Verify if the following limit exists (by the formal definition): $\lim_{(x,y)\to(0,0)}\frac{x^3-y^3}{x^2+y^2}$ First, I tried approaching the point through a few curves: ($x=0,\,y=0,\, y=m\cdot x,\, y=x^2,\, x=y^2)$ and to all of those I got $0$ for my limit. Since this isn't enough to prove that the limit actually exists, I need to move to the formal definition.
The definition:
If $0<\sqrt{(x-x_0)^2+(y-y_0)^2}<\delta$ then $\left|f(x,y)-L\right|<\epsilon$
But I'm having a lot of difficulty in using this definition of limits to prove if it exists or not.
 A: Theorem: If $\lim f(x) = 0$ and $g(x)$ is bounded, then $\lim f(x)g(x) = 0$.

$$\frac{x^3-y^3}{x^2+y^2} = \frac{(x-y)(x^2+xy+y^2)}{x^2+y^2} = \frac{(x-y)xy + (x-y)(x^2+y^2)}{x^2+y^2} = (x-y)\frac{xy}{x^2+y^2}+x-y$$
Because $\begin{aligned}\lim_{(x, y)\to(0, 0)} x-y = 0\end{aligned}$ we have $\begin{aligned}\lim_{(x, y)\to(0, 0)} \frac{x^3-y^3}{x^2+y^2} = \lim_{(x, y)\to(0, 0)} (x-y)\frac{xy}{x^2+y^2}\end{aligned}$.
By $MA\ge MG$ we have $\begin{aligned}|xy|\le \frac{x^2+y^2}{2} \text{ so } \left|\frac{xy}{x^2+y^2}\right|\le \frac{1}{2} \text{ i.e. } \frac{xy}{x^2+y^2}\end{aligned}$ is bounded, therefore, as $\begin{aligned}\lim_{(x, y)\to(0, 0)} x-y = 0\end{aligned}$, because of the Theorem we shall have
$$\lim_{(x, y)\to(0, 0)} (x-y)\frac{xy}{x^2+y^2} = 0$$
A: Remark that $|x|^3=|x|x^2\le \max(|x|,|y|)\ x^2$ and similarly for $y$.
$$\dfrac{|x^3-y^3|}{x^2+y^2}\le \dfrac{|x|^3+|y|^3}{x^2+y^2}\le\dfrac{\max(|x|,|y|)x^2+\max(|x|,|y|)y^2}{x^2+y^2}=\max(|x|,|y|)\to 0$$
Therefore $f$ is continuous at origin and $f(0,0)=0$.
Notice also that $||(x,y)||_\infty=\max(|x|,|y|)\le\sqrt{x^2+y^2}=||(x,y)||_2\ $ so it does not matter which norm you choose.
A: let $\delta = \varepsilon$ then if $\sqrt{x^{2}+y^{2}}< \delta$ we have $\left | \frac{x^{3}-y^{3}}{x^{2}+y^{2}} \right |=\left |\frac{x^{3}}{x^{2}+y^{2}}- \frac{y^{2}}{x^{2}+y^{2}} \right |= \left | x\frac{x^{2}}{x^{2}+y^{2}}-y\frac{y^{2}}{x^{2}+y^{2}} \right |\leq \left | x-y \right |\leq \sqrt{x^{2}+y^{2}}= \delta $ and we have that $\left | f(x,y) \right |\leq \delta = \varepsilon $
