# (Another) Multivariable delta-epsilon proof

Q: Find the limit if it exists $$\lim \limits_{(x,y) \to (0,0)} \frac{x^3 + y^3}{x^2 + y^2}$$

What I have done so far:

• Setting y = 0 & x = 0 I find that along these axes it looks like the limit will approach $0$ therefore: $$\lim \limits_{(x,y) \to (0,0)} \frac{x^3 + y^3}{x^2 + y^2} = 0$$

$$\mid \frac{x^3 + y^3}{x^2 + y^2} - 0 \mid < \epsilon \quad \quad \quad 0 < \sqrt{x^2 + y^2} < \delta$$

Now since $x^2 + y^2 \geq 0$ we cans simplify to:

$$\mid x^3 + y^3 \mid \leq \sqrt{ (x^3 + y^3)^2 } = \quad ....$$

I do not know how to proceed; how do I modify the $\sqrt{ (x^3 + y^3)^2 }$ into a $\sqrt{x^2 + y^2}$ ?

For both $x,y\ne 0$, then \begin{align*} \dfrac{|x|^{3}+|y|^{3}}{x^{2}+y^{2}}&=\dfrac{|x|^{3}}{x^{2}+y^{2}}+\dfrac{|y|^{3}}{x^{2}+y^{2}}\\ &\leq\dfrac{|x|^{3}}{x^{2}}+\dfrac{|y|^{3}}{y^{2}}\\ &=|x|+|y|\\ &\leq\sqrt{x^{2}+y^{2}}+\sqrt{x^{2}+y^{2}}\\ &=2\sqrt{x^{2}+y^{2}}, \end{align*} the above inequality still holds for either $x=0$ or $y=0$, so we conclude that \begin{align*} \dfrac{|x|^{3}+|y|^{3}}{x^{2}+y^{2}}\leq 2\sqrt{x^{2}+y^{2}},~~~~(x,y)\ne(0,0). \end{align*}

If $\sqrt{x^2+y^2}=r$, then $|x|,|y|\leqslant r$. Therefore $|x^3+y^3|\leqslant|x|^3+|y|^3\leqslant2r^3$ and so$$\left|\frac{x^3+y^3}{x^2+y^2}\right|\leqslant2r.$$So, for each $\varepsilon>0$, you can take $\delta=\frac\varepsilon2$.

• How does one identify when to substitute? – Andre Fu Mar 27 '18 at 21:20
• @AndreFu I don't understand your question. – José Carlos Santos Mar 27 '18 at 21:21

hint: Observe that $0 \le \dfrac{|x^3+y^3|}{x^2+y^2} \le |x|+|y| \le \sqrt{2(x^2+y^2)}$