<s>[INST] Is there a neater proof for $\lim_{x \rightarrow 1} 1/x = 1$? What I need to prove:

For all $\epsilon>0$, there exist a $\delta>0$, such that for all $x$ satisfying $0<|x-1|<\delta$, we have $|1/x-1|<\epsilon$.

My proof:
Let $\delta=\min \{ 1/2 ,\epsilon/2 \}$. Then when $\epsilon\geq 1$,
$$
|x-1| < \frac{1}{2} \\
(*)  \frac{1}{2} < x < \frac{3}{2}  \\
\frac{2}{3} < \frac{1}{x} < 2 \\
-\frac{1}{3} < \ \frac{1}{x}-1 < 1 
$$
and since $-1<-1/3$, we have $|1/x-1|<1\leq \epsilon$.
When $\epsilon<1$,
$$
|x-1| < \frac{\epsilon}{2} \\
\frac{2-\epsilon}{2} < x < \frac{2+\epsilon}{2} \\
-\frac{\epsilon}{2+\epsilon} < \frac{1}{x}-1 < \frac{\epsilon}{2-\epsilon}
$$
and since $\epsilon>0$, we have $2+\epsilon > 2-\epsilon$. So
$$
-\frac{\epsilon}{2-\epsilon} < -\frac{\epsilon}{2+\epsilon} \implies \left| \frac{1}{x}-1 \right|< \frac{\epsilon}{2-\epsilon} < \epsilon.
$$
What I'm looking for is a proof that doesn't "deconstruct" the absolute values as in the starred inequality, ie uses only absolute value identities such as the triangle inequality and its variations.
 [/INST] If $|x - 1| \leqslant 1/2$, then $1 - x \leqslant |1 - x| = |x - 1| \leqslant 1/2$, therefore $x \geqslant 1/2$.
Since $x > 0$, the expression $1/x$ is well-defined.
Also because $x > 0$, we have $|x| = x \geqslant 1/2$.
If you must use the Triangle Inequality, you can argue (using it in the form $|u - v| \geqslant ||u| - |v||$):
$$
|x| = |1 - (1 - x)| \geqslant |1 - |1 - x|| = |1 - |x - 1|| \geqslant 1/2 \quad
(\text{because } 1 - |x - 1| \geqslant 1/2 > 0),
$$
but that only seems to make the argument more complicated, not less.
(It looks a bit less unnatural, however, if you start with the substitution $x = 1 - h$, or $x = 1 + h$.)
Therefore, if $|x - 1| \leqslant 1/2$,
$$
\left\lvert\frac{1}{x} - 1\right\rvert =
\left\lvert\frac{1 - x}{x}\right\rvert =
\frac{|1 - x|}{|x|} =
\frac{|x - 1|}{|x|}
\leqslant 2|x - 1|.
$$
Therefore, for all $\epsilon > 0$,
$$
\text{if } 0 < |x - 1| < \frac{\min\left\{1, \epsilon\right\}}{2},
\text{ then } \left\lvert\frac{1}{x} - 1\right\rvert < \epsilon.
$$
So, just as you found, we can take:
$$
\delta = \frac{\min\left\{1, \epsilon\right\}}{2}.
$$

If you were asked to prove, more generally,
$$
a \ne 0 \implies \lim_{x \to a} \frac{1}{x} = \frac{1}{a}
$$
(directly, that is, rather than deducing it from the case $a = 1$, via a result about limits of constant multiples of functions), there would be a bit more scope for manipulating absolute values using the Triangle Inequality:
If $|x - a| \leqslant |a|/2$, then
$$
|a| - |x| \leqslant ||a| - |x|| \leqslant |a - x| = |x - a| \leqslant |a|/2,
$$
therefore $|x| \geqslant |a|/2$, therefore
$$
\left\lvert\frac{1}{x} - \frac{1}{a}\right\rvert =
\left\lvert\frac{a - x}{xa}\right\rvert =
\frac{|a - x|}{|xa|} =
\frac{|x - a|}{|x|\cdot|a|}
\leqslant 2\frac{|x - a|}{|a|^2}.
$$
Therefore, for all $\epsilon > 0$,
$$
\text{if } 0 < |x - a| < \frac{|a|\cdot\min\left\{1,\, |a|\cdot\epsilon\right\}}{2},
\text{ then } \left\lvert\frac{1}{x} - \frac{1}{a}\right\rvert < \epsilon,
$$
which isn't much more complicated than the special case $a = 1$.
I see that Michael Spivak, Calculus, 3rd ed. (1994), p.101, does it this way - so at least it's not wrong!
</s>