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Evaluate

$$\underset{(x,y) \rightarrow (0,0)}{\text{lim}} \frac{xy}{y-x^3}$$

My attempt: I've tried to use polar coordinates $x=r\cos \theta, \; y = r \sin \theta$: $$\underset{(x,y) \rightarrow (0,0)}{\text{lim}} \frac{xy}{y-x^3} =\underset{r \rightarrow 0}{\text{lim}} \frac{r^2 \sin \theta \cos \theta}{r(\sin \theta - r^2 \cos \theta)} = 0 $$

but the book says it doesn't exists. What am I doing wrong?

Giiovanna
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1 Answers1

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Hint: See what happens if $(x,y)$ approaches $(0,0)$ along the curve $y=x^3+x^4$.

Remark: As to what you are doing wrong, implicitly you are assuming that $\theta$ is constant as $(x,y)$ approaches $(0,0)$, that we are approaching $(0,0)$ along a ray.

André Nicolas
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  • Nice answer. Could you give some insights how you figured out this curve? Is there any algorithmic method to use? – MrYouMath Sep 19 '15 at 10:30
  • Well, I saw that the bottom had to be "smaller than $x^3$" to avoid the whole thing going to $0$. The "algorithm" is just thinking about the sizes of things. But (full disclosure) I have been around a pretty long time, have taught this stuff, probably assigned problems that depended on this sort of idea. So even though it feels like figuring out, there may be a significant element of recalling. – André Nicolas Sep 19 '15 at 15:32