Evaluating $\lim_{(x, y) \to (0, 0)} \frac{xy^2}{x^2-y^2}$ What is the value of
$$\lim_{(x, y) \to (0, 0)} \frac{xy^2}{x^2-y^2}$$
I have already tried the two paths result and I think now that the limit does exist and it is equals to 0. But i really cannot prove this.
 A: 
$$L=\lim_{(x, y) \to (0, 0)} \frac{xy^2}{x^2-y^2}$$

$$L=\lim_{(x, y) \to (0, 0)} \frac{xy^2-x^3+x^3}{x^2-y^2}=\lim_{x \to0} (-x)+\lim_{(x, y) \to (0, 0)} \frac{x^3}{x^2-y^2}=\lim_{(x, y) \to (0, 0)} \frac{x^3}{x^2-y^2}$$
Let $y=x-ax^2$
$$L=\lim_{x \to 0} \frac{x^3}{2ax^3-a^2x^4}=\lim_{x \to 0} \frac{1}{2a-a^2x}=\frac{1}{2a}$$
So the limit depends on the value of $a$, hence the limit doesn't exist.
A: Setting $x = r\cos\theta$ and $y = r\sin\theta$, we can calculate that the expression following the limit is
$$
r \frac{\cos\theta \sin^2\theta}{\cos^2\theta - \sin^2\theta} = r \frac{\cos\theta}{\cot^2\theta - 1}
$$
We can make $r$ as small as we want as $(x,y) \to (0,0)$, but we can also make $\frac{\cos\theta}{\cot^2 \theta - 1}$ as big as we want no matter how small $r$ is (how?). Can you see how to use this to prove that the limit does not exist?
A: In Polar Co-ordinates , we get $x = r\cos(\theta)$ & $y = r\sin(\theta)$
Plugging that into given limit , we get $r^3$ term in $xy^2$ & $r^2$ term in $x^2-y^2$
Then the limit is $r[f(\theta)]$ where $f$ is function of $\theta$ , in terms of $\cos$ & $\sin$
Here $f(\theta) = \frac{\cos(\theta)\sin^2(\theta)}{\cos^2(\theta)-\sin^2(\theta)}$
When $\theta=0,\pi/2,\pi$ , we have $f=0$
When $\theta=\pi/4$ , we have $\cos^2(\theta) = \sin^2(\theta)$ , hence $f$ is unbounded ($1/0$)
Depending on the Path to (0,0) , we get various values of the term. Hence , the limit will not exist !
Thanks to user @CharlesHudgins & user @MathFail who Pointed out the right way.
