Prove that if $|x-2|<0.001$, then $|\frac{1}{x}-\frac{1}{2}|<3\times 10^{-3}$

Question

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, $|\frac{1}{x}-\frac{1}{2}|<3\times10^{-3}$

Answer 1

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

$$\left|\frac1x-\frac12\right|=\frac{|2-x|}{|2x|}<\frac{0.001}{3.998}<3\cdot 10^{-3}$$

Answer 2

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.