Single Variable Calculus, Chapter 2, 2.4, Section 2.4, Problem 36

Show that the statement $\displaystyle\lim\limits_{x \to 2} \frac{1}{x} = \frac{1}{2}$ is correct using the $\varepsilon$, $\delta$ definition of limit.

Based from the defintion,


$
\begin{equation}
\begin{aligned}

\phantom{x} \text{if } & 0 < |x - a| < \delta
\qquad \text{ then } \qquad
|f(x) - L| < \varepsilon\\

\phantom{x} \text{if } & 0 < |x-2| < \delta
\qquad \text{ then } \qquad
\left|\frac{1}{x} - \frac{1}{2}\right| < \varepsilon\\

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
& \text{But, } \\
& \phantom{x} & \left|
\frac{1}{x} - \frac{1}{2}
\right|
= \left|
\frac{2-x}{2x}
\right|
= \left|
\frac{-(x-2)}{2x}
\right|
=
\frac{1}{2x}
|x-2|
\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
& \text{So, we want}\\
& \phantom{x} & \text{ if } 0 < |x-2| < \delta \qquad \text{ then } \qquad \frac{1}{2x}|x-2| < \varepsilon\\
& \text{That is,}\\
& \phantom{x} & \text{ if } 0 < |x-2| < \delta \qquad \text{ then } \qquad |x-2| < 2 \varepsilon x\\
\end{aligned}
\end{equation}
$


The statement suggests that we should choose $\displaystyle \delta = 2 \varepsilon x$

By proving that the assumed value of $\delta$ will fit the definition...



$
\begin{equation}
\begin{aligned}
\text{if } 0 < |x-2| < \delta \text{ then, }\\
& \phantom{x} & \left|
\frac{1}{x} - \frac{1}{2}
\right|
= \left|
\frac{2-x}{2z}
\right|
= \left|
\frac{-(x-2)}{2x}
\right|
=
\frac{1}{2x}
|x-2|
<
\frac{\delta}{2x}
= \left(
\frac{2 \varepsilon x}{2x}
\right)
=
\varepsilon
\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

& \text{Thus, }\\
& \phantom{x} \quad\text{if } 0 < |x-2| < \delta \qquad \text{ then } \qquad \left|\frac{1}{x} - \frac{1}{2}\right| < \varepsilon\\
& \text{Therefore, by the definition of a limit}\\
& \phantom{x} \qquad \lim\limits_{x \to 2}\frac{1}{x} = \frac{1}{2}


\end{aligned}
\end{equation}
$