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...



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



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