How does exactly the $\epsilon$-$\delta$ method tells me I am right?

Solution 1:

You have put it correctly in an update to the question:

At this point I need to find a $\delta$ such that $|x - 1| < \delta$ I then subtract $1$: $$\sqrt{1 - \epsilon} - 1 < x - 1 < \sqrt{1 + \epsilon} - 1.$$

But here's the reason why you wouldn't want to replace $\sqrt{1 + \epsilon}$ by $1 + \epsilon$ in this inequality: it's precisely because $\sqrt{1 + \epsilon} < 1 + \epsilon,$ which means that by making this replacement you allow a greater upper bound of $x$ and you therefore allow $x$ to take on values that $\sqrt{1 - \epsilon} - 1 < x - 1 < \sqrt{1 + \epsilon} - 1$ would not allow it to take. But $\sqrt{1 - \epsilon} - 1 < x - 1 < \sqrt{1 + \epsilon} - 1$ was carefully crafted to give you exactly the set of $x$ for which $\lvert f(x) - L\rvert < \epsilon$ in your example; as soon as you let any other value of $x$ be possible, you are allowing $x$ to take at least one value such that $\lvert f(x) - L\rvert \geq \epsilon$ and you will no longer have a proof.

As an aside, I don't particularly like the strategy that says you must deduce the largest possible value of $\delta$ that you can find that will still allow the proof to work (as strategies like the above seem designed to do). But you very well can replace expressions like $\sqrt{1+\epsilon}$ with something simpler if you like; the thing is just that you can only lower the upper bound and only raise the lower bound. To put it another way, "$\delta$ can never be too small."

So I might work it in this direction instead: I am going to pick some value of $\delta$ (I haven't decided which, yet), after which I'm going to prove something implied by $$ 0 < \lvert x - 1\rvert < \delta. $$ I know that if I ensure $\lvert x - 1\rvert < 1$ (which I can do just by making sure that the value of $\delta$ I choose is not greater than $1$), then $x$ is in the open interval $(0, 2)$, $x^2 + 1$ is increasing over the whole interval, and the worst-case scenario for $\lvert f(x) - 2\rvert$ is either at $x = 1 + \delta$ or $x = 1 - \delta.$ Specifically, $$ \lvert f(1 + \delta) - 2\rvert = (1 + \delta)^2 + 1 - 2 = 2\delta + \delta^2 $$ and $$ \lvert f(1 - \delta) - 2\rvert = 2 - ((1 - \delta)^2 + 1) = 2\delta - \delta^2. $$ Comparing the two, it's clear the worst case (the greatest $\lvert f(x) - 2\rvert$) is $2\delta + \delta^2,$ so I just need to ensure that $$ 2\delta + \delta^2 < \epsilon. $$

Since I already decided I would not choose $\delta > 1,$ I can be sure that $2 + \delta \leq 3$. So rather than trying to solve the a quadratic equation in $\delta$ I'm just going to try to set it up so that $$ 2\delta + \delta^2 = (2 + \delta)\delta \leq 3\delta < \epsilon. $$ Let's try $\delta = \min\left(\frac14 \epsilon, 1\right)$; the $1$ comes from the fact that I already said I will not let $\delta$ be greater than $1,$ and the $\frac14 \epsilon$ is a relatively simple way of ensuring that $3\delta < \epsilon.$

Now it might seem I made $\delta$ smaller than necessary. And I did make $\delta$ smaller than necessary; I don't care. The definition of the limit merely says there has to exist a $\delta$ satisfying certain conditions; it makes no requirement for me to tell you what the largest possible $\delta$ value is. It only matters that you will be able to prove that if $1 - \delta < x < 1$ or $1 < x < 1 + \delta$ then $-\epsilon < (x^2 + 1) - 2 < \epsilon.$

Solution 2:

Here is the standard way to do it.

$|(x^2 + 1) - 2 | < \epsilon\\ |x^2 - 1| < \epsilon\\ |(x+1)(x-1)| < \epsilon$

Let's say that $\delta \le 1$

$|x-1| < \delta \implies |x+1| < 3$

$|(x+1)(x-1)| < 3\delta \le \epsilon$

let $\delta = \min (1,\frac \epsilon 3)$

As for your work above.

$\bullet 2 - \delta > \sqrt{1-\epsilon}\\ \bullet 2 + \delta > \sqrt{1+\epsilon}$

Seems to be coming out of nowhere.