Single Variable Calculus, Chapter 2, 2.4, Section 2.4, Problem 17

Show that the statement $\lim\limits_{x \to -3} (1-4x) = 13$ is correct using the
$\varepsilon$, $\delta$ definition of limit and illustrate its graph.






Based from the defintion,


$\begin{equation} \begin{aligned} \phantom{x} \text{if } & 0 < |x - a| < \delta \qquad \text{ then } \qquad |f(x) - L| < \varepsilon\\ \phantom{x} \text{if } & 0 < |x - (-3)| < \delta \qquad \text{ then } \qquad |(1-4x)-13| < \varepsilon\\ \end{aligned} \end{equation}$



$\begin{equation} \begin{aligned} & \text{But, } \\ & \phantom{x} & |1-4x-13| = |-4x-12| = |-4(x+3)| = 4|x+3| \\ & \text{So, we want}\\ & \phantom{x} & \text{ if } 0 < |x+3| < \delta \qquad \text{ then } \qquad 4|x+3| < \varepsilon\\ & \text{That is,} \\ & \phantom{x} & \text{ if } 0 < |x+3| < \delta \qquad \text{ then } \qquad |x+3| < \frac{\varepsilon}{4}\\ \end{aligned} \end{equation}$


The statement suggests that we should choose $\displaystyle \delta = \frac{\varepsilon}{4}$

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



$\begin{equation} \begin{aligned} \text{if } 0 < |x+3| < \delta \text{ then, }\\ |(1-4x)-13| & = |1-4x-13| = |-4x-12| = |-4(x+3)| = 4|x+3| < 4 \delta = \cancel{4} \left(\frac{\varepsilon}{\cancel{4}}\right) = \varepsilon \end{aligned} \end{equation}$



$\begin{equation} \begin{aligned} & \text{Thus, }\\ & \phantom{x} \quad\text{if } 0 < |x-(-3)| < \delta \qquad \text{ then } \qquad |(1-4x)-13| < \varepsilon\\ & \text{Therefore, by the definition of a limit}\\ & \phantom{x} \qquad \lim\limits_{x \to -3} (1-4x) = 13 \end{aligned} \end{equation}$