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.
 A: 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 $\delta$. 
A: This is the jist of it. You are required to bound the value of $ | \dfrac 1 x - \dfrac 1 2 |  $. But you are only allowed to bound $ |x  - 2| $. So by putting in a restriction on $ |x - 2| \lt \delta$ you need to achieve the  restriction $ | \dfrac 1 x - \dfrac 1 2  | \lt \epsilon$. And more importantly you need to prove that such a $\delta$ exists for every $\epsilon \gt 0$. So we begin with $ |f(x) - L| $. 
$$  | \dfrac 1 x - \dfrac 1 2 | = \frac{|x - 2|}{2|x|} ---(1)  $$
Since we can bound $|x - 2|$ as severely as we want we may as well allow $$ |x - 2| \lt 1 \iff x \in ( 1, 3 ) \implies |x| \gt 1  \implies \frac{1}{|x|} \lt 1 $$
Using this on $(1)$ we have, 
$$ | \dfrac 1 x - \dfrac 1 2  | \lt \dfrac{|x - 2|}{2} \;\; \text{as long as } |x - 2| \lt 1 $$
Now we have a bound on $  | \dfrac 1 x - \dfrac 1 2  |  $ which is a multiple of $ |x - 2| $. So as soon as we restrict $ |x - 2| $ we automatically restrict $ | \dfrac 1 x - \dfrac 1 2  |  $. 
All we need to do is to choose $ \delta = 2 \epsilon $ and we will be done. 
But not so fast. we used the additional assumption that $ |x - 2| \lt 1$. So if we let $ \delta = \min \{2 \epsilon, 1\} $ then all our required constraints will be satisfied. 
The most important thing to notice is that such a $\delta \;$ exists for any $\epsilon \gt 0$. 
