Question

Proving a limit using epsilon delta definition
Picture below--

Answer 1

\[\lim_{x\to2}x^3=8\]
You have to show that for any \(\epsilon>0\), you can find \(\delta\) such that \(0<|x-2|<\delta\) implies \(|x^3-8|<\epsilon\).
\[\begin{align*}|x^3-8|&=|x-2||x^2+2x+4|\\
&=|x-2||x^2+2x+1+3|\\
&=|x-2||(x+1)^2+3|\\
&<|x-2|\bigg(|x+1|^2+|3|\bigg)\\
&<|x-2|\bigg(|x+1|^2+3\bigg)\end{align*}\]
If we agree to set \(\delta\le1\), then
\[|x-2|<1~~\iff~~-1<x-2<1~~\iff~~2<x+1<4~~\iff~~|x+1|<4\]
so we have
\[\begin{align*}|x^3-8|&<|x-2|\bigg(4^2+3\bigg)\\
&<19|x-2|
\end{align*}\]
So, if we expect to have \(|x^3-8|<\epsilon\), we could set \(\delta=\dfrac{\epsilon}{19}\), since that gives us
\[|x^3-8|<19|x-2|<19\delta=19\left(\frac{\epsilon}{19}\right)=\epsilon\]
Keep in mind that we agree to set \(\delta\le1\) so in actuality we would be using \(\delta=\min\left\{1,\dfrac{\epsilon}{19}\right\}\).