Verification $\epsilon-\delta$ of this (false) limit
Actually, we do have$$\lim_{x\to8}x^2-2x+1=49,$$because, given $\varepsilon>0$, we have\begin{align}x^2-2x+1-49&=x^2-64-2(x-8)\\&=(x-8)(x+8)-2(x-8)\\&=(x-8)(x+6).\end{align}So, if $|x-8|<1$, $|x+6|>13$ , and therefore, if $\delta=\min\left\{1,\frac\delta{13}\right\}$, then$$|x-8|<\delta\implies|x^2-2x+1-49|<\varepsilon.$$