Limit proof for rational function $\frac{1}{x}$ A while ago I posted another one like this with a incorrect approach, please see this one!
Is this an accurate proof for limits for the function $\frac{1}{x}$
$\displaystyle \lim_{x\to1} \frac{1}{x} = \frac{1}{1} = 1$
Using $\epsilon-\delta$  So,
$\displaystyle \frac{|x-1|}{|x|} < \epsilon$ for some $\displaystyle |x-1| < \delta$                   
Lets assume $|x - 1| < \frac{1}{3}$
$\displaystyle \frac{2}{3} < |x| < \frac{4}{3}$     
$= \frac{3}{2} > \frac{1}{|x|} > \frac{3}{4}$
So we have,
$|x - 1| < \delta$
$\frac{1}{x} < \frac{4}{3}$ 
Therefore we get: $\frac{|x-1|}{|x|} < \frac{3\delta}{4}$
It is the equation: $\frac{3\delta}{4} = \epsilon$, which is possible only if, $\delta_1 = \frac{4\epsilon}{3}$
Therefore, $\delta = \min(\frac{1}{3},\frac{4\epsilon}{3})$

I think you have the right idea, but it is a bit unorganized, separate the scratch work from the proof. This is how I would approach it.
Let $\epsilon>0$ be arbitrary, let $\delta=\min(1/2,\epsilon/2)$ such that $\vert x−1\vert<\delta$,   then $$\left\vert\frac{1}{x}−1\right\vert=  \frac{\vert x−1\vert}{\vert x\vert}< 2\vert x−1\vert <2\epsilon/2=\epsilon.$$
