Prove that if $|x-2|<0.001$, then $|\frac{1}{x}-\frac{1}{2}|<3\times 10^{-3}$ I still have difficulties with absolute value, and even if I manage to solve questions and problems, I do that awkwardly.
So, please show me if this is the way to answer this question.
Thank you in advance.
$|x-2|<0.001 \iff -0.001<x-2$ or $x-2<0.001$
$$\begin{align}\\-0.001<x-2 & \iff 2-0.001<x\\  &\iff1.999<x\\  &\iff\frac{1}{x}<\frac{1}{1.999}\\  &\iff\frac{1}{x}-\frac{1}{2}<\frac{0.001}{3.998}\\ &\iff \frac{1}{x}-\frac{1}{2}<3\times 10^{-3}\end{align}$$
$$\begin{align}\\ x-2<0.001 &\iff x<2.001\\  &\iff\frac{1}{x}>\frac{1}{2.001}\\  &\iff\frac{1}{x}-\frac{1}{2}>\frac{3.002}{2}\\  &\iff\frac{1}{x}-\frac{1}{2}>-3\times10^{-3}\end{align}$$
We have:  $-3\times 10^{-3}<\frac{1}{x}-\frac{1}{2}<3\times10^{-3}$
Then, $|\frac{1}{x}-\frac{1}{2}|<3\times10^{-3}$
 A: Your work is correct. Also, you can simply write:
We have by the triangle inequality $$2-|x|\le |2-x|<0.001\implies 3.998<|2x|$$
so
$$\left|\frac1x-\frac12\right|=\frac{|2-x|}{|2x|}<\frac{0.001}{3.998}<3\cdot 10^{-3}$$
A: The way I would answer
(writing out every step):
$\begin{array}\\
|x-2|<0.001 
&\iff -0.001<x-2 < .001\\
&\iff 2-0.001<x < 2+.001\\
&\iff 1.999<x < 2.001\\
&\iff \frac1{1.999}>\frac1{x} > \frac1{2.001}\\
&\iff \frac1{1.999}-\frac12 >\frac1{x}-\frac12 > \frac1{2.001}-\frac12\\
&\iff \frac{2-1.999}{2\cdot1.999} >\frac1{x}-\frac12 > \frac{2-2.001}{2\cdot 2.001}\\
&\iff \frac{.001}{3.998} >\frac1{x}-\frac12 > \frac{-.001}{4.002}\\
&\implies  |\frac1{x}-\frac12| < \frac{.001}{3.998}\\
\end{array}
$
Note how this generalizes,
using copy, paste, and edit:
If $0 < c < a$,
$\begin{array}\\
|x-a|<c
&\iff c<x-a < c\\
&\iff a-c<x < a+c\\
&\iff \frac1{a-c}>\frac1{x} > \frac1{a+c}\\
&\iff \frac1{a-c}-\frac1{a} >\frac1{x}-\frac1{a} > \frac1{a+c}-\frac1{a}\\
&\iff \frac{a-(a-c)}{a(a-c)} >\frac1{x}-\frac1{a} > \frac{a-(a+c)}{a(a+c)}\\
&\iff \frac{c}{a(a-c)} >\frac1{x}-\frac1{a} > \frac{-c}{a(a+c)}\\
&\implies  |\frac1{x}-\frac1{a}| < \frac{c}{a(a-c)}
\quad\text{since }|\frac{c}{a(a+c)}| < |\frac{c}{a(a-c)}|\\
\end{array}
$
To make
$|\frac1{x}-\frac1{a}|
< d
$,
we can choose
$d
> \frac{c}{a(a-c)}
$
or
$c 
< da(a-c)
=da^2-cda
$
or
$c(1+da)
< da^2
$
or
$c
<\frac{da^2}{1+da}
$.
Note that
$\frac{da^2}{1+da}
< a
$,
so that
$c < a$
is automatically satisfied.
