For $\lim_{x→1}(2−1/x)=1$, finding δ, such that if $0<|x−1|<δ, then |f(x)−1|<0.1$

Question

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 $0<|x-1|<\delta$, then $|f(x)-1|<0.1$

Here is what I did:

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

The answer sheet says $\delta=1/11$. I don't understand it.

Answer 1

$\delta$ must be a constant determined by $.1$ and not dependent on the value of $x$.

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

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

To find that $|1 - \frac 1x| < .1$ so

$.9 < \frac 1x < 1.1$ so

$\frac {10}{11} < x < \frac {10}{9}$

$1 - \frac 1{11} < x < 1 + \frac 1{9}$

$-\frac 1{11} < x -1 < \frac 1{9}$

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

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

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

Answer 2

The answer means:

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

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

\begin{align} f(x) - 1 &= 1 - 1/x < 1 - 11/12 = 1/12 < 1/ 10 \\ f(x) - 1 &= 1 - 1/x > 1 - 11/10 = - 1/10. \end{align}

So $|f(x) - 1| < 1/10$.