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

## Homework Statement

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

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Well, I worked a little bit more on the problem and polar coordinates wasn't necessary. Taking a curve like ##\gamma (t) = (\sqrt{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).

pasmith
Homework Helper

## Homework Statement

Evaluate ##lim_{(x,y)\to(0,0)\frac{xy}{y-x^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?).
You don't need to resort to L'Hopital's theorem when the numerator is constantly zero: the limit will be zero.

I've suspected that the limit exists and its value is 0, then I've tried to use polar coordinates:
The obvious problem here - which you do not mention - is that $f(x,y) = \frac{xy}{y - x^3}$ is not defined on the curve $y = x^3$, which of course passes through the origin.

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

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

What can you then conclude about the existence of the limit of $f$ at the origin?

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?
Yes.

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?