So, I just finished this proof, but I can't find any place where I can confirm if it's correctly done or not. There I start:
$\lim\limits_{x\to 0}x^3=0$
$ 0 \lt |x-0| < \delta$ $\implies$ $|x^3-0| \lt \varepsilon%$
I started with
$|x^3-0|$ $=$ |$x^3$| = $|x^2|$ $\cdot$|$x$| $\lt$ $|x^2|$ $\cdot$ $\delta$ $=$ $\varepsilon$
Then I set $\delta \lt 1$:
$|x| \lt 1$
Multiply everything by $|x|$
$|x^2| \lt$ $|x|$
But knowing that $|x^2|$ $\lt$ $|x|$, I get
$|x^2|$ $\lt$ $|x|$ $\lt$ $\delta$, I go back to the $\varepsilon$ inequality:
$|x^3-0|$ $=$ |$x^3$| = $|x^2|$ $\cdot$|$x$| $\lt$ $|x^2|$ $\cdot$ $\delta$ $\lt$ $\delta$ $\cdot$ $\delta$ $=$ $\delta^2$ $=$ $\varepsilon$, so $\delta$ $=$ $\sqrt{\varepsilon}$
Proof:
let $\delta$ $=$ min{$1$,$\sqrt{\varepsilon}$}
$|x| \lt 1$
$|x^2| \lt |x| \lt \delta$, so
$|x^3|$ $=$ $|x^2|$ $\cdot$|$x$| $\lt$ $|x^2|$ $\cdot$ $\delta$ $\lt$ $\delta$ $\cdot$ $\delta$ $=$ $(\sqrt{\varepsilon})^2$ $=$ $\varepsilon$
Is this one a valid proof? Again, I really can't find this problem anywhere to get a feedback. Thank you!