Krdytkyu

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:

$|(x^3-y^3)|=$

$|(x-y)|(x^2+xy+y^2)|le$

$|x-y||x^2+y^2| +|xy|| le$

$|x-y||x^2+y^2| +|x|^2+|y|^2) le$

$2|x-y||x^2+y^2| le$

$ 2(|x|+|y|) (x^2+y^2)le$

$4 sqrt{x^2+y^2}(x^2+y^2)$.

Finally

$dfrac{|x^3-y^3|}{x^2+y^2} le 4sqrt{x^2+y^2}.$

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:

$$r < g(y) implies left|frac{(rcos t)^3 - (rsin t)^3}{(rcos t)^2 + (rsin t)^2}right| < y$$

Can you find the $g$ now? One more hint if you need it:

$-1 le cos t, sin t le 1$