ϵ\epsilon-δ\delta proof that lim\lim\limits_{x \to 1} \frac{1}{x} = 1.

I’m starting Spivak’s Calculus and finally decided to learn how to write epsilon-delta proofs.

I have been working on chapter 5, number 3(ii). The problem, in essence, asks to prove that

Here’s how I started my proof,

I haven't made any further progress past this point. Is it possible to salvage this proof? Should I try an alternate approach?

Update 2/19/2018: It appears that this answer has received a lot of attention, which I'm very glad to know about. When you're reading through this answer and you're trying to learn about $\delta$-$\epsilon$ proofs for the first time, I would recommend skipping the sections labeled Addendum. on your first read. Please let me know of any other clarifications that you would like with this answer.

Whenever I am doing a $\delta$-$\epsilon$ proof, I do some scratch work (note, this is NOT part of the proof) to figure out what to choose for $\delta$. I always tell students to think about the following:

1. What are you given?

2. What do you want to show?

In the definition of the limit, you are given an arbitrary $\epsilon > 0$ and you want to find $\delta$ such that implies You have control over what to choose for your $\delta$ in this case. The idea of this $\delta$-$\epsilon$ proof is to work with the expression $|x - 1| < \delta$ and get $\left|\dfrac{1}{x} - 1 \right| < \epsilon$ at the end.

Let's do some scratch work (again, NOT part of the proof).

Scratch Work

Let's start with what we want to show for our scratch work (starting with what you want to show is bad to do $100\%$ of the time when you're doing proofs - again, this is scratch work and not actually part of the proof).

We want to show that $\left|\dfrac{1}{x} - 1 \right| < \epsilon$. Let's work backwards and try to turn the expression $\left|\dfrac{1}{x} - 1 \right|$ into some form of $|x-1|$.

So, note that since $|y|=|-y|$ for all $y$ in $\mathbb{R}$.

The last expression can be rewritten as $\dfrac{\left|x-1 \right|}{\left| x \right|}$. Looking at this expression, we do have $|x-1|$ in the numerator, which is good. But we have that pesky $|x|$ in the denominator.

Since we do have control of what $|x-1|$ is less than (this is our $\delta$), let's choose a really convenient, small number to work with that is greater than $0$. Let's say $\delta = \dfrac{1}{2}$.

Well, if $|x - 1| < \dfrac{1}{2}$, then So if we choose $\delta = \dfrac{1}{2}$, $\dfrac{1}{2} <|x| < \dfrac{3}{2}$.

Addendum. In many examples, $\delta$ is usually chosen to be $1$. Why did we elect not to do that in this case?

It's because it wouldn't work.

Intuitively, here's why it doesn't: when you consider the neighborhood of radius $1$ centered around $x = 1$, you get the interval $(0, 2)$. $f(x) = \dfrac{1}{x}$ doesn't have a finite limit at $x = 0$, so this makes $\delta = 1$ a bad choice.

This isn't the case if $\delta = 1/2$. The neighborhood of radius $1/2$ around $x = 1$ is $(1/2, 3/2)$. $f$ has limits at every $x$-value in the interval $(1/2, 3/2)$, including the endpoints.

In terms of the algebra, if we had chosen $\delta = 1$, the algebra wouldn't have worked out. We would've gotten $0 < x < 1$ and would not have been able to obtain a finite upper bound for $\dfrac{1}{x}$. That is,

We do not have a finite upper bound for $\dfrac{1}{x}$ in this case, and hence why $\delta = 1$ will not work for this purpose.

If $\dfrac{1}{2} <|x| < \dfrac{3}{2}$, then and Now we have control over what $|x-1|$ is less than. So to get $\epsilon$, we choose $\delta = \dfrac{\epsilon}{2}$.

But, wait - didn't I say that we chose $\delta = \dfrac{1}{2}$ earlier? A simple solution would be to minimize $\delta$, i.e., make $\delta = \min\left(\dfrac{\epsilon}{2} , \dfrac{1}{2} \right)$.

Addendum. To see why $\delta = \min\left(\dfrac{\epsilon}{2} , \dfrac{1}{2} \right)$ works, suppose $\dfrac{\epsilon}{2} > \dfrac{1}{2}$, so that $\delta = \dfrac{1}{2}$. Then $\epsilon > 1$. Then

Now suppose $\dfrac{\epsilon}{2} \leq \dfrac{1}{2}$, so that $\delta = \dfrac{\epsilon}{2}$.

Then

In both cases, we have $\dfrac{\left|x-1 \right|}{\left| x \right|} < \epsilon$, as desired.

So now we've found our $\delta$ and can use this to write out the proof.

The Proof

Proof. Let $\epsilon > 0$ be given. Choose $\delta := \min\left(\dfrac{\epsilon}{2} , \dfrac{1}{2} \right)$. Then

(since if $|x - 1| < \dfrac{1}{2}$, $\dfrac{1}{|x|} < 2$) and

Success at last.

Addendum. Note that the end goal above was achieved, namely to show that

In the step textbooks
usually omit the step with the $\delta$ and just write

Addendum. It may seem that the note "(since if $|x - 1| < \dfrac{1}{2}$, $\dfrac{1}{|x|} < 2$)" may be an additional assumption added to the problem - i.e., that $\delta$ has to be $\dfrac{1}{2}$. This is not the case for the following reason: given $|x-1| < \delta$, we have

Obviously, if $\epsilon \geq 1$, we end up with $|x - 1| < \dfrac{1}{2}$, as stated above. But let's suppose that $\epsilon < 1$. Then

and you end up with $|x - 1| < \dfrac{1}{2}$, so the $\dfrac{1}{|x|} < 2$ implication holds in either case.