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 $\epsilon < |x - 2|$?
 A: 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.
A: You've encountered a very common problem with proofs of this sort, namely that the logical order of the steps is different from the order in which one does the work of finding the proof.
First, here's the proof for your question, in logical order. According to the definition of "limit", we're supposed to prove that, for every $\epsilon>0$, there exists some $\delta>0$ such that whenever $|x-2|<\delta$ then $|\frac{x-2}{1+x^2}-0|<\epsilon$. So consider an arbitrary $\epsilon>0$. We need to produce an appropriate $\delta$, and we choose $\delta=\epsilon$. [This choice looks completely arbitrary here, and in other problems the formula for $\delta$ would be more complicated.] Now we need to prove that this choice of $\delta$ works, i.e., that whenever $|x-2|<\delta$ then $|\frac{x-2}{1+x^2}|<\epsilon$.So consider any $x$ that satisfies $|x-2|<\delta$ and compute (as in your question) that $|\frac{x-2}{1+x^2}|<\delta=\epsilon$.  So our arbitrary-looking choice of $\delta$ has succeeded and the proof is complete.
It's reasonable to ask, though, where that arbitrary choice of $\delta$ came from. (It's even more reasonable to ask that in other problems, where the formula for $\delta$ can be a complicated mess rather than a nice simple $\delta=\epsilon$.) It came from the computation at the end of the logical proof. To figure out the right choice of $\delta$, one does the computation first, finding that when $|x-2|<\delta$ then $|\frac{x-2}{1+x^2}|<\delta$, and then saying "That would be what I want, namely $|\frac{x-2}{1+x^2}|<\epsilon$, if $\delta$ were equal to $\epsilon$." So then one can go back and fill in the part of the proof that says "choose $\delta=\epsilon$."
After going through this process in several examples, people often just write the computation that leads to $|\frac{x-2}{1+x^2}|<\delta$, and then say we're done if $\delta=\epsilon$", without bothering to say that, in logical order, this formula for $\delta$ should have been way earlier, near the beginning of the proof. That's what apparently happened in the proof that you quoted.
