# 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.