Prove that $\lim_{x\to 2} \frac{1}{x} = \frac{1}{2}$ I want to prove this limit by using $(\epsilon,\delta)$ definition
$$\lim_{x\to 2} \frac{1}{x} = \frac{1}{2}$$
Here is what I have done
$$|\frac{1}{x} - \frac{1}{2}|<\epsilon \Leftrightarrow -\epsilon < \frac{1}{x} - \frac{1}{2} < \epsilon$$
$$\Leftrightarrow -\epsilon + \frac{1}{2} < \frac{1}{x} < \epsilon + \frac{1}{2}$$
$$\Leftrightarrow \frac{1}{-\epsilon + \frac{1}{2}} > x > \frac{1}{\epsilon + \frac{1}{2}}$$
I have problem at this point by the definition, $\epsilon$ can be any positive arbitrary number so $\frac{1}{-\epsilon + \frac{1}{2}}$ can be negative number so $x$ can be both smaller than $\frac{1}{-\epsilon + \frac{1}{2}}$ and bigger than $\frac{1}{\epsilon + \frac{1}{2}}$ then the domain of $x$ will be $[-\infty,\frac{1}{-\epsilon + \frac{1}{2}}]\cup [\frac{1}{\epsilon + \frac{1}{2}},\infty]$ But also by the definition, when $f(x)$ approach the limit, it is very small and we can chose $\epsilon$ to be sufficient small to make $\frac{1}{-\epsilon + \frac{1}{2}}>0$. So which way should I do. By the former, the thing will be a mess because the domain of x must be something like $[a,b]$ not $[a,b]\cup[c,d]$
So what should I do to prove this limit.

Let $\epsilon >0$. Then you have to bound $|\frac{1}{x}-\frac{1}{2}|=\frac{1}{2}\frac{|x-2|}{|x|}$. So, by choosing $ \delta=min\{1,2\epsilon\} $ you have $-1<x-2<1\Longrightarrow 1<x<3\Longrightarrow \frac{1}{x}<1$ and $|x-2|<2\epsilon$ therefore 
$ |\frac{1}{x}-\frac{1}{2}|=\frac{1}{2}\frac{|x-2|}{|x|}<\epsilon$
