$\varepsilon$-$\delta$ definition of a limit [Calculus]

Question

$$\lim_{x\to 2} x^2 - 8x + 8= -4 $$ 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.

Answer 1

No, it is not correct. If you choose $\delta\leqslant-\frac\varepsilon3$, then $\delta<0$. And it is part of the $\varepsilon-\delta$ definition of limite that $\delta>0$.

Note that the inequality $\bigl|f(x)-l\bigr|\leqslant|x-2|^2-4|x-2|$ is false, since it leads to situations in which $\bigl|f(x)-l\bigr|<0$.

Answer 2

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 $$\lim_{x\to2} x^2-8x+8 = -4$$