Why does this epsilon-delta proof for $\lim_{x\to 1}\frac1x=1$ not work?

Question

I do not understand what is wrong with this $\epsilon-\delta$ proof that is supposed to show $\displaystyle\lim_{x\to 1}\frac{1}{x} = 1$

We have to show that for every $\epsilon > 0$ there exists a $\delta$ such that for all $x$ that satisfy $0 < \left| x - 1\right| <\delta$ the following is implied: $\left|\dfrac{1}{x} - 1\right| < \epsilon$

Here is how I started: $$ \left|\dfrac{1}{x} - 1\right| < \epsilon \implies\dfrac{\left|x-1\right|}{\left|x\right|} < \epsilon\\ $$ To bound the $\left|x\right|$, let $\delta = 1$ then $$\begin{align*} & \left|x - 1\right| < 1\\ \implies& |x| - |1| \leq \left|x - 1\right| < 1\\ \implies& |x| < 2\\[1em] \implies& |x-1|<\epsilon|x|<2\epsilon \end{align*}$$ So if we let $\delta = \min\left(1,\, 2\epsilon\right)$, all $x$ that satisfy $0<|x - 1|< \delta$ should also satisfy $\left|\dfrac{1}{x} - 1\right| < \epsilon$ for all $\epsilon > 0$, but that is not the case if you take a look at a graph.

Answer 1

$\left|\dfrac{1}{x} - 1\right| < \epsilon \implies\dfrac{\left|x-1\right|}{\left|x\right|} < \epsilon$

Great up to here

To bound the $|x|$, let $δ=1$ uh oh...look at what follows

$|x−1|<1\\ 0< x<1\\ \frac {|x-1|}{x} < \infty$

Not going to work. You need a tighter bound for $\delta$

let $\delta = \frac 12$

$|x|> \frac 12\\ |\frac {1-x}{x}| < 2|1-x|<2\delta<\epsilon\\ \delta = \min(\frac 12, \frac \epsilon 2)$

Answer 2

The inequality $$\vert x - 1 \vert < \epsilon \vert x \vert$$ doesn't follow from the previous line. You need an upper bound on $\vert x \vert$. To get that, just pick $\delta = \min(\epsilon/2,1/2)$. Then $\vert x - 1 \vert < \delta$ implies that $\vert x \vert > 1/2$, so we have $$\frac{\vert x - 1 \vert}{\vert x \vert} < \frac{\epsilon/2}{1/2} = \epsilon.$$