A problem about $\epsilon-\delta$ definition of a limit where the target is to find $\delta$ The fuction is:
$f(x)=2-\frac{1}{x}$
It requires to find $\delta$ such that $0<\mid x-1 \mid<\delta$ else, $\mid f(x)-1 \mid<0.1$ 
My process so far, has been trying to compare the stuff that I have, because I found the $\delta$ like that before, but I don't know how to continue with this especific problem.
What I have:
$\mid 2-\frac{1}{x}-1\mid<0.1$ 
$\mid 1-\frac{1}{x}\mid<0.1$ 
$\mid\frac{x-1}{x}\mid<0.1$
$\frac{\mid x-1\mid}{\mid x\mid}<0.1$
$\mid x-1 \mid<0.1\mid x \mid$ 
From here, I don't know how to proceed. The answer in the book says that $\delta=\frac{1}{11}$
 A: First, choose $\delta<1$ so that we have
\begin{eqnarray}
|x-1|&<&\delta\\
-\delta<x-1&<&\delta\\
1-\delta&<&x<1+\delta\\
0&<&x
\end{eqnarray}
Then we have
\begin{eqnarray}
\left\vert f(x)-1\right\vert&<&\frac{1}{10}\quad\text{ iff}\\
\left\vert 2-\frac{1}{x}-1\right\vert&<&\frac{1}{10}\quad\text{ iff}\\
\left\vert 1-\frac{1}{x}\right\vert&<&\frac{1}{10}\quad\text{ iff}\\
-\frac{1}{10}<1-\frac{1}{x}&<&\frac{1}{10}\quad\text{ iff}\\
-\frac{1}{10}-1<-\frac{1}{x}&<&\frac{1}{10}-1\quad\text{ iff}\\
-\frac{11}{10}<-\frac{1}{x}&<&-\frac{9}{10}\quad\text{ iff}\\
\frac{11}{10}>\frac{1}{x}&>&\frac{9}{10}\quad\text{ iff}\\
\frac{9}{10}<\frac{1}{x}&<&\frac{11}{10}
\end{eqnarray}
which is equivalent to the previous inequality.
So the reciprocals are related in the reverse order
$$ \frac{10}{11}<x<\frac{10}{9} $$
Subtract $1$ to get
$$ -\frac{1}{11}<x-1<\frac{1}{9}$$
So we should choose $|x-1|<\dfrac{1}{11}$
This gives us
$$ -\frac{1}{11}<x-1<\frac{1}{11} $$
And since $\dfrac{1}{11}<\dfrac{1}{9}$ we have
$$-\frac{1}{11}<\frac{1}{x}<\frac{1}{9}$$
which is equivalent to
$$ \left\vert f(x)-1\right\vert<\frac{1}{10} $$
provided it is true that
$$|x-1|<\dfrac{1}{11}$$
