Finding limits using $\epsilon$ - $\delta$ method

Question

How do I prove the following function is continuous at $(0,0)$ using epsilon-delta approach?

$$\frac{x^5-y^5}{x^2+y^2}$$

Answer 1

As long as $(x,y)\to 0$, $$\left|\frac{x^5-y^5}{x^2+y^2}\right|\leq (|x|^3+|y|^3)\left(\frac{|x|^2}{x^2+y^2}+\frac{|y|^2}{x^2+y^2}\right)=|x|^3+|y|^3\to 0.$$

Answer 2

I suppose your function $$f(x,y) = \frac{x^5-y^5}{x^2+y^2}$$ is defined as $0$ for $(x,y) = (0,0)$. Then it is continuous if for every $\epsilon > 0$ a $\delta > 0$ exists, so that for $||(x,y)|| < \delta \implies ||f(x,y)|| < \epsilon$. Choose $$\delta := \frac{\sqrt[3]{\epsilon}}{\sqrt[3]{2}},$$ because of: $$0 \leq |\frac{x^5-y^5}{x^2+y^2}| \leq |\frac{|x^5|+|y^5|}{x^2+y^2}| = |\frac{|x^5|}{x^2+y^2} + \frac{|y^5|}{x^2+y^2}| \leq |\frac{|x^5|}{x^2} + \frac{|y^5|}{y^2}| = ||x^3| +|y^3||, $$ and $$||x^3| +|y^3|| < |\delta^3 + \delta^3| = |(\frac{\sqrt[3]{\epsilon}}{\sqrt[3]{2}})^3 + (\frac{\sqrt[3]{\epsilon}}{\sqrt[3]{2}})^3| = |\frac{\epsilon}{2} + \frac{\epsilon}{2}| = \epsilon. $$