Give a delta-epsilon proof that $\lim_{x \to \frac{1}{2}} \frac{1}{x} = 2$, is my answer right? One of my first $\delta$-$\epsilon$ proofs, but I don't feel that I'm heading in the right direction. I want to proof that:
$$
    \lim_{x \to \frac{1}{2}} \frac{1}{x} = 2
$$
By the limit definition, For every $\epsilon > 0$, I must find a $\delta$, such that for every $x \in \operatorname{Dom}(f)$:
$$
    \left| x - \frac{1}{2}\! \right| < \delta \Rightarrow \left|\! \frac{1}{x} - 2\right| < \epsilon
$$
I assume $f(x) = \frac{1}{x}$ is a function $f : \mathbb{R}\backslash \{0\} \rightarrow \mathbb{R}$, since $\frac{1}{x}$ is a reciprocal.
Now,
$$ \begin{align}
    \left|\!\frac{1}{x} - 2\right| &= \left|\!\frac{1}{x} - 2\right|\\
    &= \frac{|1-2x|}{|x|} \\
    &= \frac{|\!\frac{1}{2}-x|}{2|x|} \end{align}
$$
Since $|x-y| = |y-x|$ and therefore,
$$
    < \frac{\delta}{2|x|} \\
$$
I recognise that I have to take $\delta<\frac{1}{2}$ to prevent dividing by zero. But finding an estimate in terms of $\delta$ for $\frac{1}{2|x|}$ seems difficult. My attempt:
$$
    2|x| = 2\left(x - \frac{1}{2} + \frac{1}{2}\right) < 2\left|x-\frac{1}{2}\!\right|+1 = 2\delta+1
$$
Concluding,
$$
    \left|\!\frac{1}{x} - 2\right| < \frac{\delta}{2\delta+1} < \epsilon
$$
And therefore,
$$
    \delta < \min\left(\frac{1}{2}, \frac{1}{\frac{1}{\epsilon} - 2}\right)
$$
Is this conceptual proof right? I have the feeling that I'm missing something.

The one equation should be
\begin{equation}
\frac{|1-2x|}{|x|} = \frac{2|\frac{1}{2}-x|}{|x|}\ .
\end{equation}
So you have that 
\begin{equation}
d(x,\frac{1}{2}) < \delta \Rightarrow d(\frac{1}{x},2) < \frac{2 \delta}{|x|}\ .
\end{equation}
Then, the second problem I see is that you take $2|x| < 2\delta +1$ and conclude that
\begin{equation}
\frac{\delta}{2|x|} < \frac{\delta}{2\delta +1} 
\end{equation}
The last inequality has to be $>$ since you devide by something larger.
