Given $f(x)=\frac{1}{x-1}$, find $δ$, such as if $0<|x-2|<δ$, then $|f(x)-1|<0.01$

Question

Answer: Given $f(x)=\frac{1}{x-1}$, find $δ$, such as if $0<|x-2|<δ$, then $|f(x)-1|<0.01$

So,I begin by expressing the aforementioned as a limit:

$(\lim\limits_{x \to 2} [\frac{1}{x-1}]=1) \equiv (∀ε|ε>0, ∃!δ|δ>0 : [0<|x-2|<δ] → [|f(x)-1|<0.01]) $

I take this : $[0<|x-2|<δ] → [|f(x)-1|<0.01]$, and try to find a relation between the antecedent and the consecuent:

$[0<|x-2|<δ] → [|\frac{1}{x-1}-1|<0.01]$

$[0<|x-2|<δ] → [|\frac{1-(x-1)}{x-1}|<0.01]$

$[0<|x-2|<δ] → [|\frac{-(x-2)}{x-1}|<0.01]$

$[0<|\color{red}{(x-2)}|<δ] → [|\color{fuchsia}{\frac{-1}{x-1}}||\color{red}{(x-2)}|<0.01]$

Here's where I get stuck. I know that if $\color{fuchsia}{\frac{-1}{x-1}}$ was expressed as a single number, let's say $Λ|Λ∈ℝ$, i could just devide both sides of the inequation by $Λ$, and would have:

$[0<|\color{red}{(x-2)}|<δ] → [\color{red}{(x-2)}|<\frac{0.01}{Λ}$

$∴ δ=\frac{0.01}{Λ}$

But that's not the case; i don't know what to do when that term that term that goes together with the common term to both inequations, has a variable, and my book (Larson Calculus), doesn't explain it too clearly.

How should I proceed?

Thanks in advance.

Answer 1

First let $0 < |x-2| < 0.1\implies 2-|x| < |x-2| < 0.1\implies |x| > 1.9\implies |x| - 1> 0.9\implies |f(x) - 1| = \left|\dfrac{1}{x-1}-1\right|= \dfrac{|x-2|}{|x-1|}\le \dfrac{|x-2|}{|x|-1}< \dfrac{|x-2|}{0.9}< 0.01 \implies |x-2| < 0.009\implies \delta = 0.009$.