Single Variable Calculus, Chapter 2, 2.4, Section 2.4, Problem 18

Using the definition of limit of $\delta$, $\varepsilon$ prove $\lim\limits_{x \to 4} (7-3x) = -5$ and graph




Based from the defintion,


$\begin{equation} \begin{aligned} \phantom{x} \text{if } & 0 < |x - a| < \delta \qquad \text{ then } \qquad |f(x) - L| < \varepsilon\\ \phantom{x} \text{if } & 0 < |x - 4| < \delta \qquad \text{ then } \qquad |(7-3x)-(-5)| < \varepsilon\\ \end{aligned} \end{equation}$



$\begin{equation} \begin{aligned} & \text{But, } \\ & \phantom{x} & |(7x-13x)-(-5)| = |7-3x+5| = |-3x+12| = |-3(x-4)| =3|x-4| \\ & \text{So, we want}\\ & \phantom{x} & \text{ if } 0 < |x-4| < \delta \qquad \text{ then } \qquad 3|x-4| < \varepsilon\\ & \text{That is,} \\ & \phantom{x} & \text{ if } 0 < |x-4| < \delta \qquad \text{ then } \qquad |x-4| < \frac{\varepsilon}{3}\\ \end{aligned} \end{equation}$


The statement suggests that we should choose $\displaystyle \delta = \frac{\varepsilon}{3}$

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



$\begin{equation} \begin{aligned} \text{if } 0 < |x-4| < \delta \text{ then, }\\ |(7-3x)-(-5)| & = |7-3x+5| = |-3x+12| = |-3(x-4)| = 3|x-4| < 3 \delta = 3 \left(\frac{\varepsilon}{3}\right) = \varepsilon \end{aligned} \end{equation}$



$\begin{equation} \begin{aligned} & \text{Thus, }\\ & \phantom{x} \quad\text{if } 0 < |x-4| < \delta \qquad \text{ then } \qquad |(7x-3x)-(-5)| < \varepsilon\\ & \text{Therefore, by the definition of a limit}\\ & \phantom{x} \qquad \lim\limits_{x \to 4}(7x-3x)= -5 \end{aligned} \end{equation}$