, 2015 #2
    Well, I worked a little bit more on the problem and polar coordinates wasn't necessary. Taking a curve like ##\gamma (t) = (\sqrt[3]{t-t^2},t)## is enough.
    About my second question: now is quite clear that if I choose a curve that goes through the origin, then I have to study what happens when t -> 0 to conclude something about the behavior of the original function when (x,y) -> (0,0).
     
  4. Oct 10, 2015 #3

    pasmith

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    Homework Helper

    You don't need to resort to L'Hopital's theorem when the numerator is constantly zero: the limit will be zero.

    The obvious problem here - which you do not mention - is that [itex]f(x,y) = \frac{xy}{y - x^3}[/itex] is not defined on the curve [itex]y = x^3[/itex], which of course passes through the origin.

    It isn't: For each [itex]r[/itex], it is unbounded when [itex]\sin k = r^2 \cos^3 k[/itex]. But what you actually want is the limit of [itex]r[/itex] times that expression. That might be bounded as [itex]r \to 0[/itex], even if [itex]\frac{cos(k)sin(k)}{sin(k)-r^2 cos^3 (k)}[/itex] is not. But even that would not be sufficient to conclude that the limit of [itex]f[/itex] exists: it only considers what happens as you approach along straight lines, and there are other ways of approaching the origin.

    Exercise: Set [itex]\delta > 0[/itex] and consider the line [itex]y = \delta - x[/itex] for [itex]0 < x < \delta[/itex]. This lies inside the open ball [itex]\|(x,y)\| < \delta[/itex]. Look at [itex]f(x,\delta - x)[/itex]. What happens to the sign of the denominator between [itex]x = 0[/itex] and [itex]x = \delta[/itex]?

    What can you then conclude about the existence of the limit of [itex]f[/itex] at the origin?

    Yes.
     
  5. Oct 10, 2015 #4
    Thank you for your insightful answer. About the exercise, I'm not sure, but it seems that ##y## goes through (0,##\delta##) and (##\delta##,0). Then ##x^3## (which also lies on the circle) will be greater than ##\delta - x## at some point, so the sign of the denominator becomes negative. ##y## won't pass at the origin, but the sign will always change no matter how small delta is. So, can I conclude the limit of ##f## at origin doesn't exist precisely because of that behavior?
     
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