As always, I demand some clarification and / or to right my wrongs...
$$\lim_{x\to 8} x^2 - 2x + 1 = 49$$
The limit is obviously false, but let's verify it. This is where I faced some troubles. For example:
$$|x^2 - 2x - 48 | < \epsilon$$
$$48 - \epsilon < x^2 - 2x < 48 + \epsilon$$
How could I manage this expression? I have several questions about this. I know I should split them but I think it's not the case so:
Question 1.
Could I write something like
$$x^2 - 2x \leq x^2$$ and then proceed in this sense? If I can: why can I? And If I cannot, why?
Also, it would be possible instead to use? $$x^2 - 2x \geq 2x$$
This makes me confused about this: when should I use a majorisation $\leq$ and when $\geq$?
For example, following the first case (if possible) I would obtain
$$\sqrt{48 - \epsilon} < x < \sqrt{48 + \epsilon}$$
Which for sure guarantees the neighbourhood of $x$ is not $8$, as wanted. Is this acceptable?
Question 2.
Suppose instead I proceed this way:
$$|(x-8)(x+6)| < \epsilon$$
$$|x-8| < \frac{\epsilon}{|x+6|}$$
Now since I know $x\to 8$, I can for sure tell $7 < x < 9$, that is adding $6$
$$13 < x+6 < 15$$ So $$\frac{1}{x+6} > \frac{1}{15}$$
And then I could write
$$|x-8| < \frac{\epsilon}{15}$$
So apparently I have managed to find a control $\delta_{epsilon}$. The question is: how this would make me understand the limits is wrong?
Note the two approaches are different and rise different questions, and I needed to write down them all here because once someone clarified these to me, I guess I would be able to reason by myself having understood some finicky rules which now put me on hold.
Thank you so much!
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$$|x-8|<\delta\implies|x^2-2x+1-49|<\varepsilon.$$