Single Variable Calculus, Chapter 2, 2.4, Section 2.4, Problem 27

Show that the statement $\displaystyle\lim\limits_{x \to 0} |x| = 0$ 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-0| < \delta
\qquad \text{ then } \qquad
||x|-0| < \varepsilon\\

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
& \text{That is,}\\
& \phantom{x} & \text{ if } 0 < |x| < \delta \qquad \text{ then } \qquad ||x|| < \varepsilon\\
\end{aligned}
\end{equation}
$

But according to the the property of absolute value, we write
$\quad ||x|| < \varepsilon \Rightarrow |x| < \varepsilon \qquad$ (Idempotence, where the absolute value of the absolute value is equal to the absolute value)

So,
$\quad \text{if } \, 0 < |x| < \delta \quad \text{ then } \quad |x| < \varepsilon$

Therefore,
$\quad \lim\limits_{x \to 0} |x| = 0$