I know that I have to find a $\epsilon > 0$ s.t. |$\frac{x^3-y^3}{x^2+y^2}$|<$\epsilon$ whenever $0<\sqrt{x^2+y^2}<\delta$ but what are the steps to proving this?
I'm looking for an answer which shows the steps as well as explains the intuition.
Thanks
Numerator:
$|(x^3-y^3)|=$
$|(x-y)|\,|(x^2+xy+y^2)|\le$
$|x-y|\,\big(|x^2+y^2| +|xy|\big) \le$
$|x-y|\,\big(|x^2|+|y^2| +|x|^2+|y|^2\big) \le$
$2\,|x-y|\,|x^2+y^2| \le$
$ 2(|x|+|y|) (x^2+y^2)\le$
$4 \sqrt{x^2+y^2}(x^2+y^2)$.
Finally
$\dfrac{|x^3-y^3|}{x^2+y^2} \le 4\sqrt{x^2+y^2}.$
Choose $\delta =\epsilon/4.$
Used: $x^2+y^2 \ge 2|xy|;$
$|x| =\sqrt{x^2} \le \sqrt{x^2+y^2},\;\;$ $|y| =\sqrt{y^2} \le \sqrt{x^2+y^2}$.
$$(\forall y > 0 ~ \exists r > 0 ~ \forall \bar x )~ 0 < | \bar x - \bar c | < r \implies |f \bar x - L| < y$$
$(\forall y > 0 ~ \exists r > 0 ~ \forall \bar x )$ is equivalent to asking for the existence of a function $g$ that, given a maximum deviation from $L$, outputs the neighborhood $r$ that satisfies that maximum deviation. In other words, find the disc centered at $\bar c$ with radius $g(y)$ such that every value of $f$ within that region is within $y$ of $L$. Symbolically:
$$(\exists g : \mathbb R \to \mathbb R ~\forall y > 0 ~ \forall \bar x) ~ 0 < | \bar x - \bar c | < g(y) \implies |f \bar x - L| < y$$
Putting in the values for your problem:
$$(\exists g : \mathbb R \to \mathbb R ~\forall y > 0 ~ \forall \bar x) ~ 0 < | \bar x | < g(y) \implies \left|\frac{x_1{}^3 - x_2{}^3}{x_1{}^2 + x_2{}^2}\right| < y$$
or to put more tersely, find $g : \mathbb R \to \mathbb R$ such that for $\bar x \ne (0, 0)$ $$|\bar x| < g(y) \implies \left|\frac{x_1{}^3 - x_2{}^3}{x_1{}^2 + x_2{}^2}\right| < y$$
A polar substitution helps:
$$r < g(y) \implies \left|\frac{(r\cos t)^3 - (r\sin t)^3}{(r\cos t)^2 + (r\sin t)^2}\right| < y$$
Can you find the $g$ now? One more hint if you need it:
$-1 \le \cos t, \sin t \le 1$
