The way I did it is that $$\left|\frac{x^3 y}{x^2 - y}\right|\leq \left|\frac{x^3}{x^2 - y^2}\right|$$ for small values of $x$. Hence, if the limit of the expression on the right exists, then the limit of the expression on the left is the same by the Squeeze Theorem.

Since the limit goes to $(0,0)$, we can use a polar coordinate substitution. Thus, let $x=r\cos \theta$ and $y = r\sin \theta$. Then \begin{align*} \lim_{(x,y)\to (0,0)}{\left|\frac{x^3}{x^2 - y^2}\right|} &= \lim_{r\to 0}{\left|\frac{r^3 \cos^3 \theta}{r^2\left(\cos^2 \theta - \sin^2 \theta\right)}\right|} \\ &=\frac{\cos^3 \theta}{\cos 2\theta}\cdot \lim_{r\to 0}{r} \\ &=0, \end{align*} meaning the original limit (in the title) is also $0$, by the Squeeze Theorem.

As far as I can tell, this works. However, I wanted to know if there is a simpler way to do this problem, especially since the "small $x$" criterion does not have a 'nice' boundary, such as $x\in (-1,1)$.

So any advice on alternative ways of doing this problem, maybe with the limit laws, would be appreciated.

  • $\begingroup$ See my comment, following the answer of Wang YeFei. $\endgroup$ Oct 6, 2022 at 1:33

1 Answer 1


The first line of your argument is flawed. That inequality isn't true. In fact, there is no limit. Take the path $x = t, y = \dfrac{2t^2}{t^3+2}$. Along this path toward $(0,0)$ we have: $\dfrac{x^3y}{x^2-y} = 2 \to 2$. Take another path: $x = t, y = 2t^2$. On this path to $(0,0)$, $\dfrac{x^3y}{x^2-y} \to 0$. So we have two different limits and thus there is no limit or the limit does not exist.

  • $\begingroup$ +1 : Very nice answer. I posted an answer that made the analytical mistake of assuming that if neither $(x)$ nor $(y)$ is held constant at $(0)$, that there must exist some non-zero $(k)$ such that $(x,y)$ is approaching $(0,0)$ from the direction $x = yk$. Under this invalid assumption, the limit does in fact exist and equals $(0)$. However, your (very nice) answer clearly demonstrates that my assumption is invalid. This compelled me to delete my invalid answer. $\endgroup$ Oct 6, 2022 at 1:32
  • $\begingroup$ Thanks @user2661923. $\endgroup$
    – Wang YeFei
    Oct 6, 2022 at 1:35
  • $\begingroup$ First of all, thank you for your answer. I had two follow-up questions: (1) I checked the function in MATLAB, and according to it my first line is correct. So, why is the inequality not true? (2) How do people come up with these paths for disproving limit existence? As in, I would probably have to think on this question for a substantial amount of time before coming up with the path you posted, yet this was meant to be a 10 - 15 minute exam question. Is there some kind of trick to finding paths I'm not aware of? Sorry if these questions are too long. Thank you again for your answer. $\endgroup$
    – Trisztan
    Oct 6, 2022 at 2:12
  • $\begingroup$ @Trisztan: I simply set it equal to $2$ and solve for $y$. Luckily the denominator is not $0$ when $t$ goes to $0$. So we're good. If not I had to look for another path. $\endgroup$
    – Wang YeFei
    Oct 6, 2022 at 2:48
  • $\begingroup$ But how did you come up $2$? Was it just an arbitrary value? Or is there something about this function in particular that implies a choice like $2$ is adequate? $\endgroup$
    – Trisztan
    Oct 6, 2022 at 3:24

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