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

Question

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.

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.