$\lim_{(x,y)\rightarrow (0,0)}\dfrac{xy^2}{x^2-y^2}$

Question

I want to solve the following question. $$\lim_{(x,y)\rightarrow (0,0)}\dfrac{xy^2}{x^2-y^2}$$

I am going to use polar coordinates. $x=r\cos\theta$ and $y=r\sin\theta$

$$\lim_{(x,y)\rightarrow (0,0)}\dfrac{xy^2}{x^2-y^2}=\lim_{r\rightarrow 0^+}\frac{r^3\cos\theta\sin\theta}{r^2(\cos^2\theta-\sin^2\theta)}=\lim_{r\rightarrow 0^+}\dfrac{r\cos\theta\sin^2\theta}{\cos^2\theta-\sin^2\theta}=0$$

if $\theta\neq \dfrac{\pi}{4},\dfrac{3\pi}{4},\dfrac{5\pi}{4}, \dfrac{7\pi}{4}.$

For values of $\theta$ we have the lines $y=x$ and $y=-x$, but they are bot in the domain of the function.

Wolfram says it does not exist. Why?

Answer 1

Suppose $f(x,y),g(x,y)$ are continuous, $f$ is never $0$ in the first quadrant, and $g$ vanishes along a curve in the first quadrant tending to $(0,0).$ We are interested in the behavior of $f(x,y)/g(x,y).$ Of course we should throw the curve out of the domain of this quotient, otherwise we divide by $0.$ But throwing the curve out doesn't help much, because you can see $f/g$ blows up near any point on the curve. Thus $f/g$ is unbounded in any deleted neighborhood of $(0,0),$ hence there is no chance for a limit. We are in such a situation here.