Calculus limit epsilon delta Prove using only the epsilon , delta - definition
$\displaystyle\lim_{x\to 2}\dfrac{1}{x} = 0.5$
Given $\epsilon > 0 $, there exists a delta such that 
$ 0<|x-2|< \delta$  implies $|(1/x) – 0.5| < \epsilon$.
Therefore, $$\begin{align}|(1/x) – 0.5| & = |(2-x)/2x| \\ 
& =|\frac{-(x-2)}{2(x-2+2)}| \\ & < |\frac{(x-2)}{2(x-2)} |=\frac{|x-2|}{2|x-2|}=\delta/2\delta=0.5
\end{align}$$
But epsilon is $> 0$. My $0.5$ is causing issue.

Hint: You want the following to hold for all $x$ such that $2-\delta < x < 2+\delta$: 
$\left|\dfrac{1}{x} - \dfrac{1}{2}\right| < \epsilon$
$-\epsilon < \dfrac{1}{x} - \dfrac{1}{2} < \epsilon$
$\dfrac{1}{2} -\epsilon < \dfrac{1}{x}< \dfrac{1}{2} + \epsilon$
For what range of values for $x$ is this inequality true? That will help you pick $\delta$. 
