<s>[INST] 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.
 [/INST] 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
$$\lim_{(x, y)\to(0, 0)} (x-y)\frac{xy}{x^2+y^2} = 0$$
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