How to compute this multivariable limit? How do I evaluate $$\lim_{x \to 0 ,\, y \to 0} \frac{x^3y-xy^3}{(x^2+y^2)^{3/2}}$$ 
I tried using squeeze theorem and writing it in polar coordinates, but I got stuck. Can anyone give me a hint?
 A: Hint: $$\frac{|xy(x^{2}-y^{2})|}{(x^{2}+y^{2})^{3/2}}\leq |xy|\frac{|x|^{2}+|y|^{2}}{(x^{2}+y^{2})^{3/2}} \leq \frac{x^{2}+y^{2}}{2}\cdot\frac{2(x^{2}+y^{2})}{(x^{2}+y^{2})^{3/2}}=\sqrt{x^{2}+y^{2}}$$ ere I've used the following elementary estimates $$|x|\leq \sqrt{x^{2}+y^{2}}$$ $$|y|\leq \sqrt{x^{2}+y^{2}}$$ $$|xy|\leq \frac{x^{2}+y^{2}}{2}$$
This proves that the limit is zero
A: Using polar coordinates does work:
$$\biggl\lvert\frac{x^3y-xy^3}{(x^2+y^2)^{3/2}}\biggr\rvert=\frac{r^4
\lvert\cos^3\theta\sin\theta-\cos\theta\sin^3\theta\rvert}{r^3}=r\lvert\,\sin\theta\cos\theta\cos2\theta\,\rvert\le r,$$
so the limit is $\,0$.
A: Hint
$$\left|\dfrac{xy(x^2-y^2)}{(x^2+y^2)^{\frac{3}{2}}}\right|\le \dfrac{1}{2}\dfrac{(x^2+y^2)(x^2+y^2)}{(x^2+y^2)^{\frac{3}{2}}}=\dfrac{1}{2}\sqrt{x^2+y^2}\to 0$$
A: $$\lim\limits_{(x, y) \to (0, 0)} \frac{x^3y-xy^3}{(x^2+y^2)^{\frac32}}$$
Using polar coordinates, we have 
$$\lim\limits_{r\to 0^+} \frac{r^4\cos^3\phi \sin\phi-r^4\cos\phi \sin^3\phi}{\sqrt{(r^2\cos^2\phi+r^2\sin^2\phi)^3}}$$
$$=\lim\limits_{r\to 0^+} \frac{r^4(\cos^3\phi \sin\phi-\cos\phi \sin^3\phi)}{\sqrt{(r^2(\cos^2\phi+\sin^2\phi))^3}}$$
$$=\lim\limits_{r\to 0^+} \frac{r^4\cos\phi (\cos^2\phi \sin\phi- \sin^3\phi)}{\sqrt{r^6}}$$
$$=\lim\limits_{r\to 0^+} \frac{r^4\sin\phi\cos\phi(\cos^2\phi - \sin^2\phi)}{r^3}$$
$$=\lim\limits_{r\to 0^+} r\sin\phi\cos\phi\cos(2\phi)$$
$$=\frac12\lim\limits_{r\to 0^+} r\sin(2\phi)\cos(2\phi)$$
$$=\frac14\lim\limits_{r\to 0^+} r\sin(4\phi)$$
Now let's attempt to find bounds that are independent of $\phi$
$$ \left|\sin(4\phi)\right| \leq 1 $$
$$ r\left|\sin(4\phi)\right| \leq r$$
$$ \frac14\lim\limits_{r\to 0^+} r\left|\sin(4\phi)\right| \leq \frac14\lim\limits_{r\to 0^+}r $$
$$ \frac14\lim\limits_{r\to 0^+} r\left|\sin(4\phi)\right| \leq 0 $$
Therefore by the squeeze theorem, we have
$$ \frac14\lim\limits_{r\to 0^+} r\sin(4\phi) =0$$
Hence
$$\lim\limits_{(x, y) \to (0, 0)} \frac{x^3y-xy^3}{(x^2+y^2)^{\frac32}}=0$$
A: You can replace x = rcosθ, y = rsinθ and solve the problem
