<s>[INST] Prove that $\lim_{x\to0, y\to0}\frac{x^3+y^3}{x^2+y^2}=0$ Prove that $\lim_{x\to0, y\to0}\frac{x^3+y^3}{x^2+y^2}=0$
My Try

 [/INST] HINT
Notice that
\begin{align*}
|x|^{3} = |x|\times x^{2} \leq |x|(x^{2} + y^{2}) \Rightarrow \left|\frac{x^{3}}{x^{2} + y^{2}}\right| \leq |x|
\end{align*}
Based on the same procedure, you can prove that
\begin{align*}
|y|^{3} = |y|\times y^{2} \leq |y|(x^{2} + y^{2}) \Rightarrow \left|\frac{y^{3}}{x^{2} + y^{2}}\right| \leq |y|
\end{align*}
Then apply the squeeze theorem.
Can you take from here?
</s>