<s>[INST] $\epsilon$-$\delta$ 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,

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

I haven't made any further progress past this point. Is it possible to salvage this proof? Should I try an alternate approach?
 [/INST] $\tag 1 |\frac{1}{x} - 1| < \varepsilon \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$
$ \text{ iff } \; -\varepsilon < \frac{1}{x} - 1 < \varepsilon \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$
$ \text{ iff } \; 1 -\varepsilon < \frac{1}{x} < 1 + \varepsilon \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$
$ \text{ iff } \; 1 -\varepsilon < \frac{1}{x} < 1 + \varepsilon \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$
$ \text{ iff } \; \frac{1}{1 + \varepsilon} < x < \frac{1}{1 - \varepsilon} \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$
$ \text{ iff } \; \frac{1}{1 + \varepsilon} < x < \frac{1}{1 - \varepsilon} \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$
$ \text{ iff } \; \frac{-\varepsilon}{1 + \varepsilon} < x - 1 < \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.
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