How to Prove $\lim_{(x, y) \rightarrow (0, 0)}\frac{x^3-y^3}{x^2+y^2}=0$ I wanted to prove how
$$
\lim_{(x, y) \to (0, 0)} \frac{x^3-y^3}{x^2+y^2} = 0.
$$
Specifically, I want to use the Squeeze Theorem for multivariable calculus. Then I know that I should pick an arbitrary function $g(x, y)$ where $|f(x, y)-L| \leq g(x, y)$ and the limit of $g(x, y)$ is $0$. Since the limit of $f(x, y) =0$, I just have to make
$$
\Biggl\lvert \frac{x^3-y^3}{x^2+y^2} \Biggr\rvert \leq g(x, y),
$$
where the limit of $g(x,y)=0$. I was given a hint that $|x^3-y^3| \leq |x|^3+|y|^3$. So I divided both sides by $|x^2+y^2|$,
$$
\frac{\bigl\lvert x^3-y^3 \bigr\rvert}{\bigl\lvert x^2+y^2 \bigr\rvert} 
\leq \frac{\lvert x\rvert^3+ \lvert y\rvert^3}{\bigl\lvert x^2+y^2 \bigr\rvert},
$$
meaning that
$$
\frac{\bigl\lvert x^3-y^3 \bigr\rvert}{\bigl\lvert x^2+y^2 \bigr\rvert}  
\leq \lvert x\rvert + \lvert y\rvert. 
$$
What do I do next?

Here it is a similar way to approach it:
\begin{align*}
0\leq x^{2} \leq x^{2} + y^{2} & \Rightarrow 0\leq \frac{x^{2}}{x^{2} + y^{2}}\leq 1\\\\
& \Rightarrow \left|\frac{x^{2}}{x^{2} + y^{2}}\right| \leq 1\\\\
& \Rightarrow \left|\frac{x^{3}}{x^{2} + y^{2}}\right|\leq |x|
\end{align*}
Then you can apply the squeeze theorem as $(x,y)\to(0,0)$.
Similarly, we do also have that
\begin{align*}
\left|\frac{y^{3}}{x^{2} + y^{2}}\right| \leq |y|
\end{align*}
where we can apply the squeeze theorem as well.
Along with the triangle inequality, the desired result holds.
