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


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


Here's how I started my proof,


  $$\left| f(x)-l \right|=\left| \frac{1}{x} - 1 \right| =\left| \frac{1}{x} \right| \left| x - 1\right| &lt; \epsilon \implies \left| x-1 \right| &lt; \epsilon |x|$$


I haven't made any further progress past this point. Is it possible to salvage this proof? Should I try an alternate approach?
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 $\epsilon &gt; 0$ and you want to find $\delta$ such that  $$0 &lt; |x - 1| &lt; \delta$$ implies $$\left|\dfrac{1}{x} - 1 \right| &lt; \epsilon\text{.}$$ 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 $|x - 1| &lt; \delta$ and get $\left|\dfrac{1}{x} - 1 \right| &lt; \epsilon$ 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 $\left|\dfrac{1}{x} - 1 \right| &lt; \epsilon$. 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 $|x - 1| &lt; \dfrac{1}{2}$, then $$-\dfrac{1}{2} &lt; x-1 &lt; \dfrac{1}{2} \implies \dfrac{1}{2} &lt; x &lt; \dfrac{3}{2} \implies \dfrac{1}{2} &lt; |x| &lt; \dfrac{3}{2}\text{.}$$ So if we choose $\delta = \dfrac{1}{2}$, $\dfrac{1}{2} &lt;|x| &lt; \dfrac{3}{2}$.


  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 $0 &lt; x &lt; 1$ and would not have been able to obtain a finite upper bound for $\dfrac{1}{x}$. That is, 
  $$0 &lt; x &lt; 1 \implies 1 &lt; \dfrac{1}{x} &lt; \infty\text{.}$$
  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 $\dfrac{1}{2} &lt;|x| &lt; \dfrac{3}{2}$, then $$\dfrac{2}{3} &lt;\dfrac{1}{|x|} &lt; 2$$ and $$\dfrac{1}{|x|} &lt; 2 \implies \dfrac{\left|x-1 \right|}{\left| x \right|} &lt; 2\left| x-1 \right|\text{.}$$ 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 $\dfrac{\epsilon}{2} &gt; \dfrac{1}{2}$, so that $\delta = \dfrac{1}{2}$. Then $\epsilon &gt; 1$. Then $$\dfrac{\left|x-1 \right|}{\left| x \right|} &lt; 2|x-1| &lt; 2 \cdot \dfrac{1}{2} = 1 &lt; \epsilon\text{.}$$
  
  Now suppose  $\dfrac{\epsilon}{2} \leq \dfrac{1}{2}$, so that $\delta
 = \dfrac{\epsilon}{2}$. 
  
  Then $$\dfrac{\left|x-1 \right|}{\left| x \right|} &lt; 2|x-1| &lt; 2 \cdot \dfrac{\epsilon}{2} = \epsilon\text{.}$$
  
  In both cases, we have $\dfrac{\left|x-1 \right|}{\left| x \right|}  &lt; \epsilon$, 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 $\epsilon &gt; 0$ be given. Choose $\delta := \min\left(\dfrac{\epsilon}{2} , \dfrac{1}{2} \right)$. Then
  $$\left|\dfrac{1}{x} - 1 \right| = \left|\dfrac{x-1}{x} \right| =
 \dfrac{\left|x-1 \right|}{\left| x \right|} &lt; 2\left| x-1 \right|$$
  (since if $|x - 1| &lt; \dfrac{1}{2}$, $\dfrac{1}{|x|} &lt; 2$) and $$2\left| x-1 \right| &lt; 2\delta \leq 2\left(\dfrac{\epsilon}{2}\right) = \epsilon\text{. }\square$$


Success at last.


  Addendum. Note that the end goal above was achieved, namely to show that $$\left|\dfrac{1}{x}-1\right| &lt; \epsilon\text{.}$$
  
  In the step $$2\left| x-1 \right| &lt; 2\delta \leq
 2\left(\dfrac{\epsilon}{2}\right) = \epsilon\text{,}$$ textbooks
  usually omit the step with the $\delta$ and just write $$2\left| x-1
 \right| &lt; 2\left(\dfrac{\epsilon}{2}\right) =
 \epsilon\text{.}$$
  
  Addendum. It may seem that the note "(since if $|x - 1| &lt; \dfrac{1}{2}$, $\dfrac{1}{|x|} &lt; 2$)" 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 $|x-1| &lt; \delta$, we have
  $$|x-1| &lt; \min\left(\dfrac{\epsilon}{2}, \dfrac{1}{2}\right)\text{.} $$ 
  Obviously, if $\epsilon \geq 1$, we end up with $|x - 1| &lt; \dfrac{1}{2}$, as stated above. But let's suppose that $\epsilon &lt; 1$. Then
  $$|x - 1| &lt; \dfrac{\epsilon}{2} &lt; \dfrac{1}{2}$$
  and you end up with $|x - 1| &lt; \dfrac{1}{2}$, so the $\dfrac{1}{|x|} &lt; 2$ implication holds in either case.
$\tag 1 |\frac{1}{x} - 1| &lt; \varepsilon \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$

$ \text{ iff } \; -\varepsilon &lt; \frac{1}{x} - 1 &lt; \varepsilon \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$

$ \text{ iff } \; 1 -\varepsilon &lt; \frac{1}{x} &lt; 1 + \varepsilon \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$

$ \text{ iff } \; 1 -\varepsilon &lt; \frac{1}{x} &lt; 1 + \varepsilon \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$

$ \text{ iff } \; \frac{1}{1 + \varepsilon} &lt; x &lt; \frac{1}{1 - \varepsilon} \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$

$ \text{ iff } \; \frac{1}{1 + \varepsilon} &lt; x &lt; \frac{1}{1 - \varepsilon} \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$

$ \text{ iff } \; \frac{-\varepsilon}{1 + \varepsilon} &lt; x - 1 &lt; \frac{+\varepsilon}{1 - \varepsilon} \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$

We are trying to find our $\delta \gt 0$ 'setup', $|x -1| \lt \delta$ contrained by

$\tag 2 \frac{-\varepsilon}{1 + \varepsilon} \le -\delta \lt x - 1 \lt +\delta \le  \frac{+\varepsilon}{1 - \varepsilon} \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$

so that $\text{(2)}$ implies $\text{(1)}$ (we can use our developed "$\text{iff }$-logic-chain" in reverse).

Setting

$\quad \delta = \frac{\varepsilon}{1 + \varepsilon}$

sets up the left side of $\text{(2)}$, but also takes care of the rights side since

$\quad \delta = \frac{\varepsilon}{1 + \varepsilon} \le \frac{\varepsilon}{1 - \varepsilon}$

Note: Here we don't use the 'min delta' approach. Using the same mechanical technique found here, we simply 'turn-the-crank'. As we work things out we realize that we want to constrain $\varepsilon$. 

Instructional Example 1: You are challenged with an $\varepsilon$ that is greater than or equal to $1/2$.
Then set $\delta = 1/3$.

Note that if $\varepsilon = 1/500$, we take $\delta = 1/501$, a 'looser' number than the $\varepsilon/2 = 1/1000$ found in other (excellent) answers in this thread.