When you use the epsilon delta method for proving limits and you face a problem of expressing $\delta$ as a function of $\epsilon$ only, I know that you have to put a restriction on $\vert{x-c}\vert$ to attain an inequality $\vert{x-c}\vert<1$ (1 is arbitrary). This will get us an expression relating $\delta$ with $\epsilon$. and We choose the smaller one of the two.
I understand this procedure, but not quite the reason why we use $\vert{x-c}\vert<1$ as the restriction in the first place. I guess we can use any other values, but wouldn't that result in different answers?
Also, what if we chose a restriction such as $\vert{x-c}\vert<0.0001$ (a small number). Would this still be okay?
For example, given $\epsilon>0$, we want to show that if $\vert{x-3}\vert<\delta$, then $\vert{\frac{2x+3}{4x-9}-3}\vert<1$
$\vert{\frac{2x+3}{4x-9}-3}\vert=\vert{\frac{10(x-3)}{4x-9}}\vert=\frac{10}{\vert{4x-9}\vert}\vert{x-3}\vert<\epsilon$
So if we place a restriction such that $\vert{x-3}\vert<1$, then $2<x<4$. Thus, $1<\vert{4x-9}\vert<7$. So if $\vert{x-3}\vert<1$, then $\vert{4x-9}\vert>1$. Thus $\delta=min(1,\frac{\epsilon}{10}$).
However, if we chose the restriction $\vert{x-3}\vert<1/10$, then $13/5<\vert{4x-9}\vert<17/5$. Meaning that we would get $\vert{x-3}\vert<13\epsilon/50$. So $\delta=min(1,\frac{13\epsilon}{50}$)