Show that the function is continuous To show that the function $f: \mathbb{R}^2 \rightarrow\mathbb{R}$ with $f=\left\{\begin{matrix} 
\frac{x^3-y^3}{x^2+y^2} & , (x,y) \neq (0,0)\\ 
0 & , (x,y)=(0,0) 
\end{matrix}\right.$ is continuous on $(0,0)$ we have to show that $|f(x,y)-f(x_0,y_0)| \leq L ||(x,y)-(x_0,y_0)||$ so we have to show that $\left |\frac{x^3-y^3}{x^2+y^2}\right | \leq L \sqrt{x^2+y^2}$. 
I have done the following:  
$$\left |\frac{x^3-y^3}{x^2+y^2}\right |=\frac{|x-y||x^2+xy+y^2|}{x^2+y^2}\leq \dfrac{(|x-y|)(x^2+y^2+|xy|)}{x^2+y^2} \leq 2\dfrac{(|x-y|)(x^2+y^2)}{x^2+y^2} = 2(|x-y|)\leq 2(|x|+|y|)\leq 2\sqrt{(|x|+|y|)^2}=2\sqrt{|x|^2+|y|^2+2|xy|} \leq 2\sqrt{|x|^2+|y|^2+|x|^2+|y|^2}=2\sqrt{2(|x|^2+|y|^2)}=2\sqrt{2}\sqrt{|x|^2+|y|^2}=2\sqrt{2}\sqrt{x^2+y^2}=2\sqrt{2}||(x,y)||$$ 
Is this correct? 
 A: Hint for a different, but elegant, approach: Try polar coordinates.
A: Continue by observing that for any $\varepsilon > 0$, all values in the origin-less open disc of radius $\delta = \varepsilon/(2\sqrt{2})$ have absolute value $< \varepsilon$; this, in conjunction with $f(0, 0) = 0$, establishes the continuity of $f$ at $(0, 0)$.
A: You can take also another approach. Since $x^3-y^3=(x-y)(x^2+y^2+xy)$ you need show that $\frac{xy(x-y)}{x^2+y^2}$ goes to $0$ when $(x,y)\to 0$. So, call $x=r\cos \theta$, $y=\sin \theta$ and $r^2=x^2+y^2$. We have $\frac{xy(x-y)}{x^2+y^2}=\frac{r^3(\cos^2 \theta \sin \theta-\cos \theta \sin^2\theta)}{r^2}$ and this goes to zero when $r\to 0$.
A: Note that$$\frac{x^3}{x^2 + y^2} = x\cdot\frac{x^2}{x^2 + y^2}$$ In absolute value, this is $\le |x|.$ As $(x,y)\to (0,0),|x| \to 0.$ Same thing for $y^3/(x^2 + y^2).$ So $f(x,y)$ is the difference of two functions that both $\to 0$ as $(x,y) \to (0,0);$  hence $\lim_{(x,y)\to (0,0)}f(x,y) = 0.$ This shows $f$ is continuous at $(0,0).$
