Answer
Here, the given function, $f(x) = 2- \dfrac{1}{x}$, for $x \ne 0$.
Consider the following
$|f(x) - 1| = \Big|2 - \dfrac{1}{x} - 1\Big| = \Big|1 - \dfrac{1}{x} \Big| = \Big| \dfrac{x-1}{x} \Big| = \dfrac{|x-1|}{|x|}$
[Using property of modulus]
Now, if $|f(x) - 1|< 0.1$ , then,
$\dfrac{|x-1|}{|x|} < 0.1$ [ Inequality 1 ]
As, $x \neq 0$ , we have $|x|> 0$, hence,
$|x| = |(x- 1) + 1| \leq |x-1| + 1 $ [ Using the triangle inequality]
$\Rightarrow \dfrac{1}{|x-1|+1} \leq \dfrac{1}{|x|}$
$\Rightarrow \dfrac{|x-1|}{|x-1|+1} \leq \dfrac{|x-1|}{|x|} < 0.1$
Thus, we get
$\Rightarrow \dfrac{|x-1|}{|x-1|+1} < 0.1$
$\Rightarrow |x-1| < 0.1|x-1| + 0.1$
$\Rightarrow |x-1| - 0.1|x-1| < 0.1$
$\Rightarrow 0.9|x-1| < 0.1$
$\Rightarrow |x-1| < \dfrac{0.1}{0.9}$
Therefore, for $\delta = \dfrac{0.1}{0.9},$ we get $0$
Work Step by Step
Understanding how to find appropriate $\delta$ for a given $\epsilon$ is a very important aspect to start the pure mathematics.
In practice for a given $\epsilon$ > 0, if we need $\delta$ such that $0 < |x-c|< \delta$ whenever $|f(x)-L|< \epsilon$.
Then, we have to follow the given steps.
$\textbf{Step I}$: Find the relation between $|f(x) -L|$ and $|x-c|$ using the properties of modulus, generally used ones are as follows.
(a) $|x \cdot y| = |x|\cdot|y|$
(b) $\Big|\dfrac{x}{y}\Big| = \dfrac{|x|}{|y|}$, for $y \neq 0$
$\textbf{Step II}$: In the inequality $|f(x)-L|< \epsilon$, replace $|f(x) -L|$ using the relation obtained in step I.
$\textbf{Step III}$: At last find the bound for $|x-c|$ using the inequality obtained in step II. This can be done using the properties of inequalities, generally used properties are as follows.
(a) $|x + y| \leq |x| + |y|$ [ Triangle Inequality]
(b) $||x| - |y|| \leq |x - y|$
(c) If $a< b$ , then, $\dfrac{1}{b} < \dfrac{1}{a}$ , where $a,b \neq 0$
(d) $|x|^{2} = |x^{2}|$
Check the numerical use in the answer to the problem.
