<s>[INST] 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?
 [/INST] Now,
$$
0\leqq\left|\frac{x^3-y^3 }{x^2+y^2} \right| 
\leq \lvert x\rvert + \lvert y\rvert$$
Letting $(x,y)\to(0,0)$, you get
$$
\lim_{(x,y)\to(0,0)} \ \ \ 0\leq\lim_{(x,y)\to(0,0)}\ \ \left|\frac{x^3-y^3 }{x^2+y^2} \right| 
\leq \lim_{(x,y)\to(0,0)}\lvert x\rvert + \lvert y\rvert$$
The LHS and RHS are $0$ so due to the squeeze theorem, you get $$\lim_{(x,y)\to(0,0)}\ \ \left|\frac{x^3-y^3 }{x^2+y^2} \right| =0.$$
This means $$\lim_{(x,y)\to(0,0)}\ \ \ \frac{x^3-y^3 }{x^2+y^2}  =0.$$

But I think that you should show why $\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}
$ means $
\frac{\bigl\lvert x^3-y^3 \bigr\rvert}{\bigl\lvert x^2+y^2 \bigr\rvert}  
\leq \lvert x\rvert + \lvert y\rvert. 
$
</s>