I’m starting

Spivak’s Calculusand 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

\lim\limits_{x \to 1} \frac{1}{x} = 1.

Here’s how I started my proof,

\left| f(x)-l \right|=\left| \frac{1}{x} – 1 \right| =\left| \frac{1}{x} \right| \left| x – 1\right| < \epsilon \implies \left| x-1 \right| < \epsilon |x|

I haven't made any further progress past this point.

Is it possible to salvage this proof?Should I try an alternate approach?

**Answer**

**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:

What are you given?

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 0 < |x - 1| < \delta implies \left|\dfrac{1}{x} - 1 \right| < \epsilon\text{.} **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 \left|\dfrac{1}{x} - 1 \right| =\left|\dfrac{1-x}{x} \right| = \left|\dfrac{-(x-1)}{x} \right| = \left|\dfrac{x-1}{x} \right| 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 -\dfrac{1}{2} < x-1 < \dfrac{1}{2} \implies \dfrac{1}{2} < x < \dfrac{3}{2} \implies \dfrac{1}{2} < |x| < \dfrac{3}{2}\text{.} 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,

0 < x < 1 \implies 1 < \dfrac{1}{x} < \infty\text{.}

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 \dfrac{2}{3} <\dfrac{1}{|x|} < 2 and \dfrac{1}{|x|} < 2 \implies \dfrac{\left|x-1 \right|}{\left| x \right|} < 2\left| x-1 \right|\text{.} 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 \dfrac{\left|x-1 \right|}{\left| x \right|} < 2|x-1| < 2 \cdot \dfrac{1}{2} = 1 < \epsilon\text{.}Now suppose \dfrac{\epsilon}{2} \leq \dfrac{1}{2}, so that \delta

= \dfrac{\epsilon}{2}.Then \dfrac{\left|x-1 \right|}{\left| x \right|} < 2|x-1| < 2 \cdot \dfrac{\epsilon}{2} = \epsilon\text{.}

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

See also Why do we need min to choose \delta?.

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

\left|\dfrac{1}{x} - 1 \right| = \left|\dfrac{x-1}{x} \right| =

\dfrac{\left|x-1 \right|}{\left| x \right|} < 2\left| x-1 \right|

(since if |x - 1| < \dfrac{1}{2}, \dfrac{1}{|x|} < 2) and 2\left| x-1 \right| < 2\delta \leq 2\left(\dfrac{\epsilon}{2}\right) = \epsilon\text{. }\square

Success at last.

Addendum. Note that the end goal above was achieved, namely to show that \left|\dfrac{1}{x}-1\right| < \epsilon\text{.}In the step 2\left| x-1 \right| < 2\delta \leq

2\left(\dfrac{\epsilon}{2}\right) = \epsilon\text{,} textbooks

usually omit the step with the \delta and just write 2\left| x-1

\right| < 2\left(\dfrac{\epsilon}{2}\right) =

\epsilon\text{.}

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 \deltahasto be \dfrac{1}{2}. This is not the case for the following reason: given |x-1| < \delta, we have

|x-1| < \min\left(\dfrac{\epsilon}{2}, \dfrac{1}{2}\right)\text{.}

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

|x - 1| < \dfrac{\epsilon}{2} < \dfrac{1}{2}

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

**Attribution***Source : Link , Question Author : Gamma Function , Answer Author : Clarinetist*