Single Variable Calculus, Chapter 2, 2.4, Section 2.4, Problem 19

Show that the statement $\displaystyle\lim\limits_{x \to 3} \frac{x}{5} = \frac{3}{5}$ is correct using the $\varepsilon$, $\delta$ definition of limit.

Based from the defintion,


$\begin{equation} \begin{aligned} \phantom{x} \text{if } & 0 < |x - a| < \delta \qquad \text{ then } \qquad |f(x) - L| < \varepsilon\\ \phantom{x} \text{if } & 0 < |x-3| < \delta \qquad \text{ then } \qquad \left|\frac{x}{5} - \frac{3}{5}\right| < \varepsilon\\ \end{aligned} \end{equation}$



$\begin{equation} \begin{aligned} & \text{But, } \\ & \phantom{x} & \left|\frac{x}{5} - \frac{3}{5}\right| = \left|\frac{1}{5} (x-3)\right| = \frac{1}{5}|x-3| \\ & \text{So, we want}\\ & \phantom{x} & \text{ if } 0 < |x-3| < \delta \qquad \text{ then } \qquad \frac{1}{5}|x-3| < \varepsilon\\ & \text{That is,} \\ & \phantom{x} & \text{ if } 0 < |x-3| < \delta \qquad \text{ then } \qquad |x-3| < 5\varepsilon\\ \end{aligned} \end{equation}$


The statement suggests that we should choose $\displaystyle \delta = 5 \varepsilon$

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



$\begin{equation} \begin{aligned} \text{if } 0 < |x-3| < \delta \text{ then, }\\ \left|\frac{x}{5} - \frac{3}{5}\right| & = \left| \frac{1}{5} (x-3) \right| = \frac{1}{5}|x-3| < \frac{\delta}{5} = \frac{5 \varepsilon}{5 } = \varepsilon \end{aligned} \end{equation}$



$\begin{equation} \begin{aligned} & \text{Thus, }\\ & \phantom{x} \quad\text{if } 0 < |x-3| < \delta \qquad \text{ then } \qquad \left|\frac{x}{5} - \frac{3}{5}\right| < \varepsilon\\ & \text{Therefore, by the definition of a limit}\\ & \phantom{x} \qquad \lim\limits_{x \to 3}\frac{x}{5} = \frac{3}{5} \end{aligned} \end{equation}$