We had to prove that
$\lim_{x \to 1} x^3 = 1$
Starting with
$|x^3 - 1| < \epsilon$
By breaking the inequality I got
$\sqrt[3]{1 - \epsilon} < x < \sqrt[3]{1 + \epsilon}$
I then put my delta as
$\delta = \min({\sqrt[3]{1 - \epsilon} - 1, \sqrt[3]{1 + \epsilon} - 1})$
This is the mistake on my part that cost me my mark, since the left expression should actually be
$1 -\sqrt[3]{1 - \epsilon}$
Because of this mistake, the left side is always negative, so the minimum will be always negative, so the proof wouldn't make sense, but if you use the right side you can derive the correct epsilon equation.
My question is, since $|x - 1| < \delta$, delta can't be negative because it's greater than an absolute value, so wouldn't the left expression never be true? Is this still a correct, if messy, epsilon delta proof?