Prove $\lim_{ (x,y) \to (0,0)} \frac{x^3-y^3}{x^2+y^2}=0$

Question

Can some one prove the limit using epsilon delta method to prove that the limit exists $$ \lim_{(x,y) \to (0,0)} \frac{x^3-y^3}{x^2+y^2}=0 $$

Answer 1

Let $x=r\cos\theta$ and $y=r\sin\theta$.

We have $$\lim_{(x,y) \to (0,0)} \frac {x^3-y^3}{x^2+y^2} = \lim_{r \to 0} \frac {r^3\cos^3\theta-r^3\sin^3\theta}{r^2\cos^2\theta+r^2\sin^2\theta} \\=\lim_{r \to 0} \frac {r^3\cos^3\theta-r^3\sin^3\theta}{r^2} =\lim_{r \to 0} r(\cos^3\theta-\sin^3\theta).$$

Can you complete?

Answer 2

Let $(x, y) \not = (0, 0);$

$\dfrac{|xx ^2-yy^2¦}{x^2+y^2}| \le\dfrac{|x|x^2+|y|y^2}{x^2+y^2}\le|x|+|y|\le 2\sqrt{x^2+y^2}.$

$\delta =? $

P. S. Tito' s comment.