# Evaluate $\lim_{(x,y)\to (0,0)}{\frac{x^3 y}{x^2 - y}}$.

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.

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

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.
• +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. Oct 6, 2022 at 1:32
• @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. Oct 6, 2022 at 2:48
• 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? Oct 6, 2022 at 3:24