$\varepsilon$-$\delta$ definition of a limit [Calculus] $$\lim_{x\to 2} x^2 - 8x + 8= -4 $$
Now to express $ f(x) - l $ it as $x - x_0$ , we write $ x = (x - x_0) + x_0$.
So \begin{align}
f(x) - l &= x^2 - 8x + 8 + 4\\
&= (x - 2 + 2)^2 - 8(x - 2 + 2)+ 12\\
&= (x - 2)^2 +4(x-2) + 4 - 8(x - 2) - 16+ 12 \\
&= (x - 2)^2 - 4(x-2)\\
&= |f(x) - l|\\
&\le |x - 2|^2 -4|x-2|
\end{align}
We choose $\delta$ such that $\delta^2 -4\delta \le \varepsilon$.
For $\delta \le 1$,  $$\delta -4\delta \le \varepsilon$$
$$\delta \le -\frac\varepsilon3$$
Is my $\delta$ correct ? Could someone guide me through this? Only then I guess I can verify this limit by $\varepsilon$-$\delta$ definition. Thanks.

Let $\varepsilon >0$. For $|x - 2| < \delta = \sqrt{4+\varepsilon} - 2$ we have:
\begin{align}
\left|x^2-8x+8+4\right| &= \left|x^2 - 8x + 12\right|\\
&= \left|(x-2)^2 - 4(x-2)\right|\\
&= \left|x-2\right|\left|x-2-4\right|\\
&\le |x-2|\big(|x - 2| + 4\big)\\
&< \delta(\delta + 4)\\
&=\varepsilon
\end{align}
Therefore
$$\lim_{x\to2} x^2-8x+8 = -4$$
