A confusion in the $\epsilon-\delta$ proof of $\lim_{x\to 2}\frac{x - 2}{1 + x ^ 2} = 0$ I want to prove
$$
\lim_{x\to 2}\frac{x - 2}{1 + x ^ 2} = 0
$$
The proof is: Let $\epsilon > 0$ be given. Then
$$\left|\frac{x - 2}{1 + x ^ 2} - 0\right| = \frac{|x - 2|}{1 + x ^ 2} \leqslant |x - 2| < \epsilon$$
provided $|x - 2| < \delta = \epsilon$.
My question is we have
$$\left|\frac{x - 2}{1 + x ^ 2} - 0\right| < \epsilon \text{ and }\frac{|x - 2|}{1 + x ^ 2} \leqslant |x - 2|.$$
How do we conclude
$$\left|\frac{x - 2}{1 + x ^ 2} - 0\right| = \frac{|x - 2|}{1 + x ^ 2} \leqslant |x - 2| < \epsilon$$
from these two inequalities? What if $\epsilon < |x - 2|$?

You have made an incorrect statement which I think needs to be corrected:
it is not true that you "have"
$$ \left\lvert \frac{x - 2}{1 + x ^ 2} - 0\right\rvert < \epsilon  \tag1 $$
at the point in the proof you are questioning.
Instead, this inequality is the thing you have to prove at that point.
The real point of difficulty, however, seems to be how to recognize the implications that are used in the proof, and to understand which one follows from which.
Note that when we write, "Blah blah blah provided $|x - 2| < \delta = \epsilon$,"
it means

If $\delta = \epsilon$ and $|x - 2| < \delta,$ then blah blah blah.

So the steps of the proof are:

*

*For whatever $\epsilon$ you are given, choose $\delta = \epsilon.$

*Now you know that if $|x - 2| < \delta,$ then $|x - 2| < \epsilon.$

*As separate facts, we see that $\left\lvert \frac{x - 2}{1 + x ^ 2} - 0\right\rvert = \frac{\left\lvert x - 2\right\rvert}{1 + x ^ 2}$ and that $\frac{\left\lvert x - 2\right\rvert}{1 + x ^ 2} \leq \left\lvert x - 2\right\rvert.$

*Putting these facts together, if $|x - 2| < \delta,$ then $\left\lvert \frac{x - 2}{1 + x ^ 2} - 0\right\rvert = \frac{\left\lvert x - 2\right\rvert}{1 + x ^ 2} \leq \left\lvert x - 2\right\rvert < \epsilon.$

*Notice that if $\left\lvert \frac{x - 2}{1 + x ^ 2} - 0\right\rvert = \frac{\left\lvert x - 2\right\rvert}{1 + x ^ 2} \leq \left\lvert x - 2\right\rvert < \epsilon$ then $\left\lvert \frac{x - 2}{1 + x ^ 2} - 0\right\rvert < \epsilon.$

*From the two implications in the previous two steps, you can conclude that if $|x - 2| < \delta,$ then $\left\lvert \frac{x - 2}{1 + x ^ 2} - 0\right\rvert < \epsilon,$ which is what you need to show for the critical step of your delta-epsilon proof.

In the proof as you saw it, many of the steps above are rephrased or condensed, which might cause confusion.
You may well ask, "What if $\left\lvert x - 2\right\rvert > \epsilon$?"
In that case, since you have already said that $\delta = \epsilon,$
if $\left\lvert x - 2\right\rvert > \epsilon$ then
$\left\lvert x - 2\right\rvert > \delta$
and the "if $\left\lvert x - 2\right\rvert < \delta$" condition is false.
For a delta-epsilon proof where $x\to 2$ you only need to show what happens if
$\left\lvert x - 2\right\rvert < \delta$ is true;
it does not matter what happens when $\left\lvert x - 2\right\rvert < \delta$ is false.
So if $\left\lvert x - 2\right\rvert > \epsilon$ then we are exploring cases that are completely irrelevant to whether the limit exists and what it is.
