$\lim\limits_{x \to 2} x^2 = 4$
$\lvert x-2\rvert \lt \delta$
$\lvert x-2 \rvert \lvert x+2\rvert \lt \varepsilon$
Assuming $\delta$ $\leqslant$ $1$ we find an upper bound for $\lvert x+2 \rvert$ which is $5$ $\implies$ $\lvert x+2\rvert$$\lvert x-2\rvert$ $\lt$ $5$$\lvert x-2\rvert$ $\lt$ $\varepsilon$ $\implies$ $\lvert x-2\rvert$ $\lt$ $\frac \varepsilon 5$
And with two restrictions on $\lvert x-2\rvert$ we arrive to $\delta$=min$\{$$1$, $\frac \varepsilon 5$$\}$
I dont understand $\lvert x+2\rvert$$\lvert x-2\rvert$ < $5$$\lvert x-2\rvert$ $\lt$ $\varepsilon$ part, why cant it be $\lvert x+2\rvert$$\lvert x-2\rvert$ $\lt$ $\varepsilon$ $\lt$ $5$$\lvert x-2\rvert$ then everything else wouldnt work.
If $\delta\leqslant1$, then\begin{align}|x-2|<\delta&\implies|x-2|<1\\&\implies|x+2|=|x-2+4|\leqslant|x-2|+4<5,\end{align}and therefore $|x+2||x-2|<5|x-2|$. So, if $|x-2|<\frac\varepsilon5$, then $|x+2||x-2|<\varepsilon$.