Prove that $\lim_{x\to0, y\to0}\frac{x^3+y^3}{x^2+y^2}=0$

Question

Prove that $\lim_{x\to0, y\to0}\frac{x^3+y^3}{x^2+y^2}=0$

My Try

Answer 1

Hint

$$\left|\frac{x^3+y^3}{x^2+y^2}\right|\leq |x|+|y|.$$

Answer 2

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?