Single Variable Calculus, Chapter 2, 2.4, Section 2.4, Problem 23

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

\end{aligned}
\end{equation}
$


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

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



$
\begin{equation}
\begin{aligned}
\text{if } 0 < |x-a| < \delta \text{ then, }\\
& \phantom{x}
|x-a| < \delta = \varepsilon
\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

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


\end{aligned}
\end{equation}
$