For $\lim_{x→1}(2−1/x)=1$, finding δ, such that if $0<|x−1|<δ, then |f(x)−1|<0.1$ I don't know what I did wrong. Can anyone point out my mistake? The problem is:


  For $\lim\limits_{x\to1}(2-1/x)=1$, finding $\delta$, such that if $0&lt;|x-1|&lt;\delta$, then $|f(x)-1|&lt;0.1$


Here is what I did:

Since $|f(x)-1|=|2-1/x-1|=|1-1/x|=|x-1|/x&lt;0.1$, then $|x-1|&lt;0.1x$. 
We know that  $0&lt;|x−1|&lt;δ$, so $\delta=0.1x=x/10$

The answer sheet says $\delta=1/11$. I don't understand it.
$\delta$ must be a constant determined by $.1$ and not dependent on the value of $x$.

(How could you chose an $x$ so that $|x -1| &lt;\frac x{10}$ is we don't know what $x$ is in the first place?)

So you need to find a constant $\delta$ so that $|x - 1| &lt; \delta \implies |2-\frac 1x-1|=|1-\frac 1x| &lt; .1$.

To find that $|1 - \frac 1x| &lt; .1$ so

$.9 &lt; \frac 1x &lt; 1.1$ so 

$\frac {10}{11} &lt; x &lt; \frac {10}{9}$

$1 - \frac 1{11} &lt; x &lt;  1 + \frac 1{9}$

$-\frac 1{11} &lt; x -1 &lt; \frac 1{9}$

$|x - 1| &lt; \min(\frac 1{11},\frac 1{9}) = \frac 1{11}$.

So that is a (and the maximum possible) delta.

NOW it's easy to chose and $x$ so that $|x-1| &lt; \frac 1{11}$.  And when we do, it will logically (by following all those steps backwards) must be that $|2 - \frac 1x - 1| &lt; .1$
The answer means:

Claim.  If $|x - 1| &lt; 1/11$, then $|f(x) - 1| &lt; 1/10$.

Now we prove this. If $|x - 1| &lt; 1/11$, then $10/11 &lt; x &lt; 12/11$. It means

\begin{align}
f(x) - 1 &amp;= 1 - 1/x &lt; 1 - 11/12 = 1/12 &lt; 1/ 10 \\
f(x) - 1 &amp;= 1 - 1/x &gt; 1 - 11/10 = - 1/10.
\end{align}

So $|f(x) - 1| &lt; 1/10$.